UNIT_7.PDF Engg Math

115Line and Surface Integrals UNIT 7 LINE AND SURFACE INTEGRALS Structure 7.1 Introduction Objectives 7.2 Integration of a Vector 7.2.1 Line Integral 7.2.2 Types of Line Integral 7.2.3 Evaluation of Line Integral 7.2.4 Path Independence – Conservative Fields 7.3 Double Integrals 7.3.1 Properties of Double Integrals 7.3.2 Evaluation of Double Integrals 7.3.3 Applications of Double Integrals 7.3.4 Change of Variables in Double Integrals 7.4 Transformation of Double Integrals into Line Integrals – Green’s Theorem 7.5 Surface Integrals 7.6 Transformation of Surface Integrals into Line Integrals – Stoke’s Theorem 7.7 Triple Integral 7.7.1 Definition 7.7.2 Properties of Triple Integrals 7.7.3 Volume 7.7.4 Evaluation of Triple Integrals 7.7.5 Physical Applications in Three Dimensions 7.8 Transformation of Volume Integrals into Surface Integrals 7.8.1 Gauss Divergence Theorem 7.8.2 Consequences and Applications of Divergence Theorem 7.8.3 Integral Definitions of Gradient, Divergence Theorem 7.8.4 Physical Interpretation of Divergence and Curl 7.8.5 Modelling of Heat Flow 7.8.6 Green’s Theorem and Green’s Formula 7.8.7 A Basic Property of Solutions of Laplace’s Equation 7.9 Solenoidal and Irrorational Vector Fields Revisited 7.10 Summary 7.11 Answers to SAQs 7.1 INTRODUCTION In Unit 6, you have learnt about the differentiation of vectors and arrived at the important concepts of directional derivative, gradient, divergence and curl, in this unit, we shall talk about integration of vector functions. In this unit, we shall begin our discussion in section 7.2 with the definition of a line integral and see that it is a natural generalization of a definite integral. We shall discuss the concepts involving double integrals and their evaluation, (iii) plane region may be transformed in double integrals in Section 7.3. Double integrals over a plane region may be transformed into the line integrals over the boundary of the region and conversely. This is achieved through the Green’s theorem in the plane and will form the subject of our discussion in Section 7.4. Section 7.5 is devoted to the discussion of surface integrals. 116 Engineering Mathematics The transformation of surface integrals into line integrals and conversely is done through Stoke’s Theorem, which will form our subject matter for Section 7.6. In this section, we shall also take up some applications of the Stoke’s theorem and give another physical interpretation of Curl. You have studied about line integrals, double integrals and surface integrals. In the process you have learnt to transform double integrals and surface integral into line integrals. You had learnt that line integrals are the generalisation of a single integral and a surface integral is a sort of generalization of a double integral. In this unit, we shall give another generalization of double integral called triple integrals or volume integrals. We shall first of all define triple integrals in Section 7.2, wherein we shall also give the properties and evaluation of such integrals. This section will be closed with some physical applications of triple integrals. We had made use of integral transform theorems-Green’s Theorem and Stoke’s Theorem in the last unit. In this unit, we shall discuss another important integral transform theorem, known as Gauss Divergence Theorem, which helps in the transformation of volume integral to surface integral and conversely. Divergence Theorem has many important consequences and various applications, some of which have been discussed in Section 7.3. We had earlier discussed solenoidal vector fields and irrotational vector fields in Unit 6. With our knowledge of vector calculus, we have revisited these concepts in Section 7.4, where we have now given integral form conditions for vector fields to be solenoidal and irrotational. The summary of the results discussed in this unit is presented at the end of this unit. Objectives After going through this unit, you should be able to • integrate a vector with respect to a scalar and solve problems based on it, • define a line integral, state various types of it and evaluate it, • compute double integrals and state the conditions under which the order to integration can be changed, • explain the need of change of variables in double integrals and evaluation of double integrals by change of variables, • outline method/conditions of transformation of double integral into line integral and conversely, • solve problems based on Green’s theorem, • define a surface integral and transform it into line integral, • outline the representation of curl in terms of line integral, • use Green’s theorem and Stoke’s theorem to relevant physical situations and solve problems based on them, • define and evaluate triple integral, and • learn the method and conditions under which a volume integral can be transformed into a surface integral. 7.2 INTEGRATION OF A VECTOR In Unit 6, you have learnt to differentiate a vector w.r. to scalar, You also know that integration is the inverse process of differentiation. If F (t) and f (t) be two vector functions of a scalar variable t, connected by the relation ( ) ( ) d t t dt = F f . . . (7.1) 117 Line and Surface Integrals then we say that ( ) t F is integral of ( ) t f and write ( ) ( ) f t dt t = ∫ F . . . (7.2) this gives an indefinite integral of ( ) t f . We can also add an arbitrary constant vector C on the right hand side of Eq. (7.2) and write it as ( ) ( ) t dt t = + ∫ f F C This is because ( ( ) ) ( ) d t t dt + = F C f 0 d dt ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ ∵ C While solving problems, this constant C can be determined using the given initial conditions. Also, the dimension of C is that of ( ). t F You also know from calculus, that for scalar function ( ), f x the definite integral ( ) b a f x dx ∫ can be defined as the limit of a sum. Here we integrate along axis x-axis from a to b and the integral f is a function defined at each point between a and b. In the same manner, we can define for a scalar variable t, the definite integral of vector function ( ) t F as the limit of a sum. We can write 1 1 2 2 0 ( ) lim ( ) ( ) ( ) , 1, 2, , i b n n t a n x dt t t t t t t i n δ → →∞ ⎡ ⎤ = δ + δ + + δ = ⎣ ⎦ ∫ … … F F F F where the ranges of integration a t b ≤ ≤ has been divided into n sub ranges corresponding to increments 1 2 δ , δ , , δ n t t t … of the variable t and 1 2 , , , n t t t … are values of t lying respectively in these sub-ranges. Further in the limit , n → ∞ as the number of sub-ranges increases indefinitely, each of the increment t δ tends to zero. The sum on the right hand side which is, of course, a vector sum tends to a definite limit, which is the definite integral of function ( ) t F . A simple example of vector integration is illustrated by a particle of unit mass moving under gravity with constant acceleration g. Let r denote the position-vector of the particle at time t. Then from Newton’s second law of motion, the equation of motion is 2 2 d dt = r g . . . (7.3) Now if we wish to find the trajectory of the particle its motion, i.e., if we wish to find r in terms of t then we integrate Eq. (7.3) and obtain d dt = 0 r gt +V . . . (7.4) where 0 V is the initial velocity (corresponding to 0 t = ). A second integration yields 2 1 2 t t = + + 0 0 r g V r . . . (7.5) where 0 r is the initial position of the particle. Thus by interpretation of a vector with respect to scalar, we could determine the trajectory of the particle when equation of motion is known. 118 Engineering Mathematics You may note that in Eqs. (7.4) and (7.5), d dt r and r have the dimensions of velocity and dimension of velocity and 0 r in equation (7.5) has the dimension of position. Note that in the above example, the particle was moving with constant acceleration g. In practice, it may happen that the force acting on the particle or the system may also be function (scalar or vector) of the scalar variable. For example, a transverse electromagnetic wave propagating in x- direction may have an electric field 0 2 ˆ cos ( ) ct x = − E E k π λ . If E is the electric permitivity and μ is magnetic permeability of space, then energy flowing through a volume V per unit time is given by (ε μ ) 2 V = + 2 2 U E B If T be the time period for total energy flowing through volume V during one complete cycle of electro-magnetic oscillation, then Total energy 0 T dt = ∫ U 0 (ε μ ) 2 T V dt = + ∫ 2 2 E B 0 (ε μ ) 2 T V dt = ⋅ + ⋅ ∫ E E B B 2 2 2 2 0 0 0 0 2π 2π ε cos ( ) μ cos ( ) 2 λ 2 λ T T V V ct x dt ct x dt = − + − ∫ ∫ E B 2 2 ( and ) = ⋅ = ⋅ ∵E E E B B B 2 2 2 0 0 0 2π ε μ cos ( ) 2 2 λ T V V E B ct x dt ⎛ ⎞ = + − ⎜ ⎟ ⎝ ⎠ ∫ 2 2 0 0 0 1 π (ε μ ) 1 cos 4 ( ) 4 λ T V E B ct x dt ⎡ ⎤ = + ⋅ + − ⎢ ⎥ ⎣ ⎦ ∫ 2 2 0 0 0 π sin 4 ( ) 1 λ (ε μ ) π 4 4 λ T ct x V E B t c − = + ⋅ + 2 2 0 0 1 λ π 4 λ 4π (ε μ ) sin ( ) sin 4 4π λ 4π λ V T cT x x c c ⎡ ⎤ ⎧ ⎫ ⎛ ⎞ = + + − + ⎨ ⎬ ⎜ ⎟ ⎢ ⎥ ⎩ ⎭ ⎝ ⎠ ⎣ ⎦ E B Note that in the above integration for the calculation of total energy, we have used 2 E E E, = ⋅ the dot product of vectors. It may also happen that the integration of vectors may involve cross product of vectors. We may wish to evaluate dr × ∫ A from 0 t = to 1 t = where 2 ˆ ˆ ˆ xy z x = − + A i j k and 2 3 , 2 , . x t y t z t = = = Here 2 3 ˆ ˆ ˆ 2 t t t = + + r i j k ∴ 2 ˆ ˆ ˆ (2 2 3 ) t t dt = + + dr i j k and 2 ˆ ˆ ˆ xy z x = − + A i j k 119 Line and Surface Integrals 3 3 4 ˆ ˆ ˆ 2t t t = − + i j k ∴ 3 3 4 2 ˆ ˆ ˆ 2 2 2 3 d t t t t dt dt t dt × = − i j k A r 5 4 5 3 4 ˆ ˆ ˆ (3 2 ) 4 (4 2 ) t t dt t dt t t dt = − + − + + i j k ∴ 1 1 1 5 4 5 3 4 0 0 0 ˆ ˆ ˆ (3 2 ) 4 (4 2 ) dr t t dt t dt t t dt × = − + + − + + ∫ ∫ ∫ ∫ A i j k 1 1 1 6 5 6 4 5 0 0 0 2 ˆ ˆ ˆ 3 2 4 6 5 6 5 t t t t t ⎛ ⎞ = − + − + + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ i j k 1 2 4 7 ˆ ˆ ˆ 2 5 6 5 ⎛ ⎞ = − + − + ⎜ ⎟ ⎝ ⎠ i j k 9 2 7 ˆ ˆ ˆ 10 3 5 = − − + i j k From the above example you must have observed that vector integration involving dot product or cross product of vectors, essentially reduces to determining the integral of a scalar with respect to scalar. Recall that in a definite integral ( ) , b a f x dx ∫ we integrate along x-axis from a to b and the integral f is a function defined at each point between a and b. However, in many physical problems, you may observe that a particle may not be moving along a line but along a curve in space, e.g., if we wish to calculate the work done by a force in moving a particle along the curve C from position A to position B, then the ordinary integration (indefinite or definite) involving vectors will not provide the result. Such problems lead us to the consideration of line integrals or curve integrals. 7.2.1 Line Integral The concept of a line integral is a simple and natural generalization of the concept of a definite integral ( ) b a f x dx ∫ . . . (7.6) In the line integral we do not integrate the integral along x-axis from a to b; instead, we integrate along a curve in plane or in space and the integrand is a function defined at the points of that curve. We can define a line integral in a way similar to that of a definite integral. Consider a curve C in space having A and B as the initial and terminal points. We may represent C in parametric from as ˆ ˆ ˆ ( ) ( ) ( ) ( ) , s x s y s z s = + + r i j k ( ) a s b ≤ ≤ where s is the arc length of C and A and B correspond to s = a and s = b respectively (Figure 7.1). 120 Engineering Mathematics Figure 7.1 : The Directed Curve C from A to B We assume that ( ) s r is continuous and has a continuous first derivative different from the zero vector for all s under consideration. The C has a unique tangent at each of its point. Such a curve C is called a smooth curve. Let ( , , ) w x y z be a scalar function which defined at each point of C, and is a continuous function of s. We divide the curve C, from A to B, in an arbitrary manner into n portions. Let 0 1 1 ( ), , , , ( ) n n P A P P P B − = = … be the end points of these portions and let 0 1 2 ( ) ( ) n s a s s s b = < < < < = … be the corresponding values of s (Figure 7.2) Let the length of each of these portion is 1 2 3 δ , δ , δ , . . . , δ n s s s s Then we evaluate w at each point ( , , ) k k k x y z in δ k s and consider the sum 1 ( , , ) δ n k k k k k x y z s = ∑ w Figure 7.2 : Sub-division of C We define the integral of ( , , ) w x y z over curve C from A to B to be limit of the sum as the subdivision of C is refined so that number of subdivisions becomes very large and the largest δ k s approaches zero, i.e., 1 ( , , ) lim ( , , ) δ n k k k k n k C w x y z ds w x y z s →∞ = = ∑ ∫ . . . (7.7) whenever the limit on the right-hand side in Eq. (7.7) exists, we call it line integral of ( , , ) w x y z along C form A to B and denote it as 121 Line and Surface Integrals ( , , ) . C w x y z ds ∫ Note that since w is continuous and C is smooth, the limit on the right hand side of Eq. (7.7) exists and is independent of the choice of subdivision of C. Moreover, if ( , , ) 1, w x y z w = is a constant function whose value is 1, then ( , , ) C w x y z ds ∫ gives the length of curve C from A to B. The concept of the line integral can be extended to vector integration and we get different types of line integrals depending upon. (i) vector element , , . . . , 1 2 δ δ δ n s s s and performing vector addition, keeping ( , , ) w x y z as a scalar function. (ii) taking function W (r) instead of scalar function and also taking vector elements , , . . . , 1 2 δ δ δ n s s s . Then products of W with δ k s can be either dot products or vector products and this in turn gives rise to two types of line integrals. We shall now take up various types of line integrals. 7.2.2 Types of Line Integral The following types of line integrals exists : (i) When ( , , ) w x y z is a scalar function defined at each point curve C from A to B, which is divided into parts , , . . . , 1 2 δ δ δ n s s s along the curve, then 1 ( , , ) lim ( , , ) n k k k n k C x y z x y z →∞ = = ∑ ∫ δ k w ds w s defines a line integral in ordinary calculus. (ii) When ( , , ) w x y z is a scalar function defined at each point of the curve C from A to B. which is divided into n parts of vector elements , , . . . , 1 2 δ δ δ n s s s along the curve C and perform vector addition, then 1 ( , , ) lim ( , , ) δ n k k k k n k C x y z x y z s →∞ = = ∑ ∫ w ds w defines a line integral, where value is a vector. (iii) When ( , , ) x y z W is a vector function defined at each point of the curve C from A to B, which is divided into n parts of vector elements , , . . . , 1 2 δ δ δ n s s s along the curve C and the product of ( , , ) k k k x y z W in δ k s in dot product of vector, then 1 ( , , ) . lim ( , , ) . n k k k n k C x y z x y z →∞ = = ∑ ∫ δ k W ds W s defines a line integral, whose value is a scalar. Moreover, if r is a position vector of a point on curve C, then ds ~ dr and in the limit when each sub-interval tends to be zero, ds = dr ; therefore the last three line integrals may also be written as 122 Engineering Mathematics ( , , ) , ( , , ) C C x y z x y z ∫ ∫ w dr W dr and ( , , ) C x y z × ∫ W dr respectively. Figure 7.3 : Formula (c) In all the above line integrals, it is assumed that the path of integration is piecewise smooth. For a line integral over a closed path C, the symbol (instead of ) C C ∫ ∫ is sometimes used in the literature. From the definition of line integral, it follows that the following properties are valid for line integrals : (a) C C k wds k wds = ∫ ∫ (k Constant) (b) ( ) C C C f g ds f ds g ds + = + ∫ ∫ ∫ (orientation of C is the same in all the three integrals). (c) 1 2 , C C C wds wds wds = + ∫ ∫ ∫ where the path C is subdivided into two arcs 1 C and 2 C , which have the same orientation as C. Note that if the sense of integration along a curve C is reversed, the value of the line integral is multiplied by – 1. The question is now arises is – How is a line integral evaluated? We shall now answer this question. 7.2.3 Evaluation of Line Integral A line integral can be evaluated by reducing it to a definite integral. This reduction is quite simple and is done by means of representation of the path of integration C as follows : If C is represented by ˆ ˆ ˆ ( ) ( ) ( ) ( ) , , t x s y s z s a s b = + + ≤ ≤ r i j k where s is the arc length of C, then we can immediately write ( , , ) [ ( ), ( ), ( )] , b a f x y z ds f x s y s z s ds = ∫ ∫ C . . . (7.8) the integral on the right being a definite integral. In applications, mostly representation of C is the form 123 Line and Surface Integrals ˆ ˆ ˆ ( ) ( ) ( ) ( ) , t x t y s z s = + + r i j k 0 1 , t t t ≤ ≤ where t is any parameter, or may easily be converted to this form. In this case, we can use [ ( )] ( ), s t t = r r so that [ ( )] ( ) s t t = x x and so on. Then we can use 1 0 ( , , ) [ ( ), ( ), ( )] , t t ds f x y z ds f x t y t z t dt dt = ∫ ∫ C . . . (7.9) where 2 2 2 ds y z dt = ⋅ = + + r r x . . . (7.10) Here we assume that ( ) t r and ( ) t r are continuous and ( ) 0, t ≠ r in agreement with the assumption mentioned in Section 7.2.2. We now illustrate the theory discussed so far with the help of a few examples. Example 7.1 Evaluate 2 2 2 2 ( ) , C x y z ds + + ∫ where C is the arc of circular helix ˆ ˆ ˆ ( ) cos sin 3 t t t t = + + r i j k, from (1, 0, 0) A to (1, 0, 6 ). π B Solution Here ˆ ˆ ˆ ( ) cos sin 3 t t t t = + + r i j k, ∴ ˆ ˆ ˆ ( ) sin cos 3 t t t = − + + r i j k Now 2 2 sin cos 9 10 ds t t dt = ⋅ = + + = r r On C, 2 2 2 2 2 2 2 2 2 ( ) [cos sin 9 ] (1 9 ) x y z t t t t + + = + + = + Also (1, 0, 0) A and (1, 0, 6 ) B π on C correspond to 0 2 . t π ≤ ≤ Thus, we have ( ) ( ) 2 2 2 2 2 2 2 0 1 9 C ds x y z t dt dt π + + = + ⋅ ∫ ∫ ( ) 2 4 2 0 10 1 81 18 t t dt π = + + ∫ 2 5 3 0 10 81 18 5 3 t t t π = + + 3 5 81 10 2 6(2 ) (2 ) 5 π π π ⎡ ⎤ = + + ⎢ ⎥ ⎣ ⎦ ≈ 506400 Using the representative of C, sometimes it may be possible for us to eliminate two of the three independent variables in the integrand of a line integral and then we can evaluate the resulting definite integral in which the remaining independent variable is the variable of integration. 124 Engineering Mathematics We illustrate this by the following example. Example 7.2 Evaluate the line integral 2 [ ( ) ] C x ydx x z dy xyz dz + − + ∫ where C is the arc of the parabola 2 y x = in the plane 2 z = from A (0, 0, 2) to B (1, 1, 2). Solution Since on C, 2 y x = and 2 z = (constant), ∴ on C, 2 dy x dx = and 0. dz = It follows that on C, the integral of the last term in the given integrand is zero. Figure 7.4 : Path in Example 7.2 Thus, 1 2 2 2 0 ( ) ( 2) 2 C x y dx x z dy xy z dz x x dx x x dx ⎡ ⎤ ⎡ ⎤ + − + = ⋅ + − ⋅ ⎣ ⎦ ⎣ ⎦ ∫ ∫ 2 ( and 2 on ) y x z C = = ∵ . ( ) 1 5 3 2 1 4 2 0 0 2 4 2 4 5 3 2 x x x x x x dx = + − = + − ∫ 1 2 17 2 5 3 15 = + − = − Let us now take up an example in which we consider integration of a line integral over different paths will the same end points. Example 7.3 Let C be the line segment from A (0, 0, 1) to B (1, 1) and let 2 ( , ) . f x y x y = + Evaluate ( , ) C f x y ds ∫ when (i) C is characterized by , , 0 1. x t y t t = = ≤ ≤ (ii) C has parametric representation sin , sin , 0 . 2 x t y t t π = = ≤ ≤ 125 Line and Surface Integrals Solution (i) If , , 0 1, x t y t t = = ≤ ≤ then 2 2 2 1 0 ( , ) ( ) C dx dy f x y ds t t dt dt dt ⎛ ⎞ ⎛ ⎞ = + ⋅ + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ∫ ∫ = 1 2 3 2 0 1 0 5 2 ( ) 1 1 2 2 3 6 t t t t dt + ⋅ + = + = ∫ (ii) If x = sin t, y = sin t, π 0 2 t ≤ ≤ , then 2 2 2 ( , ) ( ) C C dx dy f x y ds x y dt dt dt ⎛ ⎞ ⎛ ⎞ = + ⋅ + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ∫ ∫ π / 2 2 2 2 0 (sin sin ) cos cos t t t t dt = + + ∫ π / 2 2 0 2 (sin cos sin cos ) t t t t dt = + ∫ π / 2 2 3 0 sin sin 2 2 3 t t = + 5 2 6 = In this case, you may notice that even though paths described by C were different, the value of the line integral is the same. Is this always true? Let us examine this by considering yet another example. Example 7.4 Evaluate 2 2 2 ( ) ( ) ( ) C x y dx y z dy z x dz ⎡ ⎤ − + − + − ⎣ ⎦ ∫ where C is the curve from origin O to the point (1,1,1) A (i) along the straight line OA (ii) along the curve 2 3 , , , 0 1. x t y t z t t = = = ≤ ≤ Solution (i) Equation of line OA, joining O (0, 0, 0) and A (1, 1, 1) and . x y z = = For path OA, 2 2 2 ( ) ( ) ( ) C x y dx y z dy z x dz ⎡ ⎤ − + − + − ⎣ ⎦ ∫ 1 2 2 2 0 ( ) ( ) ( ) , x x dx x x dx x x dx ⎡ ⎤ = − + − + − ⎣ ⎦ ∫ ( and , so that and ) y x z x dy dx dz dx = = = = ∵ 1 3 2 0 1 3 3 2 2 x x = − = − 126 Engineering Mathematics (ii) Along the curve 2 3 , , x t y t z t = = = 2 , 2 , 3 . dx dt dy t dt dz t dt = = = We get 2 2 2 ( ) ( ) ( ) C x y dx y z dy z x dz − + − + − ∫ 1 2 2 4 3 6 2 0 ( ) ( ) 2 ( ) 3 t t dt t t t dt t t t dt ⎡ ⎤ = − + − + − ⎣ ⎦ ∫ 1 1 6 5 9 4 0 0 2 3 6 5 9 4 t t t t = − + − 1 1 1 1 2 3 6 5 9 4 ⎛ ⎞ ⎛ ⎞ = − + − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 29 60 = − In this example, the integrands and the end points of the paths of integration are the same, but the values of the line integrals are different. This illustrates the important fact that In general, the values of a line integrals of a given function depends not only on the end points but also on the geometric shape of the path of integration. In many applications, the integrands of the line integrals are of the form ( , , ) , ( , , ) or ( , , ) , f x y z f x y z f x y z dx dy dz ds ds ds where , dx dy ds ds and dz ds are the derivatives of the functions occurring in the parametric representation of the path of integration. Then we simply write ( , , ) ( , , ) C C f x y z ds f x y z ds = ∫ ∫ dx ds and similar expressions in the other two cases. For sums of these types of integrals along the same path C, we adopt the notation ( ) C C C C f dx g dy h dz f dx g dy h dz + = + + ∫ ∫ ∫ ∫ In many cases, the functions, f, g, h are components 1 2 3 , , f f f of a vector function 1 2 3 ˆ ˆ ˆ f f f = + + f i j k Then 1 2 3 1 2 3 , dx dy dz f dx f dy f dz f f f ds ds ds ds ⎛ ⎞ + + = + + ⎜ ⎟ ⎝ ⎠ the expression in parenthesis on the right being the dot product of the vector f and the unit tangent vector ˆ ˆ ˆ , d dx dy dz ds ds ds ds = + + r i j k where r (s) represents the path of the integration of the line integral. Therefore, 1 2 3 ( ) C C d f dx f dy f dz f ds ds + + = ⋅ ∫ ∫ r . . . (7.11) 127 Line and Surface Integrals C f d = ⋅ ∫ r . . . (7.12) where ˆ ˆ ˆ . d dx dy dz = + + r i j k Now if f represents a force whose point of application moves along a curve ˆ ˆ ˆ ( ) ( ) ( ) ( ) , t x t y t z t = + + r i j k a t b ≤ ≤ from a point A to a point B in space, then C f d ⋅ ∫ r represents the work done by the force f is moving a particle from point A to point B along the curve C. The representation in Eq. (7.11), i.e., C d ds ds ⋅ ∫ r f emphasizes the fact that the work done by the force f is the value of the line integral along the curve of the tangential component of the force field f. Let us consider an example illustrating the work done by a force. Example 7.5 A variable force p acts on a particle and the particle is displaced along path C in space. Find the work done by force p in this displacement. Solution Let the given path C be characterized by the equation 0 1 ( ), . t t t t ≤ ≤ r Now the work done W by force p in any displacement along the path C is given by the line integral , C d = ⋅ ∫ W p r the integration being taken in the sense of displacement. Here d d dt dt dt = = r r v where v is the velocity of the particle. Then work W becomes 1 1 0 0 , t t t t dr dt dt dt = ⋅ = ⋅ ∫ ∫ W p p v where 0 t and 1 t are the initial and final values of t. Furthermore, by Newton’s second law , m m = ⇒ = r p p v where m is the mass of the particle. Substituting this value of p in the line integral for W, we get 1 1 0 0 t t t t d m W m dt dt dt 2 ⎛ ⎞ = ⋅ = ⋅ ⎜ ⎟ ⎝ ⎠ ∫ ∫ v v v v 1 1 0 0 2 2 | | 2 t t t t d m m dt dt 2 ⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ ∫ v v ⇒ Work done = gain in kinetic energy in moving from initial point to the final point. 128 Engineering Mathematics Thus the basic law in mechanics has been reinforced through the line integral concept. You may now try the following exercises. SAQ 1 (a) If 2 2 ˆ ˆ ˆ (3 6 ) 14 20 , x y yz xz = + − + F i i k evaluate the line integral d ⋅ ∫ F r from (0, 0, 0) to (1, 1, 1) along the following paths C : (i) 2 3 , , , x t y t z r = = = (ii) The straight line form (0, 0, 0) to (1, 0, 0); then to (1, 1, 0) and then to (1, 1, 1), and (iii) The straight line giving (0, 0, 0) to (1, 1, 1). (b) Find the work done in moving a particle once round the circle 2 2 9 x y + = in the X O Y plane if the field of force, given by 2 ˆ ˆ ˆ (2 ) ( ) (3 2 4 ) . x y z x y z x y z = − − + + − + − + F i j k In example 3 you have noticed that the value of the line integral between the same end points A and B jointed by two different paths was the same, where as in Example 4, we found that the value of line integral between the same end pints jointed by two different paths was unequal. Let us now study the condition under which the value of the line integral between two points A and B in independent if the path joining them. 7.2.4 Path Independence – Conservative Fields Let F be a vector field with components 1 2 3 , , F F F and 1 2 3 , , F F F be continuous throughout some connected region D. Consider two points A and B in D. Suppose that C is any piecewise smooth curve joining A and B given by ( ), ( ), ( ), x x t y y t z z t = = = 1 2 . t t t ≤ ≤ If there exists a differentiable function f such that ˆ ˆ ˆ , x y z ∂ ∂ ∂ = ∇ = + + ∂ ∂ ∂ f f f F f i j k then along C, [ ( ), ( ), ( )] f f x t y t z t = is a function of t and df f dx f dy f dz dt x dt y dt z dt ∂ ∂ ∂ = + + ∂ ∂ ∂ ˆ ˆ ˆ dx dy dz f dt dt dt ⎛ ⎞ = ∇ ⋅ + + ⎜ ⎟ ⎝ ⎠ i j k 129 Line and Surface Integrals d f dt = ∇ ⋅ ⋅ r We thus have d f d ⋅ = ∇ ⋅ F r r d f dt dt = ∇ ⋅ r df dt dt = Now integrating F . dr along C from A to B, we get 2 1 t t C df dr dt dt ⋅ = ∫ ∫ F 2 1 [ { ( ), ( ), ( )}] t t d f x t y t z t = ∫ 2 1 ( ( ), ( ), ( )) t t f x t y t z t = 2 2 2 1 1 1 ( ), ( ), ( ) ( ), ( ), ( ) f x t y t z t f x t y t z t ⎡ ⎤ ⎡ ⎤ = − ⎣ ⎦ ⎣ ⎦ ⇒ ( ) ( ) B C A d f d f B f A ⋅ = ∇ ⋅ = − ∫ ∫ F r r The value of the integral [ ] ( ) ( ) f B f A − does not depend on the path C at all. This result is analogue of the First Fundamental Theorem of Integral Calculus (see Block 1), viz., ( ) ( ) ( ) b a f x dx f b f a ′ = − ∫ The only difference is that we have f d ∇ ⋅ r in place of ( ) f x d ′ r . This analogy suggests that if we define a function f by the rule ( , , ) ( , , ) x y z A f x y z d ′ ′ ′ ′ ′ ′ = ⋅ ∫ ∫ F r . . . (7.13) then it will also be true that f ∇ = F . . . (7.14) This result f ∇ = F is indeed true when the right-hand side of Eq. (7.13) is path independent. Thus A necessary and sufficient condition for the integral B A d ⋅ ∫ F r to be independent of the path joining the points A and B in some connected region D is that there exists a differentiable function of such that ˆ ˆ ˆ f f f f x y z ∂ ∂ ∂ = ∇ = + + ∂ ∂ ∂ F i j k throughout D, where compounds 1 2 3 , , f f f of vector field F are continuous throughout D and then ( ) ( ) B A d f B f A ⋅ = − ∫ F r When F is a force such that the work-integral from A to B is the same for all paths, the field is said to be conservative. Using the above result, we can say that 130 Engineering Mathematics A force field F is conservative if and only if is a gradient field, i.e., , f = ∇ F for some differentiable function f. A function ( , , ) f x y z that has the property that its gradient gives the force vector F is called a potential function. Sometimes a minus sign is introduced, e.g., the electric intensity of a field is the negative of the potential gradient in the field. Let us consider the following example. Example 7.6 Find the work done when a force 2 2 ˆ ˆ ( ) (2 ) x y x xy y = − + − + F i j moves a particle in the xy-plane from (0, 0) to (1, 1) along the parabola 2 . y x = Is the work done different when the path is the straight line y =x? Solution The parabola 2 y x = has a parametric representation 2 , . y t x t = = From (0, 0) to (1, 1), variation of t is 0 1. t ≤ ≤ ∴ Work done along the parabola 2 2 ˆ ˆ ˆ ˆ ( ) (2 ) ( ) C C d x y x xy y dx dy ⎡ ⎤ = ⋅ = − + − + ⋅ + ⎣ ⎦ ∫ ∫ F r i j i j 2 2 [( ) (2 ) ] C x y x dx xy y dy = − + − + ∫ 1 4 2 2 2 0 [( ) 2 (2 ) ] t t t t dt t t t dt = − + ⋅ − ⋅ ⋅ + ∫ 1 6 4 2 0 1 1 1 2 3 2 2 3 t t t = − − = − We can similarly find the work done when the particle move form (0, 0) to (1, 1) along y = x. In this case also we find that the work done is (1,1) 3 2 2 2 (0, 0) 1 1 1 1 1 1 2 1 3 2 2 3 2 2 3 x x xy y ⎡ ⎤ = + − − = + − − = − ⎢ ⎥ ⎣ ⎦ as obtained earlier. This is because, we notice that 3 2 2 2 1 1 1 grad 3 2 2 x x xy y ⎛ ⎞ = + − − ⎜ ⎟ ⎝ ⎠ F That is, the field F is conservative and the work done does not depend on the path followed. You may now attempt the following exercise. SAQ 2 (a) If 2 2 2 , xy z x y φ = + evaluate grad , C d φ ⋅ ∫ r where C is the curve 2 3 , , x t y t z t = = = from 0 t = to 1. t = 131 Line and Surface Integrals (b) In 2 ˆ ˆ ˆ xy z x = − + A i j k and C is the curve 2 3 , 2 , x t y t z t = = = from 0 t = to 1 t = , evaluate C d × ∫ A r. (c) Suppose ˆ ˆ ˆ ( cos ) ( sin ) ( ) x x e y yz xz e y xy z = + + − + + F i j k Is F conservative? If so, find f such that f = ∇ F . So far we have performed integration along a line or a curve. But as we have mentioned earlier, there are many physical situation where we are required to find integral for areas and volumes. For example, determination of moment of inertia and coordinates of center of gravity of a continuous matter. Solution of these problems involve integrals where we have to do integration with respect to more than one variable. Such integrals are called multiple integrals. We shall, in the next section, discuss only double integrals, in which the integrals is a function of two variables. 7.3 DOUBLE INTEGRALS Consider a function ( , ) f x y which is defined for all (x, y) in a closed bounded region R of xy–plane. The definition of double integral can be given in quite in quite a similar manner as that of a definite integral of a single variable. We subdivide the region R into sub-regions (see Figure 7.5). These sub-regions can be arbitrary or can be rectangles obtained by drawing lines parallel to x and y axes. (a) Sub-division of R in Arbitrary Manner (b) Sub-division of R by Drawing Parallel to x and y axes Figure 7.5 We number these sub-division, which are within R, from 1 to n. In each sub-region, we choose a point, say ( , ) k k x y in the th k sub-region, and then form the sum 1 ( , ) , n n k k k k F f x y A = = Δ ∑ where k A Δ is the area of the th k sub-regions. In a completely independent manner, we increase the number of sub-regions such that the area of the largest sub-region tends to zero as n approaches infinity (In the case of 132 Engineering Mathematics rectangles as sub-regions, we can say that the length of the maximum diagonal of the rectangles approaches zero as n approaches infinity). In this way, we obtain a sequence of real number 1 2 7 , , . . . , , . . . f f f We know that if ( , ) f x y is continuous in R and region R is bounded by finitely many smooth curves, then this sequence converges and its limit is independent of the choice of sub-divisions and corresponding points ( , ) f x y . This limit is called the double integral of ( , ) f x y over the region R and is denoted by the symbol ( , ) or ( , ) R R f x y d f x y dx dy ∫∫ ∫∫ A Thus 1 ( , ) = lim ( , ) n k k k n k R f x y d f x y →∞ = Δ ∑ ∫∫ A A The continuity of the integrand is a sufficient condition for the existence of the double integral, but not a necessary one, and the limit in question exists for many discontinuous functions as well. From the definition, it follows that double integrals enjoy properties which are quite similar to these of definite integrals of functions of a single variable. These properties hold for the sums from which integrals are defined. We shall state some of these algebraic properties that are useful in computations and applications. 7.3.1 Properties of Double Integrals Let f and g be the function of x and y, which are defined and are continuous in a region R, then (i) ( , ) ( , ) R R k f x y d k f x y d = ∫∫ ∫∫ A A (k is a constant) (ii) [ ] ( , ) ( , ) = ( , ) ( , ) R R R f x y g x y d f x y d g x y d ± ± ∫∫ ∫∫ ∫∫ A A A (iii) ( , ) 0 R f x y d ≥ ∫∫ A if ( , ) 0 f x y ≥ on R (iv) ( , ) ( , ) R R f x y dA g x y dA ≥ ∫∫ ∫∫ if ( , ) ( , ) f x y g x y ≥ on R (v) 1 2 ( , ) ( , ) ( , ) , R R R f x y dA f x y dA f x y dA = + ∫∫ ∫∫ ∫∫ Which hold when R is the union of two non overlapping regions 1 R and 2 R as shown in Figure 7.6 and is called domain-activity property. Figure 7.6 Furthermore, there exists at least one point 0 0 ( , ) x y in R such that 133 Line and Surface Integrals (vi) 0 0 ( , ) ( , ) , R f x y dA f x y = ∫∫ A where A is area of R and this is called mean-value theorem for double integrals. Let us now take up the evaluation of the double integrals. 7.3.2 Evaluation of Double Integrals If ( , ) f x y is continuous on the rectangular region R given by , a x b ≤ ≤ , c y d ≤ ≤ then by drawing lines parallel to x and y axes, the double integrals over rectangles can always be calculated as integrated (or reported) integrals and ( , ) ( , ) ( , ) d b b d R y c x a x a y c f x y dA f x y dx dy f x y dy dx = = = = = = ∫∫ ∫ ∫ ∫ ∫ . . . (7.15) This means that we can evaluate a double integral by integrating one variable at a time (treating the other variable as constant), using the integration techniques we already know for function of a single variable. Further, we may calculate the double integral over a rectangle by integrating in either order (as per our convenience) Consider the following example. Example 7.7 Calculate ( , ) , R f x y dA ∫∫ where 2 ( , ) 1 6 f x y x y = − and : 0 2, R x ≤ ≤ 1 1. y − ≤ ≤ Solution Using Eq. (7.15), we can write 1 2 2 1 0 ( , ) (1 6 ) y x R f x y d x y dx dy =− = = − ∫∫ ∫ ∫ A 0 3 1 1 0 6 3 y x x x y dy =− = = − ∫ 1 1 (2 16 ) y dy − = − ∫ 1 2 1 2 16 2 y y − = − (2 8) ( 2 8) 4 = − − − − = Reversing the order of integration gives the same answer (you may check it for yourself). Let us now consider double integral for bounded non-rectangular regions. In such regions R, the double integral may be evaluated by two successive integration again, but the method will be as follows : Suppose first we draw lines parallel to y-axis and the that R can be described by the inequalities of the form , a x b ≤ ≤ ( ) ( ) g x y h x ≤ ≤ (Figure 7.7(a)) . . . (7.16) 134 Engineering Mathematics (a) (b) Figure 7.7 : Evaluation of a Double Integral so that ( ) y g x = and ( ) y h x = represent the boundary of R. Then ( ) ( ) ( , ) ( , ) b h x x a y g x R f x y dA f x y dy dx = = ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ ∫∫ ∫ ∫ . . . (7.17) In this case, we first integrate the line integral ( ) ( ) ( , ) h x y g x f x y dy = ∫ , keeping x fixed, i.e., treating x as a constant. The result of this integration will be a function of x, say ( ). x F Integrating ( ) x F over x form a to b, we obtain the value of the double integral in Eq. (7.17). Next, if we draw lines parallel to x-axis and the region R can be described by the inequalities of the form , ( ) ( ) c y d p y x q y ≤ ≤ ≤ ≤ (Figure 7.7 (b)) . . . (7.18) then we obtain ( ) ( ) ( , ) ( , ) d q y y c x p y R f x y dx dy f x y dx dy = = ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ ∫∫ ∫ ∫ . . . (7.19) we now integrate first over x (treating y as a constant) and then integrate with respect to y resulting the function of y from c to d. If the region R cannot be represented by inequalities of the Eqs.(7.16) or (7.18), but can be subdivided into finitely many portions which have that property, we may integrate ( , ) f x y over each position separately and add the results of these integrals, this will give us the value of the double integral of ( , ) f x y over the region R. In the equivalence of double into repeated or iterated integrals, we have assumed that ( , ) f x y is uniformly continuous and bounded over R and ( ) ( ) ( ) ( , ) h x y g x x f x y dy = = ∫ F is bounded and integrable from a to b with respect to x along with ( ) ( ) ( ) ( , ) g x x p y G y f x y dx = = ∫ is bounded and integrable form c to d with respect to y. In ( , ) f x y for discontinuities within R or on its boundary, it may happen that the two integrals given by Eq. (7.17) and (7.19) are not equal. Let us try to understand this point through the following example. Example 7.8 Show that 135 Line and Surface Integrals 1 1 1 1 3 3 0 0 0 0 ( ) ( ) x y x y dx dy dy dx x y x y − − ≠ + + ∫ ∫ ∫ ∫ Solution L.H.S. = 1 1 1 1 3 3 0 0 0 0 2 ( ) ( ) ( ) x y x x y dx dy dx dy x y x y − − + = + + ∫ ∫ ∫ ∫ 1 1 1 1 3 2 2 0 0 0 0 2 1 1 ( ) ( ) ( ) y x x dx dy dx x y x y x y x y = ⎡ ⎤ − = + = + ⎢ ⎥ + + + + ⎢ ⎥ ⎣ ⎦ ∫ ∫ ∫ 1 1 1 2 2 0 0 0 1 1 1 1 1 1 1 1 2 (1 ) ( 1) x dx dx x x x x x x ⎡ ⎤ − = + + − = = − = ⎢ ⎥ + + + + ⎢ ⎥ ⎣ ⎦ ∫ ∫ R.H.S. 1 1 1 1 1 1 3 3 2 3 0 0 0 0 0 0 ( ) 2 1 2 ( ) ( ) ( ) ( ) x y x y y y dy dx dy dx dy dx x y x y x y x y ⎡ ⎤ − + − = = = − ⎢ ⎥ + + + + ⎢ ⎥ ⎣ ⎦ ∫ ∫ ∫ ∫ ∫ ∫ 1 1 1 1 2 2 2 2 0 0 0 0 1 1 1 1 1 ( ) (1 ) (1 ) y y y dy dy dy x y y y x y y y y ⎛ ⎞ = − + = − + + − = − ⎜ ⎟ ⎜ ⎟ + + + + + ⎝ ⎠ ∫ ∫ ∫ 1 1 1 2 2 = − = − ∴ L.H.S R.H.S ≠ Sometimes it may happen that we are required to change the order of integration in a double integral for which limits are given. In such a case, first of all we ascertain from the given limits of the region R of integration. Knowing the region of integration, we then put the limits for integration in the reverse orders. We illustrate it through the following example. Example 7.9 Change the order of integration in the integral 2 2 cos 0 tan ( , ) a a x x f x y dy dx α α − ∫ ∫ Solution The given limits show that the region R of integration is bounded by the curves 0; x = cos α x a = tan α y x = and 2 2 y a x = − Now y = x tan α is a line through the origin and 2 2 y a x = − is a circle of radius a and center at origin and they intersect at the point ( cos , sin ) a a α α . Hence the region R is the shaded region OAB as shown in Figure 7.8. 136 Engineering Mathematics Figure 7.8 : Region R When we have to integrate with respect to x first, i.e., along a horizontal strip parallel to x-axis, we see the starting point of all the strips is the same line. They all start form the line 0, x = but some of the strips end at the line OA, While other end on the circular arc AB. The line of division (or demarcation) is the line CA given by sin α y a = Hence region OAB must be subdivided into two sub-regions : OAC and CAB. Now region OAC is bounded by the curves 0, x = cot α, 0 x y y = = and sin α y a = The region OAC is bounded by the curves 2 2 0, , sin α, x x a y y a = = − = and . y a = Hence on changing the order of integration, the given double integral becomes 2 2 sin α cot α 0 0 sin α 0 ( ) ( , ) a x y a a y y x y a x f x y dx dy f x y dx dy = − = = = = + + + ∫ ∫ ∫ ∫ You may now try the following exercise. SAQ 3 Describe the region of integration and evaluate the following integrals : (i) 1 2 2 2 0 0 (1 ) x x y dy dx + + ∫ ∫ (ii) 2 1 2 0 (1 ) x x xy dy dx − ∫ ∫ Double integrals have various geometrical and physical applications. We take up these application in the next sub-section. 7.3.3 Application of Double Integrals The area A of a region R in the xy-plane is given by the double integral R A dx dy = ∫∫ Recall that while defining the double integral, we had framed the sum 1 ( , ) n n k k k k F f x y A = = Δ ∑ 137 Line and Surface Integrals In this sum, if we consider a rectangular parallelepiped with base k A Δ and altitude ( , ) k k f x y , then ( , ) k k k f x y A Δ represents the volume of the volume of the parallelepiped and hence the volume V beneath the surface ( , ) ( 0) z f x y = > and above a region R in the xy-plane, as shown in Figure 7.9, is ( , ) R V f x y dx dy = ∫∫ Figure 7.9 : Double Integral as a Volume Further, if ( , ) f x y be the density (mass per unit area) of a distribution of mass in the xy-plane, if M is the total mass in region R, then ( , ) R M f x y dx dy = ∫∫ From mechanics, we know that the coordinates , x y of the center of gravity of the mass (of density ( , ) f x y in region R is given by ( , ) ( , ) and ( , ) ( , ) R R R R f x y dx dy y f x y dx dy x y f x y dx dy f x y dx dy = = ∫∫ ∫∫ ∫∫ ∫∫ Thus, the coordinates of center of gravity of a mass is another application of double integrals. Similarly, the moments of inertia x I and y I of the mass in R about x and y axes, respectively, are 2 ( , ) x R I y f x y dx dy = ∫∫ and 2 ( , ) , y R I x f x y dx dy = ∫∫ Which are again double integrals Let us now take up a few examples. Example 7.10 Find the volume of the prism whose base in the triangle in the xy-plane bounded by x-axis and the lines y x = and 1 x = and whose top lies in the plane ( , ) 3 . z f x y x y = = − − Solution For any x between 0 and 1, the variable may vary from 0 y = to y x = (parallels to y-axis yields these equalities), as shown in Figure 7.10. 138 Engineering Mathematics (a) Prism with a Triangular Base in the xy-plane (b) Integration Limits of ( , ) = ∫ ∫ = 1 =0 0 f x y dx dy y x x y Figure 7.10 Hence 1 0 0 (3 ) x x y V x y dy dx = = ⎡ ⎤ = − − ⎢ ⎥ ⎣ ⎦ ∫ ∫ 2 1 0 0 (3 2 y x y y y xy dx = = = − − ∫ 1 2 2 3 1 0 0 3 3 3 1 3 3 1 2 2 2 3 2 2 x x x x dx ⎛ ⎞ = − = − ⋅ = − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ∫ Similarly, parallels to x-axis yields that for any y between y = 0 and y = 1. The variables x may vary from 0 x = to x y = (Figure 7.11) Figure 7.11 : Integration of Limits of ( , ) = ∫ ∫ = 1 =0 0 f x y dx dy y x x y Thus, when the order of integration is reversed, the integral for volume is 1 1 0 (3 ) y x y V x y dx dy = = ⎡ ⎤ = − − ⎢ ⎥ ⎣ ⎦ ∫ ∫ 1 2 1 0 3 2 y x y x x yx dy = = = − − ∫ 2 1 2 0 1 3 3 2 2 y y y y y dy = ⎛ ⎞ = − − − + + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ∫ 1 2 0 5 3 4 2 2 y y y dy = ⎛ ⎞ = − + ⎜ ⎟ ⎝ ⎠ ∫ 1 2 3 0 5 3 4 2 2 2 3 y y y = − + 139 Line and Surface Integrals 5 4 1 1 2 2 2 = − + = Thus, the value V obtained in two ways is equal, as it should be. We now determine the area of a given region R. Example 7.11 Find the area of region R enclosed by the parabola 2 y x = and the line 2 y x = + Solution If we draw lines parallel to x- axis to determine the limits of integration; it is seen that we must divide region R into the region 1 R and 2 R as shown in the Figure 7.12(a) and then we may calculate the area as 1 2 1 4 0 1 2 y y y x y y x y R R A dA dA dx dy dx dy + = =− = = − ⎡ ⎤ ⎡ ⎤ = + = + ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ∫∫ ∫∫ ∫ ∫ ∫ ∫ . . . (7.20) (a) (b) Figure 7.12 : Area in Example 7.11 On the other hand, reversing the order of integration (by drawing lines parallel to y-axis), the required area is (see Figure 7.12(b)). 2 2 2 2 1 x x y x R A dA dy dx + =− = ⎡ ⎤ = = ⎢ ⎥ ⎣ ⎦ ∫∫ ∫ ∫ . . . (7.21) Clearly the area given by Eq. (7.21) is simpler as it is easier to calculate. In practice one would bother to write the integral only in this form. Evaluation of integral (Eq. 7.21) yields 2 2 2 3 2 2 2 2 1 1 1 ( 2 ) 2 2 3 x x x x x x A y dx x x dx x + =− =− − ⎡ ⎤ = = + − = + − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ∫ ∫ 4 8 1 1 9 4 2 . 2 3 2 3 2 ⎛ ⎞ = + − − + − = ⎜ ⎟ ⎝ ⎠ Let us take up a physical application in our next example. Example 7.12 A thin plate of uniform (constant) thickness and density δ( , ) x y is bounded by , 2 y x y x = = − and axis. x − Find the center of mass of the plate if δ( , ) 1 2 . x y x y = + + Solution 140 Engineering Mathematics The line 2 y x = − cuts axis x − at A (2, 0). The lines y x = 2 y x = − intersect at B (1, 1). The given plate is OAB as shown in Figure 7.13. Figure 7.13 : Given Plate By drawing lines parallel to x-axis, the whole plate is-characterized by : 0 2 R x ≤ ≤ and 2 x y x ≤ ≤ − Now, Mass of the plate δ( , ) R M x y dx dy = = ∫∫ 2 2 0 (1 2 ) x x y x x y dy dx − = = ⎡ ⎤ = + + ⎢ ⎥ ⎣ ⎦ ∫ ∫ 2 2 2 0 2 2 x x y x y y xy dx − = = = + + ∫ 2 2 2 2 0 (2 ) 2 2 (2 ) 2 2 2 x x x x x x x x dx = ⎡ ⎤ − = − + − + − − − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ∫ 2 2 2 2 2 0 1 2 4 2 (4 4 ) 2 2 2 x x x x x x x x x dx = ⎡ ⎤ = − + − + + − − − − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ∫ 2 2 0 (4 4 ) x x dx = = − ∫ 2 2 0 4.8 8 4 4 8 0 3 3 3 x x = − = − − = − First Moment, 2 2 0 0 δ( , ) (1 2 ) x x x y R M y x y dx dy y x y dy dx − = = ⎡ ⎤ = = + + ⎢ ⎥ ⎣ ⎦ ∫∫ ∫ ∫ 2 2 2 3 2 0 2 2 2 3 x x y x y y y x dx − = = = + + ∫ 2 2 3 2 2 3 3 0 (2 ) 1 (2 ) (2 ) 2 3 2 3 x x x x x x x x dx = ⎡ ⎤ − = + − + − − − − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ∫ 141 Line and Surface Integrals 2 3 2 2 2 3 2 3 0 1 1 (4 4 ) (4 4 ) (8 12 6 ) 2 3 2 3 x x x x x x x x x x x x dx = ⎡ ⎤ = + − + + − + − − + − − − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ∫ 2 2 2 0 14 2 2 2 3 3 x x x dx ⎛ ⎞ = − − − ⎜ ⎟ ⎝ ⎠ ∫ 2 2 4 2 0 14 2 2 3 3 3 4 x x x x = − − − 28 16 8 8 4 3 3 3 3 − = − − − = First Moment, 2 2 0 δ( , ) (1 2 ) x y x y x R M x x y dx dy x x y dy dx − = = ⎡ ⎤ = = + + ⎢ ⎥ ⎣ ⎦ ∫∫ ∫ ∫ 2 2 2 2 0 ( 2 ) 2 x x y x y x x y x dx − = = = + + ∫ 3 2 2 2 2 0 ( 2 ) (2 ) (4 4 ) ( 2 ) 2 2 x x x x x x x x x x x dx = ⎡ ⎤ = + − + + − − + − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ∫ 3 3 2 2 2 3 2 2 3 0 2 4 2 2 2 2 2 2 x x x x x x x x x x x dx = ⎡ ⎤ = − + − + + − − − − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ∫ 2 2 4 2 3 0 0 4 4 (4 4 ) 2 4 x x x x x dx = = − = − ∫ 2 4 16 8 = × − = − Hence, the center of mass is ( , ) x y where 8 3 1, 8 3 x M x M ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ = = = ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ ( 8) 3 8 3 y M y M − = = = ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ Example 7.13 Let ( , ) 1 f x y = be the density of mass in the region 2 : 0 1 , y x ≤ ≤ − R 0 1 x ≤ ≤ Find the center of gravity and the moments of 0 , , . x y I I I Solution The given region is a circle 2 3 1 x y + = in the first quadrant (Figure 7.14). ∴ Total Mass in R dx dy = ∫∫ R 2 1 1 0 0 x dy dx − ⎡ ⎤ = ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ∫ ∫ 142 Engineering Mathematics 1 2 0 1 x dx = − ∫ π / 2 2 0 π cos θ θ 4 d = = ∫ Figure 7.14 : Unit Circle in the 1 st Quadrant Now 2 1 1 1 2 0 0 0 1 4 4 1 π π x R x x dx dy x dy dx x x dx M − ⎡ ⎤ = = = − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ∫∫ ∫ ∫ ∫ Let 2 2 2 1 1 x z x z − = ⇒ − = ∴ 2 2 x dx z dz − = ∴ 0 3 0 2 1 1 4 4 4 ( ) π π 3 3π z x z dz = − = − = ∫ Since the area R is symmetrical about both the axes, ∴ 4 3π x y = = The coordinates of center of gravity are 4 4 ( , ) , 3π 3π x y ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ Now, Moment of Inertia, 2 1 1 2 2 0 0 x x R I y dx dy y dy dx − ⎡ ⎤ = = ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ∫∫ ∫ ∫ 1 2 3/ 2 0 1 (1 ) 3 x dx = − ∫ (let sin θ x = ∴ cos θ θ dx d = ) π / 2 4 0 1 π cos θ θ . 3 16 d = = ∫ By symmetry, π . 16 x y I I = = Also 0 π π π 0.3927. 16 16 8 x y I I I = + = + = ≈ You may now try the following exercise. SAQ 4 Prove that the area in the positive quadrant, bounded by the curve 2 4 , y ax = 2 2 4 , y bx xy c = = and 2 xy d = is 143 Line and Surface Integrals 2 2 1 ( ) log . 3 b d c a ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ You may recall that in the case of a definite integral ( ) , b a f x dx ∫ sometimes we have to introduce a new variable of integration u in order to simplify the integration by setting ( ) x x u = where function ( ) x u is continuous and has a continuous derivative in some interval α β u ≤ ≤ such that (α) x a = and (β) x b = [or (α) , (β) ] x b x a = = . Then β α ( ) [ ( )] . b a dx f x dx f x u du du = ∫ ∫ In the same manner we often simplify the evaluation of double integral by the introduction of a new variable. In next sub-section we shall show how this new variable is introduced. 7.3.4 Change of Variables in Double Integrals Let a region R in xy-plane be transformed into a region G (Figure 7.15) in the uv-plane by differentiable functions of the form ( , ), x f u v = ( , ) y g u v = so that each point 0 0 ( , ) u v in the region G corresponds to a point 0 0 0 0 0 0 0 0 ( , ) ( , ), ( , ) ( , ) x u v f u v y u v g u v ⎡ ⎤ = = ⎣ ⎦ in the region R and conversely; then a function , x y φ( ) defined in R can be thought of as a function [ ] , , , f u v g u v φ ( ) ( ) defined on G. Figure 7.15 : Transformation of Region R in xy-plane to Region G in uv-plane From the calculus of two variables, we have the result that if all the functions involved are continuous and have continuous first derivatives, then the integrand , x y φ( ) , of the double integral ( , ) R x y dx dy φ ∫∫ , can be expressed in terms of u and v, and dx dy 144 Engineering Mathematics replaced by du dv times the absolute value of the Jacobain of the coordinate transformation x = f (u,v), ( , ) y g u v = given by ( , ) ( , ) x x u v x y y y u v u v ∂ ∂ ∂ ∂ ∂ = = ∂ ∂ ∂ ∂ ∂ J (which is either positive through out G or negative G ). Here the integral , x y φ( ) over R and the integral of [ ] , , , f u v g u v φ ( ) ( ) over G are related by the equation ( , ) , ( , , ( , )] ( , ) R R x y x y dx dy f u v g u v du dv u v ∂ φ( ) = φ[ ) ∂ ∫∫ ∫∫ We now take up an example to illustrate how the change of variables simplifies the evaluation of a double integral. Example 7.14 Evaluate the double integral 2 2 R x + y dx dy ( ) ∫∫ where R is the square bounded by lines , , 2, 2 y x y x x y x y = = − − = + = Solution Shape of the square R of the problem is as shown in Figure 7.16. Figure 7.16 : Region R The shape of region R suggests the transformation , x y u + = x y v − = Then 1 ( ), 2 x u v = + 1 ( ) 2 y u v = − The Jacobian of transformation of coordinates is 1 1 ( , ) 1 1 1 2 2 1 1 ( , ) 4 4 2 2 2 x y J u v ∂ = = = − − = − ∂ − ∴ Absolute value of 1 . 2 J J = = Now, region R in xy-plane corresponds to the square 0 2, u ≤ ≤ 0 2 v ≤ ≤ (see Figure 7.17) 145 Line and Surface Integrals Figure 7.17 Thus, 2 2 2 2 2 2 0 0 1 1 1 ( ) ( ) ( ) 2 2 2 u v R x y dx dy u v u v du dv = = ⎡ ⎤ ⎧ ⎫ ⎧ ⎫ + = + + − − ⎨ ⎬ ⎨ ⎬ ⎢ ⎥ ⎩ ⎭ ⎩ ⎭ ⎣ ⎦ ∫∫ ∫ ∫ 2 3 2 2 2 2 2 2 0 0 0 0 1 1 ( ) 4 4 3 u v u v u v du dv u v du = = = ⎛ ⎞ = + = + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ∫ ∫ ∫ 2 3 2 2 0 0 1 8 1 2 4 4 8 2 4 3 2 3 3 2 3 3 u u u du u = ⎛ ⎞ = + = ⋅ + ⋅ = + = ⎜ ⎟ ⎝ ⎠ ∫ Notice that had we not changed the variable, evaluation of R would have to be carried out by first dividing R into two regions 1 R and 2 R and then finding the range of integrations for the two regions R 1 and R 2 which are different as is evident from the Figures 7.18(a) and (b) given below. (a) When Lines Parallel to x-axis are Drawn (b) When Lines Parallel to y-axis are Drawn Figure 7.18 Let us consider anther example. Example 7.15 Find the mass of the plate bounded by the four parabolas 2 2 2 2 4 , 4 , 4 and 4 y ax y bx x cy x dy = = = = if the density of the plate is ρ , kxy = where k is a constant. Solution 146 Engineering Mathematics In this case, the mass of the plate ( ) , A k xy dx dy = ∫ ∫ where region A is shown in Figure 7.19(a). It is quite clear from Figure 7.19(a), that if we wish to evaluate this integral, we shall have to subdivide the region A in many sub-regions (a) Region A′ in xy-plane (b) Region A′ Transformed in uv-plane Figure 7.19 However, if we assume 2 2 , , x y u v y x = = then in uv-plane the region A is transformed to 4 , 4 , 4 , 4 , v a v b u c u d = = = = which is rectangle A having sides parallel to u and v axes. 2/ 3 1/ 3 x u v = 1/ 3 2/ 3 y u v = ∴ 1/ 3 1/ 3 2/ 3 2/ 3 2/ 3 2/ 3 1/ 3 1/ 3 2 / 3 1/ 3 ( , ) 4 1 1 ( , ) 9 9 3 1/ 3 2 / 3 u v u v x y J u v u v u v − − − − ∂ = = = − = ∂ Also 2/ 3 1/ 3 1/ 3 2/ 3 ρ k xy ku v u v k uv = = ⋅ = ∴ Mass of plate 4 4 4 4 1 3 d b v c u a k uv du dv = = = ∫ ∫ 2 2 2 2 64 ( ) ( ) 3 k b a d c = − − We know that double integration is integration over an area in plane. So far we have used Cartesian coordinates ( , ) x y for a plane area. However, in some practical problems, it is convenient to use polar coordinates ( , θ) r representing a plane. You might already be knowing the relation between cartesian coordinates ( , ) x y and polar coordinates ( , θ) r . We write cos θ, sin θ x r y r = = (Figure 7.20) Then cos θ sin θ ( , ) sin θ cos θ ( , θ) r x y J r r r − ∂ = = = ∂ and ( , ) ( cos θ, sin θ) θ R G f x y dx dy f r r r dr d = ∫∫ ∫∫ Where G is the region in rθ-plane corresponding region R in xy-plane. 147 Line and Surface Integrals Figure 7.20 How the change of the cartesian coordinates to polar coordinates helps in solving problems, is illustrated in the next example. Example 7.16 Find the polar moment of inertia about the origin of a thin plane of density ρ = 1 bounded by the quarter circle 2 2 1 x y + = in the first quadrant. Solution By definition, the polar moment of inertia about the origin is given by 2 2 0 ρ( ) , R I x y dx dy = + ∫∫ where R is the quarter circle in the first quadrant (Figure 7.21). Figure 7.21 : Region in XY-plane Now, if we take cos θ, sin θ, x r y r = = the quarter circle in first quadrant is transformed to a rectangle G in rθ-plane as shown in Figure 7.22, where 0 1, r ≤ ≤ π 0 θ . 2 ≤ ≤ Figure 7.22 : Region in rθ-plane Here ( , ) ( , θ) x y J r r ∂ = = ∂ 148 Engineering Mathematics Also 2 2 2 2 2 2 2 cos θ sin θ x y r r r + = + = ∴ 2 2 2 0 ( ) θ R G I x y dx dy r r dr d = + = ⋅ ∫∫ ∫∫ 1 4 π / 2 1 π / 2 π / 2 3 θ 0 0 0 0 0 1 π θ θ θ 4 4 8 r r r dr d d d = = = = = = ∫ ∫ ∫ ∫ SAQ 5 (a) Find the volume under the plane 6 x y z + + = and above the triangle in the xy-plane bounded by 2 3 , 0 x y y = = and x = 3. (b) Find the mass of the plate between 3 y x = and 2 , x y = if density of plate is 2 2 ρ ( ). K x y = + 7.4 TRANSFORMATION OF DOUBLE INTEGRALS INTO LINE INTEGRIALS – GREEN’S THEOREM We may transform double integrals over a plane region, under suitable condition, into line integrals over the boundary of a region and conversely. This transformation is of practical interest because it makes the evaluation of an integral easier. It also helps in the theory whenever we want to switch from one type of integral to other. This transformation can be done by means of a theorem known as Green’s Theorem which is due to English mathematician, George Green (1793-1841). We shall now state this theorem. Green’s Theorem in a Plane Let R be a closed bounded region in the xy- plane, where boundary C consists of finitely many smooth curves. Let M (x, y) and N (x, y) be functions which are continuous and have continuous partial derivatives y ∂ ∂ M and x ∂ ∂ N every where in some domain containing R. Then ( ) C R dx dy dx dy x y ⎛ ⎞ ∂ ∂ + = − ⎜ ⎟ ∂ ∂ ⎝ ⎠ ∫ ∫∫ N M M N . . . (7.22) the integration being taken along the entire boundary C of R such that R is on the left as one advances in the direction of integration. We shall not be proving this theorem here as it is beyond the scope of this course. Learner interested in knowing its proof may see Appendix-I. However, we shall explain what this theorem means and illustrate it through examples, But, before that we state the theorem in vector form. Green’s Theorem in Vector Form 149 Line and Surface Integrals Let ˆ ˆ ˆ M N P = + + F i j k and ˆ ˆ , x y = + r i j then d M dr N dy ⋅ = + F r Also Curl ˆ ˆ ˆ y z M N P ∂ ∂ ∂ = ∇× = ∂ ∂ ∂ i j k F F x ˆ ˆ ˆ y z z x x y ⎛ ⎞ ⎛ ⎞ ∂ ∂ ∂ ∂ ∂ ∂ ⎛ ⎞ = − + − + − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ∂ ∂ ∂ ∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ P N M P N M i j k The component of Curi F which is normal to a region R in the xy-plane is ( ) x y ∂ ∂ ⋅ − ∂ ∂ N M F k Δ × Here, Green’s Theorem in the plane can be written in the vector form as ˆ (Curl ) ( ) C R R d dx dy d ⋅ = ⋅ = ∇ × ⋅ ∫ ∫∫ ∫∫ F r F k F A . . . (7.23) where dA = ˆ ˆ d dx dy A k = k is the vector normal to the region R in xy-plane and is of magnitude . d dx dy = A Green’s Theorem states that the integral around C of the tangential component of F is equal to the integral, over the region R bounded by C, of the component of Curl F that is normal to R. The integral over R is the flux of curl F through R. We shall later, in Section 7.6, extend this result to more general curves and surfaces in the form of a theorem known as Stoke’s Theorem. There is second vector form of Green’s theorem as follows : Let ˆ ˆ N M = − F i j and let ˆ = n Unit outward vector normal curve C ˆ ˆ cos α cos (90 α) = + + i j ˆ ˆ cos α sin α = − i j ˆ ˆ dy dx ds ds = − i j Thus ˆ ˆ ˆ ˆ ˆ ˆ ( ) dy dx ds N M ds ds ds ⎛ ⎞ ⋅ = − ⋅ − ⎜ ⎟ ⎝ ⎠ F n i j i j dy dx ds ds ds ds = + N M . dx dy = + M N Also in plane in plane ˆ ˆ ˆ ˆ (dis ) ( ) ( ) xy xy M x y − − ⎛ ⎞ ∂ ∂ = ∇⋅ = + ⋅ − ⎜ ⎟ ∂ ∂ ⎝ ⎠ F F i j Ni j = x y ∂ ∂ − ∂ ∂ N M Hence, Green’s Theorem, which says 150 Engineering Mathematics ( ) C R M dx N dy dx dy x y ⎛ ⎞ ∂ ∂ + = − ⎜ ⎟ ∂ ∂ ⎝ ⎠ ∫ ∫∫ N M gives us ˆ C R ds dx dy ⋅ = ∇⋅ ∫ ∫∫ F n F . . . (7.24) In other words, Green’s theorem also states that the normal component of any vector field F around the boundary of a region R, in which F is continuous and has continuous partial derivatives, is equal to the double integral of divergence of F over R. In next section, we shall extend result (7.24) of three-dimensional vector fields and call it a divergence theorem. We can thus say that Green’s theorem in the plane is a two dimensional from of the divergence theorem. We now take up a few applications of Green’s theorem and illustrate its importance. Area of a Plane Region as a Line Integral Over the Boundary From the Green’s theorem, we have + R C dx dy dx dy x x ∂ ∂ ⎛ ⎞ − = ⎜ ⎟ ∂ ∂ ⎝ ⎠ ∫∫ ∫ N M M N . . . (7.25) Let 0 = M and = x, N then, from Eq. (7.25), we get R C dx dy x dy = ∫∫ ∫ . . . (7.26) The integral on the left is the area A of the region R. Next, let M = − y, N = 0, then, from Eq. (7.25), we get R C A dx dy y dx = = − ∫∫ ∫ . . . (7.27) Adding Eqs. (7.26) and (7.27), we get 1 ( ) 2 C A x dy ydx = − ∫ where A is the area of region R enclosed by boundary C. Thus we have been able to express the area of region R in terms of a line integral over the boundary. This interesting formula has various applications, for example, the theory of certain plan meters (instruments measuring area) is based upon this formula. Area of a Plane Region in Polar Coordinates Let r and θ be the polar coordinates. We define cos θ, sin θ. x r y r = = Then cos θ sin θ θ dx dr r d = − and sin θ cos θ θ dy dr r d = + Eq. (7.28) reduces to 1 [ cos θ (sin θ cos θ θ) sin θ sin θ θ] 2 C A r dr r d r dr r d = + − − ∫ 151 Line and Surface Integrals 2 1 θ, 2 C r d = ∫ a formula which is well known in Calculus. As an application of this formula (7.29), we consider the cordiod (1 cos θ) r a = − where 0 θ 2π ≤ ≤ (Figure 7.23) Figure 7.23 : Cordiod r = a (1 – cos θ) We find 2 2 2π 2π 2 2 0 0 (1 cos θ) θ (1 2cos θ cos θ) θ 2 2 a a A d d = − = − + ∫ ∫ 2 2π 0 1 1 2cos θ (cos 2θ 1) θ 2 2 a d ⎡ ⎤ = − + + ⎢ ⎥ ⎣ ⎦ ∫ 2 3π 2 a = We now take up a few example to illustrate the use of Green’s Theorem. Example 7.17 Using Green’s Theorem, evaluate the integral ( ), C y dx x dy − + ∫ where C is the circumference of the circle 2 2 1 x y + = . Solution Here, , y = − M x = N ∴ 2 2 2 2 Circle Circle 1 1 ( ) ( ) ( ) (1 1) C x y x y y dx x dy x y dx dy dx dy x y + = + = ⎡ ⎤ ∂ ∂ − + = − − = + ∫ ⎢ ⎥ ∂ ∂ ⎣ ⎦ ∫∫ ∫∫ 2 2 Circle 1 2 2 Area of Circle x y dx dy = + = = × ∫∫ 2 2 π 1 = × × 2 π. = Example 7.18 Use Green’s Theorem in a plane to evaluate the integral 2 2 2 2 (2 ) ( ) , C x y dx x y dy ⎡ ⎤ + + + ∫ ⎣ ⎦ where C is the boundary of the surface in xy-plane enclosed by the x-axis and the semicircle 2 1 . y x = − 152 Engineering Mathematics Solution Here 2 2 2 , x y = + M 2 2 x y = + N Using Green’s Theorem 2 2 2 2 (2 ) ( ) C x y dx x y dy ⎡ ⎤ + + + ∫ ⎣ ⎦ R dx dy x y ⎛ ⎞ ∂ ∂ = − ⎜ ⎟ ∂ ∂ ⎝ ⎠ ∫∫ N M where R is semi-circle 2 2 1 x y + = bounded by a– axis. (2 2 ) R x y dx dy = − ∫∫ 2 1 1 1 0 (2 2 ) x x y x y dy dx − =− = ⎡ ⎤ = − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ∫ ∫ Figure 7.24 (because semi-circle 2 1 y x = − bounded by x-axis is given by 2 1 1, 0 1 x y x − ≤ ≤ ≤ ≤ − ) 2 1 1 1 2 2 2 1 1 0 2 2 1 (1 ) x x y xy y dx x x x dx − =− − = ⎡ ⎤ = − = − + − ⎢ ⎥ ⎣ ⎦ ∫ ∫ 1 2 3/ 2 3 1 (1 ) 1 1 4 1 1 3 3 3 3 3 2 x x x − − = − − + = − + − + = − 4 3 = in magnitude You may now try the following exercises. SAQ 6 (a) Use Green’s Theorem to evaluate 2 2 2 ( ) ( ) , C x xy dx x y dy ⎡ ⎤ + + + ∫ ⎣ ⎦ where C is the boundary of the square 1, y = ± 1 x = ± 153 Line and Surface Integrals (b) Evaluate 2 2 ( 2 ) ( 3) C x xy dx x y dy ⎡ ⎤ − + + ∫ ⎣ ⎦ around the boundary C of the region 2 8 , 2. y x x = = In Sections 7.3 and 7.4, we have been discussing integrals over plane areas. However, there are many situations in which we may have to consider the areas, which may not lie in a plane. For example, potential due to changes distributed on surfaces, center of gravity of a curved lamina, area of the surface out form the bottom of the paraboloid 2 2 z x y = + by the plane 1, z = etc. This gives rise to the concept of surface integrals, which we shall take up in the next section. 7.5 SURFACE INTEGRALS The concept of a surface integral is a natural generalization of the concept of a double integral considered is Section 7.3. There we integrate over a region in a plane and here we integrate over a piecewise smooth surface in space. The definition of a surface integral is parallel to that of a double integral. Here we consider a portion S of a surface. We assume that S has finite area and is simple, i.e., S has no points at which it intersects or touches itself. Let ( , , ) f x y z be a function which is defined and continuous on S. We sub-divide S into n parts 1 2 , , n S S S … of areas 1 2 , , , n A A A Δ Δ Δ … respectively. Let ( , , ) k k k P x y z be an arbitrary point in each part k S . Let us form the sum 1 ( , , ) n n k k k k x J f x y z A = = Δ ∑ . . . (7.30) Now we let n tend to infinity in such a way that the largest part out of 1 2 , , , n S S S … shrinks to a point. Then the infinite sequence 1 2 , , , n J J J … has a limit which is independent of the choice of subdivisions and points . k P This limit is called the Surface Integral of ( , , ) F x y z over S and is denoted by ( , , ) . S F x y z dS ∫∫ Thus 1 ( , , ) lim ( , , ) , n k k k k n k S F x y z dS f x y z A →∞ = = Δ ∑ ∫∫ . . . (7.31) provided the limit exists. Evaluation of Surface Integral To evaluate the surface integral ( , , ) , S f x y z dS ∫∫ we may reduce it to a double integral as follows : 154 Engineering Mathematics A surface S can be represented in the parametric form as ˆ ˆ ˆ ( , ) ( , ) ( , ) ( , ) , u v x u v y u v z u v = + + r i j k . . . (7.32) where u and v are two independent real variables, called parameters of the representation. Hence ( , ) u v r is the position vector of the points of S and its tip ranges over S as ( , ) u v varies in some region R in uv-plane (Figure 7.25). To each 0 0 ( , ) u v in R there corresponds a point of S with position vector 0 0 ( , ) u v r . Hence, region R is the image of S in uv-plane. In order that surface have certain geometric properties, we assume that ( , ) u v r is continuous and has continuous first partial derivatives u u ∂ ⎛ ⎞ = ⎜ ⎟ ∂ ⎝ ⎠ r r and v v ∂ ⎛ ⎞ = ⎜ ⎟ ∂ ⎝ ⎠ r r in a domain of uv-plane which includes the region R and R is simply connected and bounded. From Eq. (7.32), we have = . u v d du dv + r r r Figure 7.25 : Parametric Representation of a Surface Thus, linear element of surface S is given by 2 ds = d d ⋅ r r ( ) . ( ) u v u v = du dv du dv + + r r r r 2 2 2 . , u u u v v v = du du dv dv ⋅ + + ⋅ r r r r r r = 2 2 2 , E du F du dv Gdv + + . . . (7.33) which is quadratic differential from and is called the First Fundamental form of S. Here , u u u v E = r r F r r ⋅ = ⋅ and . v v G = r r ⋅ In the definition of surface integral, we are sub-dividing S into parts 1 2 , , . . . , , n S S S Let A Δ be the area corresponding to one of the parts in uv − plane. Then the smallest parallelogram in Figure 7.26 has area; (by the definition of vector product). , u v u v A u v u v Δ = Δ × Δ = × Δ Δ r r r r . . . (7.34) and it is called the element of area. 155 Line and Surface Integrals Figure 7.26 : Area Hence, if R is the region corresponding to S in uv-plane, then [ ] ( , , ) ( , ), ( , ), ( , ) S R F x y z ds f x u v y u v z u v dA = ∫∫ ∫∫ ⇒ [ ] ( , , ) ( , ), ( , ), ( , ) u v S R F x y z ds f x u v y u v z u v du dv = × ∫∫ ∫∫ r r . . . (7.35) where the right-hand side is now double integral over a plane area. We know that 2 2 2 ( ) ( ) ( ) u v u u v v u v EG F × + ⋅ ⋅ − ⋅ = − r r r r r r r r . . . (7.36) Thus we may also write ∴ [ ] 2 ( , , ) ( , ), ( , ), ( , ) S R f x y z dS f x u v y u v z u v EG F du dv = − ∫∫ ∫∫ . . . (7.37) If a surface S is given by ( , ), z g x y = We may set , x u y v = = and then parametric representation of S can be written as ˆ ˆ ˆ ( , ) ( , ) u v u v g u v = + + r i j k Thus ˆ ˆ u u g = + r i k and ˆ ˆ v v g = + r j k ∴ 2 1 , u E g = + , u v F g g = 2 1 v G g = + Hence 2 2 2 1 u v u v EG F g g × = − = + + r r Thus, if S is represented by ( , ), z g x y = then we have [ ] 2 2 ( , , ) , , ( , ) 1 , S R g g f x y z dS f x y g x y dx dy x y ⎛ ⎞ ∂ ∂ ⎛ ⎞ = + + ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ∫∫ ∫∫ . . . (7.38) where R is now the image of S in xy-plane. We can also write the surface integral (7.38) in terms of the normal to the surface, S as follows. Suppose that ∇F represents normal to the surface , ( , , ) S x y z C = F (constant). Let region area A be the projection of surface S on a plane. Let ˆ n be a normal to the region A and γ be the angle between the normal to the surface ( ) ∇F and normal to its projection ˆ ( ) n . Then ˆ ˆ cos γ ∇ ⋅ = ∇ ⋅ F n F n 156 Engineering Mathematics or ˆ 1 ˆ ˆ cos γ ∇ ∇ = = ∇ ∇ F n F F n F n Also if S ∇ is an element of surface S surrounding arbitrary point on S and A ∇ is its projection on the planes, then cos γ S A Δ = Δ ⇒ 1 cos γ S A Δ = Δ ⇒ ˆ A S Δ Δ Δ = Δ ⋅ F F n Hence the surface integral can be written in terms of double integral as [ ] ( , , ) ( , ), ( , ), ( , ) ˆ S S f x y z dS f x u v y u v z u v dA Δ = = Δ ⋅ ∫∫ ∫∫ F F n Let us now take up example from various physical situations to illustrate the evaluation of surface integrals. Example 7.19 Find the moment of inertia I of a homogeneous spherical lamina 2 2 2 2 : S x y z a + + = of mass M and z-axis. Solution Let a mass of density μ ( , , ) x y z be distributed over the surface S, then moment of inertia I. of the mass with respect to a given axis L is defined by the surface integral 2 μ , S I D ds = ∫∫ where D is the distance of the point ( , , ) x y z from axis L. Since in our case, spherical lamina is homogeneous, thus μ is constant. Also area of sphere 2 4π . S a = Here μ = mass per Unit Area μ 2 4π a = M The parametric representation of the sphere 2 2 2 2 x y z a + + = is ( , ) ˆ ˆ ˆ cos cos cos sin sin u v a u v a u v a u = + + r i j k ∴ ˆ ˆ ˆ sin cos sin sin sin u a u v a u v a u = − − + r i j k ˆ ˆ cos sin cos cos v a u v a u v = − + r i j ∴ 2 2 2 2 2 2 2 2 2 sin cos sin sin cos u v = a u v a u v a u a ⋅ = + + = E r r 2 2 sin cos sin sin sin cos cos 0 u v = a u v v a u v u v ⋅ = − = F r r 157 Line and Surface Integrals 2 2 2 2 2 2 2 2 cos sin cos cos cos v v = a u v a u v a u ⋅ = + = G r r If area A is the image of surface S in uv-plane, then u v dA du dv = × r r 2 EG F du dv = − 2 2 2 cos 0 a u a du dv = ⋅ − 2 cos a u du dv = ⋅ Further, the square of the distance of a point ( , , ) x y z from z-axis is 2 2 2 2 2 2 2 2 2 2 2 cos cos cos sin cos D x y a u v a u v a u = + = + = Hence, we obtain 2 2 2 2 μ ( , ) ( , ) 4π S r M D ds x u v y u v dA a ⎡ ⎤ = + ⎣ ⎦ ∫∫ ∫∫ π / 2 2π 2 2 2 2 π / 2 0 ( cos ) cos 4π u u M a u a u du dv a =− = = ⋅ ∫ ∫ 4 2 π / 2 π / 2 2π 3 3 0 2 π / 2 0 cos 2 cos 2 4π M a M a u v du u du a = = ⋅ ∫ ∫ 2 2 2 2 2 2 3 3 M a M a = ⋅ ⋅ = Let us now take up an example from geometry. Example 7.20 Find the area of the surface out from the bottom of the paraboloid 2 2 z x y = + by the plane 1 z = . Figure 7.27 : Area of Parabolic Surface Solution The projection of the area of the surface of paraboloid 2 2 z x y = + out from the bottom by the plane 1 z = on xy-plane is the disk 2 2 1 x y + ≤ Now surface area 158 Engineering Mathematics S dS = ∫ ∫ 2 2 1 xy x y R g g dx dy = + + ∫ ∫ 2 2 2 2 1 4 4 1 x y x y dx dy + ≤ = + + ∫ ∫ 2 2 Hence, ( , ) z g x y x y = ⎛ ⎞ ⎜ ⎟ = + ⎝ ⎠ Let cos θ, x = r cos θ y = r Then for 2 2 1, 0 1 x y r + ≤ ≤ ≤ and 0 θ 2π ≤ ≤ ∴ Required surface area 2π 1 2 0 0 4 1 θ r r dr d ⎡ ⎤ = + ⋅ ⎢ ⎥ ⎣ ⎦ ∫ ∫ 1 2 3/ 2 3/ 2 2π 2π 0 0 0 1 (4 1) (5 1) θ θ 3 8 12 2 r r d d = + − = = ∫ ∫ π (5 5 1) 6 = − You may now try the following exercises. SAQ 7 (a) The electrostatic potential at (0, 0, ) a − of a charge of constant density σ on hemisphere 2 2 2 2 : , S x y z a + + = 0 z ≥ is 2 2 2 ( ) dS x y z a σ = + + + ∫ ∫ U Evaluate U . (b) Find the area of the upper cap cut from the sphere 2 2 2 2 x y z + + = by the cylinder 2 2 1 x y + = (Figure 7.28) (Hint : Take surface as 2 2 2 , z x y = − − its projection on xy-plane as disk : 2 2 1 x y + ≤ and use polar coordinates to evaluate double integral.) Figure 7.28 159 Line and Surface Integrals In Section 7.4, we had shown that double integrals over a plane region can be transformed into line integrals over the boundary curve of the region. We shall generalize this result and we shall now also consider the corresponding problem in the case of a surface integral in the next section. 7.6 TRANSFORMATION OF SURFACE INTEGRAL INTO LINE INTEGRALS – STOKE’S THEORM Transformation of surface integrals into line integrals and conversely is done with the help of a theorem known as Stoke’s theorem, given by Gorge Gabriel Stokes (1819-1903), an Irish mathematician and physicist, who made important contribution to the theory of infinite series and several branches of theoretical physics. Stokes theorem is an extension of Green’s theorem in vector from to surfaces and curves in three dimensions. Under suitable restrictions (i) on the vector F, (ii) on the boundary curve C, and (iii) on the surface S boundary by C. Stoke’s theorem connects line integral to a surface integral. Stoke’s Theorem, we require that the surface S is orientable. By orient able, we mean that it is possible to consistently assign a unique direction, called positive, at each point of S, and that there exists a unit normal ˆ n pointing in this direction. As we move about over the surface S without touching its boundary, the direction cosines of the unit vector ˆ n should vary continuously and when we return to the straight position, ˆ n should return to its initial direction. We now state stoke’s theorem. Stoke’s Theorem Let S be a piecewise smooth oriented surface in space and let the boundary of S be a piecewise smooth simple closed curve C. Let ( , , ) x y z F be a continuous vector function which has continuous first partial derivatives in a domain in space which contains S. Then (Curl ) S C dS d ⋅ = ⋅ ∫∫ ∫ F F r , C d = ⋅ ∫ t F r . . . (7.39) where ˆ , dS dS = n where ˆ n is a unit normal vector of S, so that (Curl F). ˆ n is the component of Curl F in the direction of ˆ n ; the integration around C is taken in the direction of integration on S and t F is the component of F in the direction of the tangent vector of C. If we represent curve C in the form ( ), s = r r where the arc-lengths increases in the direction of the direction of integration, then unit tangent vector is ˆ ˆ ˆ , dr dx dy dz ds ds ds ds = + + i j k and, therefore, if 1 2 3 ˆ ˆ ˆ , F F F = + + F i J k then 1 3 3 dr dx dy dz F F F ds ds ds ds = ⋅ = + + t F F 160 Engineering Mathematics Thus 1 3 3 d dx dy dz ds F F F ds ds ds ds = ⋅ = + + t r F F Let ˆ ˆ ˆ ˆ cos cos cos α β γ = + + n i j k be the outward unit normal vector of S. If we use the representation of Curl in terms of right-handed Cartesian coordinates, then formula (7.39) in Stoke’s Theorem may be written as 3 3 2 1 2 1 cos α cosβ cos γ S F F F F F F dS y z z x x y ⎡ ⎤ ∂ ∂ ∂ ∂ ∂ ∂ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − + − + − ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ∂ ∂ ∂ ∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ∫ ∫ 1 2 3 , C F dx F dy F dz = + + ∫ . . . (7.40) where α, β, γ are direction cosines of unit normal ˆ n to surface S. We shall not prove this theorem here. Learner interested in knowing the proof of the theorem may refer to Appendix-II. There are many consequences and applications of Stoke’s theorem. We shall, however, take up a few of these consequences and applications. Consequences and Applications of Stoke’s Theorem (a) Green’s Theorem in the Plane as a Special Case of Stoke’s Theorem Let ˆ ˆ M N = + F i j be a vector function, which is continuously differentiable in a domain in the xy-plane. Let xy-plane contain a simply connected closed region S whose bounding C is a piece-wise simple smooth curve. Then ˆ k is the direction of unit outward normal to S. ˆ ˆ (Curl ) (Curl ) (Curl ) = n ∂ ∂ = ⋅ = ⋅ − ∂ ∂ N M F F n F k x y Further more, . t F ds dx dy = + M N Now formula (7.39) of Stoke’s theorem yields (Curl ) n t S C ds ds = ∫ ∫ ∫ F F Using above values in this case, we get ( ), S C dx dy dx dy y ⎛ ⎞ ∂ ∂ − = + ⎜ ⎟ ∂ ∂ ⎝ ⎠ ∫ ∫ ∫ N M M N x which is in agreement with the result of the Green’s theorem in the plane (refer Section 7.4). This shows that Green’s theorem in the plane is a special case of Stoke’s theorem. (b) Physical Interpretation of the Curl Stoke’s theorem provides a physical interpretation of the Curl. Let us first define the term Circulation. Let ( , , ) x y z V be a continuous differentiable vector function in a domain containing a surface S, bounded by a closed curve C. If V be the velocity field of a moving fluid of density ρ, then the integral ρ C d ds ds ⋅ ∫ r V or ρ C ds ∫ t V 161 Line and Surface Integrals measures the extent to which the corresponding fluid motion is a rotation around the contour C and is called the circulation. By Stoke’s theorem, circulation is also equal to the flux of curl (ρ ) V through a surface S spanning C. Thus ˆ (ρ ) (ρ ) Curl (ρ ) C C S d ds ds ds ds ⋅ = = ⋅ ∫ ∫ ∫ ∫ t r V V V n . . . (7.41) Next, let us fix a point P and a direction u at P. Let C be a circle of radius δ, with center at P, whose plane is normal to u. If curl (ρ ) V is continuous at P, then (by mean value theorem for surface and double integrals) the average value of u component of curl (ρ ) V at P as δ 0, → i.e., [ ] 2 δ 0 1 Curl (ρ ) lim Curl (ρ ) πδ P S ds → ⎡ ⎤ ⋅ = ⋅ ⎣ ⎦ ∫ ∫ V u V u . . . (7.42) Figure 7.29 : Intercept of Curl Now in our case, the plane of curve C is normal to u, area of 2 πδ , S A = = say so that . dS dA = Let ρ . = F V Thus, form Eqs. (7.41) and (7.42), we get [ ] δ 0 1 ˆ Curl ( ) lim Curl n S P dA A → = ⋅ ∫ ∫ F F n δ 0 1 lim t C ds A → = ∫ F i.e., the component of the curl in the positive normal direction can be regarded as the specific circulation (circulation per unit area) of the flow in the surface at the corresponding point. (c) Evaluation of Line Integral by Stoke’s Theorem You may notice that evaluation of line integral by Stoke’s theorem leads to a lot of simplification in many problems. Let us evaluate , t C ds ∫ F where C is the circle 2 2 4, 3 x y z + = = − oriented in the counterclockwise sense as viewed from the origin and with respect to right-handed coordinates, and 3 3 ˆ ˆ ˆ . y xz zy = + − F i j k We can take the plane circular disk 2 2 4 x y + ≤ in the plane 3 z = − as a surface S bounded by C. Then ˆ n in the Stoke’s theorem point in the positive z-direction, so that ˆ ˆ n = k. Hence, (Curl ) n F is simply the component of Curl F in the ˆ k direction. Now F with 3 z = − has the components 1 2 , 27 F y F x = = − and 3 3 3 . F y = Thus, 162 Engineering Mathematics 2 1 (Curl ) (Curl ) 27 1 28 n k F F x y ∂ ∂ = = − = − − = − ∂ ∂ F F Hence, integral over S in Stoke’s theorem equals – 28 times the area of the disk S, which is 4π. Thus, 28 4π 112π 352 t C F ds = − × = − ≈ − ∫ You will appreciate the simplification due to Stoke’s theorem in this case. The computations will be long and difficult if you evaluate the given line integral directly, starting from parametric representation of C, calculating unit tangent vector ˆ t of curve C in the direction of integration, finding ˆ t F = ⋅ F t and then integrating will respect to arc length s of C. It should be remembered that for Stoke’s theorem, the surface S is assumed to be an open surface. The significance of the statement will become clear from the following result. (d) Curl C dS ⋅ ∫ ∫ F will be Zero Over any Closed Surface S Since is any closed surface, we cut open the surface S by a plane and let 1 S and 2 S denote the lower and upper positions of S. Let C denote the common bounding curve for both these positions. Then, 1 2 Curl Curl Curl C S S dS dS dS ⋅ = ⋅ + ⋅ ∫ ∫ ∫ ∫ ∫ ∫ F F F . . . (7.43) It may be noted that the normal to lower portion 1 S and the normal to upper portion 2 S will be in opposite senses so that the direction of integration about 1 S is reverse of the direction of integration about 2 S . Using Stoke’s theorem and result (7.43) above, we get Curl 0 C C S d d d ⋅ = − ⋅ + ⋅ = ∫ ∫ ∫ ∫ F S F S F r 1 (due to surface ) S 2 (due to surface ) S which proves the required result. Let now we take up some example to illustrate the use of Stoke’s theorem in numerical problems. Example 7.21 Let S be the portion of the paraboloid 2 2 4 z x y = − − that lies above the plane 0. z = Let C be their bounding curve of intersection. If ˆ ˆ ˆ ( ) ( ) ( ) , z y z x x y = − + + − + F i j k verify Stoke’s theorem. Solution The curve C is the circle 2 2 4 x y + = in the xy-plane C, where 0 z = and ˆ ˆ dr dx dy = + i j, we have ( ) ( ) ( ) 0 d z y dx z x dy x y ⋅ = − + + − + ⋅ F r 163 Line and Surface Integrals , y dx x dy = − + on C Now, ( ) C C dr ydx xdy ⋅ = − + ∫ ∫ F on C, 2cos θ, 2sin θ, 0 θ 2π x y = = ≤ ≤ 2π 0 [ 2sin θ ( 2sin θ) θ (2cos θ) (2cos θ θ)] C dr d d ⋅ = − − + ∫ ∫ F 2π 0 4 θ 8π d = = ∫ Now for surface integral, we have ˆ ˆ ˆ ˆ ˆ ˆ Curl 2 2 2 ( ) x y z y z x x y ∂ ∂ ∂ = = − + + ∂ ∂ ∂ − + − + i j k F i j k z For the positive unit normal on the surface 2 2 : ( , , ) 4 0 S f x y z z x y = − + + = we take 2 2 ˆ ˆ ˆ 2 2 ˆ 4 4y 1 gradf x y gradf + + = = + + i j k n x The projection on S on xy-plane is the region 2 2 4 x y + ≤ and for the element of surface area dS, we take 2 2 2 2 1 4 4 1 z z dS dx dy x y dx dy x y ⎛ ⎞ ∂ ∂ ⎛ ⎞ = + + = + + ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ Thus, ˆ Curl S ⋅ ∫∫ F ndS 2 2 2 2 2 2 4 ˆ ˆ ˆ 2 2 ˆ ˆ ˆ ( 2 2 2 ) 4 4y 1 4 4y 1 . x y x x dx dy + ≤ ⎛ ⎞ + + ⎜ ⎟ = − + + + + ⎜ ⎟ + + ⎝ ⎠ ∫∫ i j k i j k x x 2 2 4 ( 4 4 2) x y x y dx dy + ≤ = − + + ∫∫ 2 2 4 2 x y dx dy + ≤ = ∫∫ (Since odd power of x or y integrate to zero over the interior of the circle) 2 2 4 2 x y dx dy + ≤ = ∫∫ = 2 × area of circle 2 2 4 x y + ≤ 2 2 π 2 8π = × ⋅ = Hence, Stoke’s theorem in verified. 164 Engineering Mathematics You may now attempt a few exercises. SAQ 8 (a) Verify Stoke’s theorem for the function 2 ˆ ˆ x xy = − F i j integrated round the square in the plane 0 z = and bounded by the lines 0, 0, x y x a = = = and . y a = (b) A fluid of constant density ρ rotates round the z–axis with velocity ˆ ˆ ( ), w x y = − V j i where w is a positive constant. If ρ , = F V find curl F and show its relation to specific circulation. (c) Let ˆ n be the outer normal of the elliptical shell 2 2 2 : 4 9 36 36, 0 S x y z z + + = ≥ and let 2 2 2 3/ 2 ˆ ˆ ˆ ( ) sin ( ) . xyz y x x y e = + + + F i j k Use Stoke’s theorem to find the value of ˆ (Curl ) S ds ⋅ ∫∫ F n (d) Let S be the region bounded by the ellipse 2 2 : 4 4 C x y + = in the plane z =1 and let ˆ ˆ = n k and 2 2 ˆ ˆ ˆ 2 . x x z = + + F i j k Use Stoke’s theorem to find the value of . C d ∫ F r 7.7 TRIPLE INTEGRALS In section 7.1, the concept of a line integral was introduced. Section 7.2 was devoted to the properties and evaluation of double integrals, in which the integrand is a function of two variables. A natural generalization of double integrals was provided in Section 7.5 in terms of surface integrals, where though the integrand may be a function of three variable, but it is defined on a surface only and it is still evaluated as a double integral. However, there are many physical and geometrical situations, where the integrand may be a function defined in a region in space and integration may have to be carried out with respect to all the three variables involved. This gives rise to triple integrals. A triple integral is again a generalization of double integral (introduced in Section 7.2). 7.7.1 Definition For defining a triple integral, consider a function ( , , ) f x y z defined on a bounded region R in space (such as a solid cube, a ball, a truncated cone or the space between two concentric spheres). We subdivide the region R into rectangular cells by planes parallel to the three coordinate planes. The parallelopiped cells may have the dimensions of , x Δ y Δ and . z Δ We number the cells inside R in same order 1 2 , , . . . , n V V V Δ Δ Δ 165 Line and Surface Integrals Figure 7.30 : Partitioning a Solid with Rectangular Cells of Volume Δ V In each such parallelopiped cell, we choose an arbitrary point, say, ( , , ) k k k x y z in the th k parallelopiped cell, and form the sum 1 ( , , ) , n n k k k k k S f x y z V = = Δ ∑ . . . (7.43) where k V Δ is the volume of th k cell. We now arbitrarily do this for larger and larger positive integer n so that the lengths of the edges of the largest parallelopiped of subdivision approaches zero as n approaches infinity. In this way, we obtain a sequence of real numbers 1 2 , , . . . n n S S If we assume that ( , , ) f x y z is continuous in a domain containing R and R is bounded by finitely many smooth curves, then as , , x y z Δ Δ Δ all approach zero with n approaching infinity, the sum n S will approach a limit [independent of choice of subdivisions corresponding points ( , , ) k k k x y z ] Lt ( , , ) , n n R S f x y z dV →∞ = ∫ ∫ ∫ . . . (7.44) which is called the triple integral of ( , , ) f x y z over the region R. It may be mentioned that the limit in Eq. (7.44) may exist for some discontinuous functions also. 7.7.2 Properties of Triple Integrals Triple integrals share may algebric properties with double and single integrals. Writing F for ( , , ) F x y z and G for ( , , ), G x y z some of the properties of triple integrals are (i) R R K F dV K F dV = ∫ ∫ ∫ ∫ ∫ (any constant) (ii) ( ) R R R F G dV F dV GdV ± = ± ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ (iii) 0 R F dV ≥ ∫ ∫ ∫ if 0 F ≥ in R (iv) R R F dV GdV ≥ ∫ ∫ ∫ ∫ ∫ ∫ if F G ≥ in R The triple integrals have the above properties because the sums that approximate them have these properties. Triple integrals also have a domain additivity property that proves useful in physics, mathematics as well as in engineering. If the domain R of a continuous function F be partitioned by smooth surfaces into a finite number of regions 1 2 , , . . . , n R R R in space, then 1 2 . . . n R R R R F dV F dV F dV F dV = + + + ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ 166 Engineering Mathematics 7.7.3 Volume If ( , , ) 1 f x y z = is the constant function whose value is one, then the sums in Eq. (7.43) reduce to 1 1 1 n n n k k k k S V V = = = ⋅ Δ = Δ ∑ ∑ . . . (7.45) As , x y Δ Δ and z Δ all approach zero, the cells k V Δ become smaller and smaller and more numerous and fill up more of R. We, therefore, define the volume of R to the triple integral of constant function ( , , ) 1 f x y z = over R. Hence, Volume of 8 1 n k n k R R Lt V dV → = = Δ = ∑ ∫ ∫ ∫ . . . (7.46) The triple integral is seldom evaluated directly from its definition as the limit of a sum. Instead, one applies a three-dimensional version to evaluate the integral by repeated single integrations. This method is explained below. 7.7.4 Evaluation of Triple Integrals To express the triple integral as a repeated integral we divide the region R into elementary cuboids by planes parallel to the three coordinate planes. The Volume R may then be considered as the sum of vertical columns extending from the lower surface of R, say 1 ( , ) z z x y = to its upper surface 2 ( , ) z z x y = The bases of these columns are the elementary area δ , r A which cover a certain region A in the xy-plane when all the columns in R are taken. Figure 7.31 : Evaluation of Triple Integral Thus, if we take the sum over the elementary cuboids in the same vertical column first, and then take the sum for all the columns in R, we can write Eq. (7.43) as ( , , ) δ k k k r r k f x y z z A ⎡ ⎤ Δ ⎢ ⎥ ⎣ ⎦ ∑ ∑ where ( , , ) k k k x y z is a point in the th k cuboid above the th r elementary area δ . r A Taking the limit when the dimensions of r A δ and z Δ tend to zero, the above sum yields 2 1 ( , ) ( , ) ( , , ) z x y z x y R f x y z dz dA ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ ∫ ∫ ∫ . . . (7.47) 167 Line and Surface Integrals Now integration with respect to z is performed first, keeping x and y constant. The remaining integration is performed as for the double integral. Therefore, if A is bounded by the curves 1 2 ( ), ( ), , , y y x y y x x a x b = = = = then triple integral in equation (7.5) may be written as 2 2 1 1 ( ) ( , ) ( ) ( , ) ( , , ) , b y x z x y a y x z x y f x y z dz dy dx ∫ ∫ ∫ where the three integrations are performed in order from the innermost to be outermost. It should, however, be noted that the region A is the projection in the xy-plane of the bounding surface of the volume bounded by R. In case the lateral surface of the system reduces to zero, as in Figure 7.32, we may find the boundary of A by eliminating z between the two equations 1 ( , ) z f x y = and 2 ( , ). z f x y = Figure 7.32 : Lateral Surface Reduces to Zero This gives 1 2 ( , ) ( , ) f x y f x y = an equation that contains no z and that defines the boundary of A in the xy-plane. To obtain the z-limits of integration in any particular instance, we may use a procedures as follows : We imagine a line L through a point (x, y) in area A and parellel to z-axis. As z increase, the line enters R at 1 ( , ) z f x y = and leaves R at 2 ( , ). z f x y = These give the lower and upper limits of the integration with respect to z. The result of this integration is a function of x and y alone, which we can integrate supplying the limits as in double integrals. Let us take up an example to illustrate the evaluation of triple integrals. Example 7.22 Find the volume enclosed between the two surface 2 2 3 z x y = + and 2 2 8 z x y = − − Solution The given two surfaces intersect on the elliptic cylinder 2 2 2 2 3 8 x y x y + = − − ⇒ 2 2 2 4 x y + = The volume projects into the area A (in the xy-plane) that is enclosed by the ellipse having the same equation. In the double integrals with respect to y and x over. A, if 168 Engineering Mathematics we integrate with respect to y, holding x fixed, then y varies from 2 (4 ) 2 x − − to 2 (4 ) 2 x − + . Then x varies form – 2 to 2. Thus, we have 2 2 2 2 2 2 (4 ) 2 8 2 2 3 (4 ) 2 x x y x z x y x y V dz dy dx − + − − =− = + − =− = ∫ ∫ ∫ 2 2 (4 ) 2 2 2 2 2 (4 ) 2 (8 2 4 ) x x x y x y dy dx − + =− − =− = − − ∫ ∫ 3/ 2 2 2 2 2 2 4 8 4 2(8 2 ) 2 3 2 x x x dx x =− ⎡ ⎤ − ⎛ ⎞ − = ⎢ ⎥ − − ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ ∫ 2 2 3/ 2 2 4 2 (4 ) 8π 2 3 x dx − = − = ∫ You will recall that there are sometimes two different orders in which the single integrations, that evaluate a double integrals, may be worked (refer Sub-section 7.3.4). For triple integrals there are sometimes (but not always) as many as six workable orders of integration. We consider an extreme case in which all six orders are possible. The volume of solid shown in Figure 7.33 (a prism) is given by each of the following six integrals : (a) 1 1 2 0 0 0 x dx dy dz − ∫ ∫ ∫ (b) 2 1 2 0 0 0 y dx dz dy − ∫ ∫ ∫ (c) 1 1 1 0 0 0 z dy dz dx − ∫ ∫ ∫ (d) 1 2 1 0 0 0 z dy dx dz − ∫ ∫ ∫ (e) 1 2 1 0 0 0 y dz dxdy − ∫ ∫ ∫ (f) 2 1 1 0 0 0 y dz dy dz − ∫ ∫ ∫ Figure 7.33 : Volume of Prism 169 Line and Surface Integrals 7.7.5 Physical Applications in Three Dimensions The triple integrals find various physical applications in three dimensions. It can be used to find the mass of any object having variable density, its first moments about the coordinate planes, the center of mass, the moments of inertia, the radius of gyration about any line, etc. If ( , , ) ρ( , , ) f x y z x y z = is the density of an object occupying a region R in space, then some useful formulae are summarized as follows : Mass ρ . R m dV = ∫ ∫ ∫ (ρ = density) First moments about coordinate planes : ρ , yz R M x dV = ∫ ∫ ∫ ρ , zx R M y dV = ∫ ∫ ∫ ρ xy R M z dV = ∫ ∫ ∫ Centre of mass : ρ ρ ρ , , x dV y dV z dV x y z m m m = = = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ Moments of Inertia (second moments) : 2 2 ρ( ) , x I y z dV = + ∫ ∫ ∫ 2 2 ρ( ) , y I z x dV = + ∫ ∫ ∫ 2 2 ρ( ) , z I x y dV = + ∫ ∫ ∫ 2 ρ , L I r dV = ∫ ∫ ∫ where ( , , ) r x y z = distance from point (x, y, z) to line L. Radius of gyration about a line : . L I L R m ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ We now take up an example to illustrate the physical applications of triple integrals. Example 7.23 Find the center of gravity of a solid of uniform density ρ bounded below by the disc 2 2 : 4 R x y + ≤ in the plane 0 z = and above by the paraboloid 2 2 4 . z x y = − − Solution The solid is shown in Figure 7.34. 170 Engineering Mathematics Figure 7.34 By symmetry 0 . x y = = To Find z we calculate 2 2 4 0 ρ z x y xy z R M z dz dy dx = − − = = ∫ ∫ ∫ 2 2 4 2 0 ρ 2 x y R z dy dx − − = ∫ ∫ 2 2 2 ρ (4 ) 2 R x y dy dx = − − ∫ ∫ 2π 2 2 2 θ 0 0 ρ (4 ) θ 2 r r r dr d = = = − ∫ ∫ (in polar coordinates) 2 2π 2 3 θ 0 0 ρ 1 θ (4 ) 2 6 r d r = = = − − ∫ 2π 0 16 π θ 3 d = ∫ 32πρ 3 = A similar calculation gives 2 2 4 0 ρ 8πρ x y z R m dz dy dx − − = = = ∫ ∫ ∫ ∴ 4 3 xy M z m = = Thus, the center of gravity of the given solid is 4 ( , , ) (0, 0, ) 3 x y z = You may now attempt the following exercises. SAQ 9 (a) Find the center of gravity of the volume cut off from the cylinder 2 2 2 0 x y ax + − = by the planes z mx = and . z nx = 171 Line and Surface Integrals (b) Evaluate 0 0 0 a x x y x y z e dz dy dx + + + ∫ ∫ ∫ and state precisely what is the region of integration. (c) Find the mass of a solid in the form of the positive octant of the sphere 2 2 2 2 , x y z a + + = if the density at any point is k xyz. We shall now show that the triple integral of the divergence of a vector function u over a region R in space, under certain conditions, can be transformed into a surface integral of the normal component of u over the boundry S of R. This can be done by means of the Divergence Theorem, which is the three – dimensional analog of the Green’s Theorem in the plane considered in Section 7.4. The divergence theorem is of basic importance in various theoretical and practical considerations. 7.8 TRANSFORMATION OF VOLUME INTEGRALS INTO SURFACE INTEGRALS AND CONVERSELY 7.8.1 Gauss Divergence Theorem Statement Let R be a closed bounded region in space whose boundary is a piece-wise smooth orient able surface S. Let u (x, y, z) be a vector function, which is continuous and has continuous first partial derivatives in some domain containing R. Then div . n R R dV u dS = ∫ ∫ ∫ ∫ ∫ u . . . (7.48) where ˆ n u = ⋅ u n is the component of u is the direction of the outward normal of S with respect of R and ˆ n is the outer unit normal vector of S. If we write u and ˆ n in terms of components, say, 1 2 3 ˆ ˆ ˆ u u u + + u = i j k and ˆ ˆ ˆ ˆ cosα cosβ cos γ = + + n i j k where α, β and γ are the angles between ˆ n and the positive directions of x, y and z axes, respectively, then Eq. (7.48) takes the form 3 1 2 1 2 3 ( cosα cosβ cos γ) R S u u u dxdy dz u u u dS x y z ∂ ∂ ∂ ⎛ ⎞ + + − + + ⎜ ⎟ ∂ ∂ ∂ ⎝ ⎠ ∫ ∫ ∫ ∫ ∫ . . . (7.49) Since S is an orient able surface, then, by definition, this means 1 1 cosα S S u dS u dy dz = ∫ ∫ ∫ ∫ . . . (7.49a) 2 2 cosβ S S u dS u dz dx = ∫ ∫ ∫ ∫ . . . (7.49b) 3 3 cos γ S S u dS u dx dy = ∫ ∫ ∫ ∫ . . . (7.49c) Then relation (7.48), with the help of relations (7.49a, b, c), may be written as 172 Engineering Mathematics 3 1 2 1 2 3 ( ) R S u u u dx dy dz u dy dz u dz dx u dx dy x dy dz ∂ ∂ ∂ ⎛ ⎞ + + = + + ⎜ ⎟ ∂ ⎝ ⎠ ∫ ∫ ∫ ∫ ∫ . . . (7.50) It is clear that relation (7.48) is true, if the following three relations hold simultaneously : 1 1 R S u dx dy dz u dy dz x ∂ = ∂ ∫ ∫ ∫ ∫ ∫ . . . (7.51) 2 2 R S u dx dy dz u dz dx y ∂ = ∂ ∫ ∫ ∫ ∫ ∫ . . . (7.52) 3 3 R S u dx dy dz u dx dy z ∂ = ∂ ∫ ∫ ∫ ∫ ∫ . . . (7.53) For Divergence Theorem, either the either the region R in space is some convex region with no holes (such as the interior of a sphere, or a cube, or an ellipsoid) or R is bounded by a piecewise smooth orient able surface S. In the later case, region R has the property that any straight line parallel to any one of the coordinate axes and intersecting R has at most one segment (or a single point) in common with R. This implies that R can be represented in the form ( , ) ( , ), f x y z h x y ≤ ≤ where (x, y) varies in the orthogonal projection of R in the xy-plane. It may be added that the divergence theorem is also true for region R which can be subdivided into finitely many regions of the type discussed above by means of auxiliary surfaces. It will be seen that the surface integral over the auxiliary surfaces cancel in pairs, and the sum of the remaining surface integrals is the surface integral over the whole boundary S of R. The proof of Divergence Theorem is given in Appendix-III and those interested in proof may look up the Appendix-III. We shall now take an example to evaluate a surface integral by the divergence theorem. Example 7.24 Evaluate 3 2 2 ( ) S I x dy dz x y dz dx x z dx dy = + + ∫ ∫ Where S is the closed surface consisting of the cylinder 2 2 2 (0 ) x y a z b + = ≤ ≤ and the circular disks 0 z = and 2 2 2 ( ). z x y a = + ≤ Solution If we compare the given surface integral with the right-hand side of relation (7.50) above for divergence theorem, 3 2 2 1 2 3 , , u x u x y u x z = = = Hence, if we use the symmetry of the region R bounded by S, the volume integral in relation (7.50) assumes the form 2 2 2 2 2 2 0 0 0 (3 ) 4.5 b a a y z y x R x x x dx dy dz x dx dy dz − = = = + + = ∫ ∫ ∫ ∫ ∫ ∫ 173 Line and Surface Integrals Thus 2 2 2 0 0 0 20 b a a y z y x I x dxdy dz − = = = = ∫ ∫ ∫ 2 2 3/ 2 0 0 ( ) 20 3 b a z y a y dy dz = = − = ∫ ∫ Let cos y a t = sin dy a t dt = − where 0, 2 y t π = = Where , 0 y a t = = 2 2 2 2 sin a y a t − = ∴ 0 2 2 3/ 2 3 3 4 0 π / 2 3π ( ) sin . sin 16 a a y dy a t a t dt a − = − = ∫ ∫ ∴ 4 4 0 π 5 20 π 16 4 b I a dz a b = = ∫ We take up another example wherein we shall verify the divergence theorem. Example 7.25 Verify divergence theorem for the region 2 2 2 1 4 x y z ≤ + + ≤ if 2 2 2 ˆ ˆ ˆ , ρ ρ x y z F x y z + + = = + + i j k Solution Here ρ ρ x x ∂ = ∂ And 2 3 3 4 3 5 ρ 1 3 ( ρ ) ρ 3 ρ ρ ρ x x x x x − − − ∂ ∂ = − = − ∂ ∂ The given region can be expressed as 1 ρ 2, ≤ ≤ (annular region between two spheres). Thus, throughout the region 1 ρ 2, ≤ ≤ all functions considered are continuous. Also 2 2 2 3 5 3 3 3 3 3 3 div ( ) 0 ρ ρ ρ ρ F x y z = − + + + = − + = Thus div 0 R F dV = ∫ ∫ ∫ . . . (7.54) On the outer sphere (ρ 2), = the unit normal is ˆ ˆ ˆ ρ x y z + + = i j k n ∴ 2 2 2 4 2 1 ˆ ρ ρ x y z dS dS dS + + ⋅ = − = − F n Hence 2 ρ 2 2 ρ 2 ρ 2 ρ 2 1 1 1 ˆ 4πρ | 4 4 ρ dS dS dS = = = = = − = − = − ∫ ∫ ∫ ∫ ∫ ∫ F n 2 ρ 2 πρ | 4π = = − = − . . . (7.55) On the inner sphere (ρ 1), = the positive unit normal points towards the origin and it is given by 174 Engineering Mathematics ˆ ˆ ( ) ˆ ρ x y z + + = i j k n Hence 2 2 2 4 2 1 ˆ ρ ρ x y z dS dS + + ⋅ = = − F n Thus 2 ρ 1 2 ρ 1 ρ 1 1 ˆ 4π ρ | 4π ρ dS dS = = = ⋅ = = = ∫ ∫ ∫ ∫ F n . . . (7.56) The sum of Eqs. (7.55) and (7.56) is the surface integral over the complete boundary of the given region, which is 4π 4π 0, − + = which agrees with Eq. (7.54) and this verifies the divergence theorem. You may now attempt the following exercises. SAQ 10 (a) Verify divergence theorem for the sphere 2 2 2 2 x y z a + + = if ˆ ˆ ˆ x y z = + + F i j k (b) Use divergence theorem to evaluate the surface integral , S dS ⋅ ∫ ∫ ∫ F where 2 2 ˆ ˆ ˆ 4 2 x y z = − + F i j k and S is the surface bounding of the region 2 2 4, 0 x y z + = = and 3. z = There is many important consequences and various applications of divergence theorem, some of which we shall take up next. 7.8.2 Consequences and Applications of Divergence Theorem The Gauss’s Divergence Theorem enables us to express a normal surface integral as a volume integral. We shall now see that with the help of this theorem, we are able to express the other two types of surface integral as volume integral, i.e., we shall show that Curl S S R n dS dS dV × = × = ∫ ∫ ∫ ∫ ∫ ∫ ∫ F F F . . . (7.57) and ˆ , S S R dS dS dV Φ = Φ× = ΔΦ ∫ ∫ ∫ ∫ ∫ ∫ ∫ n . . . (7.58) where S is the surface bounding a region R in space, satisfying the conditions of divergence theorem. We write , × f a F = Where a is a constant vector and apply divergence theorem to the vector function f. Thus, we have 175 Line and Surface Integrals ˆ ( × ) div ( × ) S R dS dV ⋅ = ∫ ∫ ∫ ∫ ∫ a F n a F . . . (7.59) But ˆ ( × ) × ⋅ = a F n a F n ⋅ . . . (7.60) And div ( ) ( × ) a × = ∇ ⋅ = − ⋅ ∇ × a F a F F . . . (7.61) From Equs. (7.59), (7.60) and (7.61), we get ˆ . S R dS dV ⋅ × = − ∇ × ∫ ∫ ∫ ∫ ∫ a F n a F ⇒ ˆ × S R dS dV ⋅ × = − ⋅ ∇ ∫ ∫ ∫ ∫ ∫ a F n a F ⇒ ˆ × 0 S R dS dV ⎡ ⎤ ⎢ ⎥ × + ∇ = ⎢ ⎥ ⎣ ⎦ ∫ ∫ ∫ ∫ ∫ a F n F . . . (7.62) Since a is an arbitrary vector, thus Eq. (7.62) yields ˆ × 0 S R dS dV × + ∇ = ∫ ∫ ∫ ∫ ∫ F n F ⇒ ˆ Curl S S dS dV × = − ∫ ∫ ∫ ∫ ∫ F n F Thus we have proved Eq. (7.57) To prove Eq. (7.58), let us write Φ = f a where a is a constant vector and Φ is a scalar function. Applying Gauss’s theorem to f, we have ˆ div S S dS dV ⋅ = ∫ ∫ ∫ ∫ ∫ f n f ⇒ ˆ div ( ) . ( ) S R R dS dV dV Φ = Φ = ∇ Φ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ a n a a ⇒ ˆ . . S R dS dV Φ = ∇ Φ ∫ ∫ ∫ ∫ ∫ a n a ⇒ ˆ . 0 S R dS dV ⎡ ⎤ ⎢ ⎥ Φ − ∇ Φ = ⎢ ⎥ ⎣ ⎦ ∫ ∫ ∫ ∫ ∫ a n Since vector a is arbitrary, thus we obtain ˆ 0 S R dS dV Φ − ∇ Φ = ∫ ∫ ∫ ∫ ∫ n ⇒ ˆ , S R dS dV Φ = ∇ Φ ∫ ∫ ∫ ∫ ∫ n which is Eq. (7.58). 7.8.3 Integral Definitions of Gradient, Divergence and Curl With the help of the results obtained above, we can state the following integral expressions for group φ, div F and Curl F. Let S be closed surface bounding a volume V 176 Engineering Mathematics in space, Φ be a scalar function and F be a vector function defined at every point of V. Then we have 0 grad , S V dS Lt V → Φ Φ= ∫ ∫ 0 div , S V dS Lt V → ⋅ = ∫ ∫ F F and 0 Curl S V dS Lt V → × = ∫ ∫ F F These expressions are often adopted as definitions of the differential operations. It becomes particularly simple to prove the divergence and Stoke’s theorem starting from these definitions. 7.8.4 Physical Interpretation of Divergence Theorem We consider the steady flow of an incompressible fluid of constant density ρ 1. = Such a flow is determined by the field of its velocity vector q. Let S be any closed surface drawn in the fluid, which is enclosing a volume V. By Gauss’s Divergence Theorem, ˆ div S V dS dV = ∫ ∫ ∫ ∫ ∫ q n q ⋅ . . . (7.63) Now ˆ q n ⋅ is the component of the velocity at any point of S in the direction of the outward drawn normal. Thus ˆ q n ⋅ dS denotes the amount of the fluid that flows out in unit time through the element δ . S Hence, the left hand side of Eq. (7.63) denotes the amount of fluid flowing across the surface S in unit time from the inside to the outside. This amount may be positive, negative of zero. Now the total amount flowing outwards must be continuously supplied so that inside the region we must have sources producing fluid. We know that div q at any point denotes the amount of fluid per unit time per unit volume that goes through any point. Thus, div q may be through of as the source-intensity of the incompressible fluid at any point. P. Hence, the right-hand side of Eq. (7.63) denotes the amount of fluid per unit time supplied by the source within S. Thus the equality in Eq. (7.63) appears intuitively evident. We know that in body heat flows in the direction of decreasing temperature. Physical experiments show that the rate of flow is proportional to the gradient of the temperature. This means that the velocity q of the heat flow in a body is of the form grad q k U = − . . . (7.64) where U (x, y, z) is the temperature, t is the time and k is called the thermal conductivity of the body – in ordinary physical circumstances k is a constant. Using this information and Gauss’s divergence theorem, we now set up the mathematical model of heat flow, the so called heat-equation. 7.8.5 Modeling of Heat Flow Let R be a region in the body and let S be its bounding surface. Then the heat leaving R per unit time is , n S q dS ∫ ∫ 177 Line and Surface Integrals where, ˆ n q = ⋅ q n is the component of q in the direction of the outward unit normal vector of S. From the Eq. (7.64) and the divergence theorem, we obtain 2 div (grad ) n S R R q dS k U dx dy dz k U dx dy dz = − = − ∇ ∫ ∫ ∫ ∫ ∫ ∫ ∫ . . . (7.65) On the other hand, the total amount of heat H in R is . R U H dxdy dz t σ ρ ∂ = ∂ ∫ ∫ ∫ where the constant σ is the specific heat of the material of the body and ρ is the density (mass per unit volume) of the material. Hence, the time rate of decrease of H is ρ . R H U dx dy dz t t σ ∂ ∂ − = − ∂ ∂ ∫ ∫ ∫ This rate of decrease of total heat must be equal to the above amount of heat leaving R. Thus from equation (7.65) we have 2 ρ . R R U dx dy dz k U dx dy dz t σ ∂ − = − ∇ ∂ ∫ ∫ ∫ ∫ ∫ ∫ ⇒ 2 0. ρ R U dx dy dz k U t σ ∂ ⎡ ⎤ − = − ∇ ⎢ ⎥ ∂ ⎣ ⎦ ∫ ∫ ∫ Since this holds for any region R in the body, thus the integrand (if continuous) must be zero everywhere, i.e., 2 ρ 0 U k U t σ ∂ − ∇ = ∂ ⇒ 2 2 , U c U t ∂ = ∇ ∂ where 2 σρ k c = The above partial differential equation is called the heat-equation and it is fundamental for heat conduction. We next take up a basic consequence of divergence theorem, known as Green’s theorem. 7.8.6 Green’s Theorem and Green’s Formula Let f and g be scalar functions such that u = f grad g satisfies the assumptions of the divergence theorem in some region R. Then 2 div div( grad ) grad grad . f g f g f g = = ∇ + ⋅ u Also ˆ ˆ ˆ ( grad ) ( grad ) f g f g = ⋅ = ⋅ u n n n ⋅ The expression ˆ grad g ⋅ n is the directional derivative of g in the direction of outward normal vector ˆ n of the surface S in the divergence theorem. If we denote this derivative by , ∂ ∂ g n then the formula in the divergence theorem becomes 2 ( gard grad ) R S g f g f g dV f dS n ∂ ∇ + ⋅ = ∂ ∫ ∫ ∫ ∫ ∫ . . . (7.66) which is called Green’s First Formula or the first form of Green’s Theorem. Interchanging f and g, we can obtain a similar formula. Substracting the two Green’ first Formulae, we shall obtain, 178 Engineering Mathematics 2 2 ( ) , R S g f f g g f dV dS f g n n ∂ ∂ ⎛ ⎞ ∇ − ∇ = − ⎜ ⎟ ∂ ∂ ⎝ ⎠ ∫ ∫ ∫ ∫ ∫ . . . (7.67) which is called Green’s Second Formula or the second form of Green’s Theorem. In Green’s Theorem, we have assumed that f and g are two continuously differentiable scalar point function such that f ∇ and g ∇ are also continuously differentiable. In particular, if we take 1 , g r = where r is the position vector of any point relative to a fixed point O, within the region R, then g is twice continuously differentiable scalar function except at O. Figure 7.35 We surround O by small sphere of radius ε. Let 1 S be the surface of this sphere and 1 R be the region bounded by S and 1 . S We apply Green’s Theorem to the region 1 , R enclosed by surface S and 1 S and taking the limit as 0, ε → we can obtain 2 1 1 1 4π (0) S S f dS f dV f f R r r ⎛ ⎞ = ⋅ − ∇ ∇ − ∇ ⎜ ⎟ ⎝ ⎠ ∫ ∫ This result is known as Green’s Formula. Another important application of divergence theorem is in terms of a basic property of solutions of Laplace’s Equation, which we take up next. 7.8.7 A Basic Property of Solutions of Laplace’s Equation Consider the formula in the divergence theorem, namely, div . n R S f dV f dS = ∫ ∫ ∫ ∫ . . . (7.68) Let us assume that f is the gradient of a scalar function, say, grad . f = Φ Then 2 div div(grad ) f = Φ = ∇ Φ Further, ˆ ˆ grad . n f f = ⋅ = ⋅ Φ n n Since the right hand side of n f represents the directional derivative of Φ in the outward direction of S, it may be denoted as n ∂Φ ⋅ ∂ Then Eq. (7.68) becomes 2 R S dV dS n ∂Φ ∇ Φ = ∂ ∫ ∫ ∫ ∫ ∫ Taking into account the assumptions under which the divergence theorem holds, we immediately obtain, form above, the following result : 179 Line and Surface Integrals Theorem 1 Let ( , , ) Φ x y z be a solution of Laplce’s Equation x y z = + + 2 2 2 2 2 2 2 ∂ Φ ∂ Φ ∂ Φ ∇ Φ ∂ ∂ ∂ = 0 in some domain D, and suppose that the second partial derivation of Φ are continuous in D. Then the integral of the normal derivative of the function Φ over any piecewise smooth closed orientable surface S in D is zero. Now if we suppose that the assumptions in above result are satisfied by Φ and is zero everywhere one piece wise smooth closed orientable surface S in D, then putting f g = = Φ in formula (7.66) in the first form of Green’s Theorem (Section 7.8) and denoting the interior of S by R, we get 2 (grad ) (grad ) | grad | 0 R R dV dV Φ ⋅ Φ = Φ = ∫ ∫ ∫ ∫ ∫ ∫ Since by assumption | grad | Φ is continuous in R and on S and in non-negative, it must be zero every where in R. Hence, 0 x y z ∂Φ ∂Φ ∂Φ = = = ∂ ∂ ∂ and Φ is constant in R and, because of continuity, equal to its value O on S. Thus, Theorem 2 If a function ( , , ) Φ x y z satisfies the assumptions of Theorem 1 and is zero at all points of a piecewise smooth orient able surface S in D, then it vanishes identically in region R bounded by S. Theorem 2 has an important consequence in that it gives the uniqueness theorem for Laplace’s Equation, which can be stated as below : Theorem 3 Let Φ be a solutionof Laplace’s Equations which has continuous second partial derivatives in a domain D, and let R be a region in D which satisfies the assumption of the divergence theorem. Then Φ is uniquely determined in R by its value on the bounding surface S of R. Let us consider the following example. Example 7.26 Show hat two functions which are harmonic in a region enclosed by a surface and take on the same values at any point of the surface are identical. Solution Let Φ and Ψ be two harmonic functions in a region bounded by a surface S and let Φ = Ψ at every point S. On applying Green’s divergence theorem to the function θ θ, ∇ we obtain 2 θ θ θ θ θ θ V V S dV dV dS ∇ + ∇ ⋅ ∇ = ∇ ⋅ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ Let us θ = Φ − Ψ so that 2 2 θ 0 ∇ = ∇ Ψ = at ever point of V and θ 0 = at every point of S. ∴ 2 ( θ) 0 V dV ∇ = ∫ ∫ ∫ 180 Engineering Mathematics ⇒ θ 0 ∇ = ⇒ θ is constant. As θ is zero at every point of S, thus θ is zero everywhere. Let us take another example. Example 7.27 If grad = Φ F and 2 4πρ, ∇ Φ = − prove that ˆ 4π ρ , S V F dS dV ⋅ = − ∫ ∫ ∫ ∫ n where the symbols have their usual meaning and assumptions of divergence theorem are satisfied. Solution From divergence theorem, we have ˆ div S V dS dV ⋅ = ∫ ∫ ∫ ∫ ∫ F n F div(grad ) V dV = Φ ∫ ∫ ∫ 2 V dV = ∇ Φ ∫ ∫ ∫ ( 4πρ) V dV = − ∫ ∫ ∫ 4π ρ . V dV = − ∫ ∫ ∫ You may now try the following exercises. SAQ 11 (a) Let ˆ ˆ ˆ ( , , ) ( , , ) ( , , ) x y z N x y z P x y z = + + F M i j k be a vector field whose components M, N and P are continuous and have continuous second partial derivatives of all kinds. Show that ˆ (curl ) 0 S dS ⋅ = ∫ ∫ F n for any surface to which the divergence theorem applies. [Hint : By direct computation, show that div (Curl F) = 0 and then apply divergence theorem to (Curl F).] (b) Let S be the spherical cap 2 2 2 2 2 , x y z a z a + + = ≥ together with its base 2 2 2 , x y a + ≤ . z a = Find the flux of 2 ˆ ˆ ˆ xz yz y = − + F i j k outward through S by applying the divergence theorem. [Hint : Flux ˆ S ds = ⋅ ∫ ∫ F n .] 181 Line and Surface Integrals (c) If ( , , ) x y z q is the velocity vector of a differentiable fluid flow through a region D in space, ρ( , , , ) x y z t is the density of the fluid element at each point (x, y, z) at time t, then taking ρ F q = and using divergence theorem obtain the continuity equation of hydrodynamics in the form ρ div(ρ ) 0. q t ∂ + = ∂ In Unit 6, Section 6. 6 and Sub-section 6.7.3, we had defined solenoidal and irrotational vector fields as divergence free and crul free vectors respectively. We shall revisit the concepts of solenoidal and irrotational vector fields in the next section and express the conditions for such vector fields in terms of integrals of vector fields and show that the two approaches are equivalent. 7.9 SOLENOIDAL AND IRROTATIONAL VECTOR FIELDS REVISITED In Section 6.6, we had defined a solenoidal vector F in a region if for all points in that region div F = 0 In Section 7.8, we had given an integral definition of divergence of a vector field F as 0 div S V F dS Lt V → ⋅ = ∫ ∫ F At any point in region of volume V bounded by surface S. Now if div F = 0, then 0 S dS ⋅ = ∫ ∫ F Also from divergence theorem, we have div S V dS dV ⋅ = ∫ ∫ ∫ ∫ ∫ F F Now if div F = 0, then above result yields 0 S dS ⋅ = ∫ ∫ F across every closed surface S. Also S dS ⋅ ∫ ∫ F represents flux of vector point function F in a simply connected region V bounded by S. Thus we can define Solenoidal Vector Field as 182 Engineering Mathematics A vector point function F is said to be solenoidal in a region of its flux across every closed surface S on that region is zero, i.e., 0 S dS ⋅ = ∫ ∫ F If F is a soenoidal vector, then there exists a vector point function f such that F = curl f. In Section 6.6.2, we had defined a vector point function F to be irritation when curl = 0 F or there exists a scalar function Φ such that = ∇Φ F By Stocke’s theorem, we have (curl ) S C F d d ⋅ = ⋅ ∫ ∫ ∫ r F S Thus, when curl F is zero, we get 0 C d ⋅ = ∫ F r The above relation provides another definition of an irrotational vector. We may now define an irrotational vector as A continuous vector point function F is said to be irrotational if its circulation along every closed contour C in the region is zero. This above definition can be useful indetermining the scalar point function Φsuch that , = ∇Φ F F being a given irrotational vector function. In this regard we have to remember that ( ) , P A P d Φ = ⋅ ∫ F r A being any fixed point. Let us take some examples to illustrate the above theory. Example 7.28 Show that the vector point function 2 ˆ ˆ ˆ ( sin sin ) ( sin 2 ) ( cos ) y z x x z yz xy z y = − + + + + F i j k is irrotational and find the corresponding scalar function Φ such that . = ∇Φ F Solution We may verify that curl . = 0 F Here 2 ˆ ˆ ˆ curl sin sin sin 2 cos x y z y z x x z yz xy z y ∂ ∂ ∂ = ∂ ∂ ∂ − + + i j k F ˆ ˆ ˆ ( cos 2 cos 2 ) ( cos cos ) (sin sin ) x z y x z y y z y z z z = + − − − − + − i j k = 0 To determine , Φ we may take fixed point as O(0, 0, 0) and general point P as (x, y, z). Then 183 Line and Surface Integrals , , 0, 0, 0 x y z d Φ = ⋅ ∫ F r ( ,0,0) ( , ,0) ( , , ) 2 (0,0,0) ( ,0,0) ( , ,0) ( sin sin ) ( sin 2 ) ( cos ) x x y x y z x x y y z x dx x z yz dy xy z y dz = − + + + + ∫ ∫ ∫ ( , , 0) ( , , ) ( , 0, 0) 2 2 (0, 0, 0) ( , 0, 0) ( , , 0) sin cos sin sin x y x y z x x x y xy z x xy z y z xy z y z = + + + + + 2 cos 1 sin x xy z y z = − + + If we omit the constant – 1, then 2 (cos sin ). x xy z y z = ∇ + + F Example 7.29 If ˆ ˆ ˆ ( ) ( ) ( ) x z z x x y = − + − + − F i j k be a continuous vector point function, verify that it is solenoidal and find the function f such that F = curl f. Solution Here, ˆ ˆ ˆ ( ) ( ) ( ) y z z x x y = − + − + − F i j k ∴ div ( ) ( ) ( ) y z z x x y x y z ∂ ∂ ∂ = − + − + − ∂ ∂ ∂ F = 0 + 0 + 0 = 0 If F = curl f, let 1 2 3 ˆ ˆ ˆ . f f f = + + f i j k We suppose that 1 0, f = then using expression of curl, we get 0 0 2 3 3 2 , ( , ), x x x x f F dx f F dx y z = = − + Φ ∫ ∫ where 0 1 0 ( , ) ( , , ) x y y z F x y z dy Φ = ∫ Let us take 0 0 0, 0 x y = = Now 2 2 0 1 ( ) 2 y f x y dx x xy = − = − ∫ 2 3 0 1 ( ) ( , ) ( , ) 2 x f z x dx y z x xz y z = − − + Φ = − + Φ ∫ and 2 0 1 ( , ) ( ) 2 y y z y z dy y yz Φ = − = − ∫ Hence 1 2 3 ˆ ˆ ˆ f f f f = + + i j k 2 2 2 1 1 ˆ ˆ ( ) ( ) 2 2 x xy x y x y z ⎛ ⎞ ⎡ ⎤ = + − + − + ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ j k The general form of f, where = Curl, F, is 2 2 2 1 1 ˆ ˆ + ( ) ( ) 2 2 x xy x y x y z ⎛ ⎞ ⎡ ⎤ = + ∇ − + − + ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ f j k g where g is any scalar point function. You may now try the following exercise to access your knowledge. 184 Engineering Mathematics SAQ 12 (a) Show that the vector function ˆ ˆ ˆ (sin cos ) ( cos sin ) ( cos sin ) y z x x y z y z x = + + + + F i j k is irrotatinal and find the corresponding scalar Φ such that . = ∇Φ F (b) Show that the following vector function F are solenoidal and find the function f such that F = curl f : (i) ˆ ˆ ˆ ( ) ( ) ( ) x y z y z x z x y = − + − + − F i j k (ii) ˆ ˆ ˆ yz zx xy = + + F i j k We now end this unit by giving a summary of what we have covered in it. 7.10 SUMMARY In this unit, we have covered the following points : • If (t) F is a vector function of a scalar variable t and ( ) ( ), d t f t dt = F then ( ) ( ) , d t dt t C dt = + f F where C is a constant vector. • If ( , , ) w x y z is a scalar or a vectro field (i) defined at each point of a curve C, characterized by ˆ ˆ ˆ ( ) ( ) ( ) ( ) , t x t y t z t = + + r i j k , a t b ≤ ≤ (ii) is continuous, and (iii) ( ), ( ), ( ) x t y t z t ′ ′ ′ are bounded and have only a finite number of discontinuties on the interval , a t b ≤ ≤ then 1 lim ( , , ) δ k n k k k S n k x y z →∞ = ∑ exists and is the line integral of ( , , ) w x y z along C from A to B (corresponding to a t b ≤ ≤ ) and we denote it as ( , , ) , C w x y z ds ∫ where curve C is assumed to be piece-wise smooth. • Line integrals can occur in the following forms : (i) ( , , ) , C w x y z ds ∫ where ( , , ) w x y z is a scalar field and it is scalar. 185 Line and Surface Integrals (ii) ( , , ) , C w x y z ds ∫ wrhere ( , , ) w x y z is a scalar field, cureve C is divided into vector elements and it is a vector. (iii) ( , , ) , C x y z ds ∫ W where ( , , ) x y z W is vector field, curve C is divided into vector elements, ( , , ) x y z W and ds are multiplied using dot product of vectors and it is a scalar. (iv) ( , , ) , C x y z × ∫ W ds where ( , , ) x y z W is a vector field, cross product is used and it is a vector. • A line integral is evaluated by reducing it to a definite integral of single variable, using parametric representation of the path of integration C. • If F represents a force moving along a curve C, ˆ ˆ ˆ ( ) ( ) ( ) ( ) , t x t y t z t = + + r i j k , a t b ≤ ≤ form a point A to a point B in space, then C d ⋅ ∫ F r represents the work done by the force F in moving a particle from point A to point B along the curve C. • A force field is conservative if its work-integral is independent of path, but depends only on the end points of the path. • A force field is conservative if and only if it is a gradient field. • If ( , ) f x y is defined and continuous for all ( , ) x y in a closed bounded region R (bounded by finitely many smooth curves), then the limit 1 lim ( , ) , n k k k n k f x y A →∞ = Δ ∑ where k A Δ is the area of th k sub-region of R exists and is the double integral of ( , ) f x y over the region R and is denoted by ( , ) R f x y dA ∫∫ or ( , ) . R f x y dx dy ∫∫ ( ) ( ) ( ) ( ) ( , ) ( , ) ( , ) h x q y b d x a y g x y c x p y R f x y dy dx f x y dy dx f x y dx dy = = = = ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ = = ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ∫ ∫ ∫ ∫ ∫∫ if ( , ) f x y is uniformaly continuous and bounded over R and ( ) ( ) ( , ) h x y g x f x y dy = ∫ bounded and integrable form a to b with respect to x along with ( ) ( ) ( , ) q y x p y f x y dx = ∫ being bounded and integrable from C to d with respect to y. • Transformation of double integrals into line integrals is based on Green’s theorem is the plane. 186 Engineering Mathematics • If ( , , ) f x y z is defined and continuous on a surface S, then the limit 1 lim ( , , ) , n k k k k n k f x y z S →∞ = Δ ∑ where k S Δ is the area of th k sub-division of S, exists and is the Surface Integral of ( , , ) f x y z over the surface S and is denote by ( , , ) . S f x y z dS ∫∫ • Transformation of surface integrals into line integrals is based on Stoke’s Theorem. • If V be the velocity field of a moving fluid of density ρ, then ρ . C dr V ds ds ∫ is the circulation of fluid around curve C. • Component of the curl in the positive normal direction can be regarded as the specific circulation (or circulation density) of the flow in the surface at the corresponding point, i.e., 1 [ ( )] lim , n t C Curl P F ds A δ →∞ = ∫ F where ρ . F V = • If f (x, y, z) is continuous in a domain containing R and R is bounded by finitely many smooth curves, then, as , , x y z Δ Δ Δ all approach zero with n approaching infinity, then limit of the sum 0 ( , , ) n n k k k k k S f x y z V = = Δ ∑ is called the triple integral of f (x, y, z) over the region R and is denoted by ( , , ) R f x y z dV ∫ ∫ ∫ • Properties of triple integral are R R k f dV k f dV = ∫ ∫ ∫ ∫ ∫ ∫ ( ) R R R f g dV f dV g dV ± = ± ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ R R F dV GdV ≥ ∫ ∫ ∫ ∫ ∫ ∫ if F G ≥ in R. 1 2 . . . n R R R R F dV F dV F dV F dV = + + + ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ • The volume of a region R in 3-dimensions is given by Volume of 1 , , 0 n k n k R x y z R Lt V dV →∞ = Δ Δ Δ → = Δ = ∑ ∫ ∫ ∫ 187 Line and Surface Integrals • To evaluate a triple integral, we divide the region R into elementary cuboids by planes parallel to say z-coordinate plane and if 1 ( , ) z z x y = and 2 ( , ) z z x y = are the lower and upper surfaces and A is the base of these columns and if A is bounded by the curves 1 2 ( ), ( ), , , y y x y y x x a x b = = = = then triple integral ( , , ) R f x y z dV ∫ ∫ ∫ may be written s repeated integrals as 2 2 1( ) 1 ( ) ( , ) ( , ) ( , , ) ( , , ) x b y x z x y a y y z z x y R f x y z dV f x y z dz dy dx = = = ∫ ∫ ∫ ∫ ∫ ∫ • The various physical quantities of any object in 3-dimensions having variable density ρ( , , ) x y z which can be evaluated with the help of triple integration are Masss = ρ R m dV = ∫ ∫ ∫ First Moments about Coordinate Planes : ρ , ρ , ρ yz zx xy R R R M x dV M y dV M z dV = = = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ Center of Mass : , , , yz xy zx M M M x y z m m m = = = Moments of Inertia (Second moments) : 2 2 ρ( ) , x R I y z dV = + ∫ ∫ ∫ 2 2 ρ( ) , y R I z x dV = + ∫ ∫ ∫ 2 2 ρ( ) , z R I x y dV = + ∫ ∫ ∫ and 2 ρ , L R I r dV = ∫ ∫ ∫ where ( , , ) r x y z = distance of (x, y, z) to line L. Radius of Gyration about a Line L : L I R m ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ • Transformation of Volume Integrals into Surface Integrals and conversely is based on Gauss Divergence Theorem, according to which “If R be a closed bounded region in space whose boundary is piecewise smooth orientable surface S and if u (x, y, z) is a vector function, which is continuous and has continuous first partial derivatives in some domain containing R, then 188 Engineering Mathematics ˆ div , R S dV dS = ⋅ ∫ ∫ ∫ ∫ ∫ u u n where ˆ n is the unit outward normal to the surface S bounding the region R”. • Under the conditions of divergence theorem, ˆ curl S R dS dV × = − ∫ ∫ ∫ ∫ ∫ F n F and ˆ S R dS dV Φ = ∇ Φ ∫ ∫ ∫ ∫ ∫ n • Integral definitions of gradient, divergence and curl are 0 grad = S V dS Lt V → Φ Φ ∫ ∫ 0 div = S V dS Lt V → ⋅ ∫ ∫ F F and curl 0 × = S V dS Lt V → ∫ ∫ F F • We can physically interpret divergence theorem as The amount of fluid flowing across the surface S in unit time from inside to outside equals the amount of fluid per unit time supplied by the sources within S. • If q k = − grad U is the velocity of heat flow in a body, then using divergence theorem, heat-flow equation is 2 2 , U c U t ∂ = ∇ ∂ where 2 , σρ k c = where σ is the specific heat of the material of the body and ρ is the density of the material. • Green’s Theorem states If f and g are two continuous functions, possessing continuous partial derivative and S is the bounding surface of a region R satisfying conditions of divergence theorem, then 2 2 ( ) ( ) . R f g f dV f g g f dS ∇ − ∇ = ∇ − ∇ ⋅ ∫ ∫ ∫ ∫ ∫ • If Φ be a solution of Laplace’s equation which has continuous second partial derivatives in a domain D, and if R be a region in D satisfying the assumptions of the divergence theorem, then Φ is uniquely determined in R by its values on the boundary surface S of R.. 189 Line and Surface Integrals • A vector point function F is said to solenoid in a region if its flux across every closed surface S in that region is zero i.e., 0. S dS ⋅ = ∫ ∫ F • A continuous vector point function F is said to irrotational if its circulation along every closed center C in the region is zero, i.e., . 0 C ds = ∫ F . 7.11 ANSWERS TO SAQs SAQ 1 (a) (i) Along the curve C 2 3 , , x t y t z t = = = ∴ 2 , 2 , 3 dx dt dy t dt dz t dt = = = Hence, 2 2 (3 6 ) 14 20 C C dr x y dx yz dy xz dz ⎡ ⎤ ⋅ = + − + ⎣ ⎦ ∫ ∫ F 1 2 2 5 6 2 0 (3 6 ) 14 2 20 3 t t dt t t dt t t t dt ⎡ ⎤ = + − − + ⋅ ⋅ ⎣ ⎦ ∫ 1 2 6 9 0 (9 28 60 ) t t t dt = − + ∫ 1 3 7 10 0 9 28 60 3 7 10 t t t ⎡ ⎤ = − + ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ = 3 – 4 + 6 = 5 (ii) Here 2 2 (3 6 ) 14 20 C C dr x y dx yz dy xz dz ⎡ ⎤ ⋅ = + − + ⎣ ⎦ ∫ ∫ F (1,0,0) 2 2 (0,0,0) (3 6 ) 14 20 x y dx yz dy xz dz ⎡ ⎤ = + − + ⎣ ⎦ ∫ (1,1,0) 2 2 (1,0,0) (3 6 ) 14 20 x y dx y z dy xz dz ⎡ ⎤ + + − + ⎣ ⎦ ∫ (1,1,1) 2 2 (1,1,0) (3 6 ) 14 20 x y dx y z dy xz dz ⎡ ⎤ + + − + ⎣ ⎦ ∫ (1,0,0) (1,1,0) 3 2 3 3 2 3 (0,0,0) (1,0,0) 20 20 6 7 6 7 3 3 x yx y z xz x yz y z xz = + − + + + − + (1,1,1) 3 2 3 (1,1,0) 20 6 7 3 x yz y z xz + + − + 20 (1 0) (1 6 1) ()1 6 7 1 6) 3 = − + + − + + − + − − 1 20 1 6 3 3 = + − = 190 Engineering Mathematics (b) Here 2 ˆ ˆ ˆ (2 ) ( ) (3 2 4 ) x y z x y z x y z = − − + + + + − + F i j k The parametric equation of circle 2 2 9 x y + = in XOY–plane is given by 3cos θ, 3sin θ, 0 x y z = = = Here 3sin θ θ, 3cos θ θ, 0 dx d dy d dz = − = = Now work done by force F in moving a particle once round the circle 2 2 9 x y + = in XOY-plane 2 (2 ) ( ) (3 2 4 ) C C x y z dx x y z dy x y z dz = ⋅ = − − + + − + − + ∫ ∫ F dr 2 [(6cos θ 3sin θ) ( 3sin θ θ) (3cos θ 3sin θ) 3cos θ θ] C d d π = − − + + ∫ 2π 2π 2 0 9 (9 9cos θsin θ) θ 9θ sin θ 18π 2 C d = − = − = ∫ SAQ 2 (a) Here 2 2 2xy z x y φ = + ∴ ˆ ˆ ˆ grad x y z ∂φ ∂φ ∂φ φ = + + ∂ ∂ ∂ i j k 2 2 2 ˆ ˆ ˆ (2 2 ) (4 ) 2 y z xy xyz x xy = + + + + i j k and ˆ ˆ ˆ d dx dy dz = + + r i j k ∴ 2 2 2 grad (2 2 ) 4 ) (2 ) C C d y z xy dx xyz x dy xy dz ⎡ ⎤ φ⋅ = + + + + ⎣ ⎦ ∫ ∫ r For curve C, 2 3 , , 0 1. x t y t z t t = = = ≤ ≤ ∴ 2 , 2 , 3 . dx dt dy t dt dz t dz = = = Here 1 5 3 2 2 5 2 0 grad (2 2 ) (4 )2 2 3 C t d t t dt t t t dt t t dt = ⎡ ⎤ φ⋅ = + + + + ⋅ ⎣ ⎦ ∫ ∫ r 1 7 5 3 0 (14 2 4 ) t t t t dt = = + + ∫ 1 8 6 4 0 14 2 4 8 6 4 t t t = + + 7 1 1 1 3 . 4 3 12 = + + = (b) Here 2 ˆ ˆ ˆ xy z x = − + A i j k ˆ ˆ ˆ d dx dy dx = + + r i j k ∴ 2 2 ˆ ˆ ˆ ( ) ( ) ( ) d zdz x dy x dx xy dz xy dy zdx × = − + + − + + A r i j k 191 Line and Surface Integrals For curve C, 2 3 , 2 , , 0 1. x t y t z t t = = = ≤ ≤ ∴ 2 2 , 2 3 . dx t dt dy dt dz t dt = = = Hence 5 5 5 5 3 4 ˆ ˆ ˆ (3 2 ) (2 6 ) (4 2 ) d t t dt t t dt t t dt × = − + + − + + A r i j k 5 5 3 4 ˆ ˆ ˆ 5 4 (4 2 ) t dt t dt t t dt = − − + + i j k ∴ 1 5 5 3 4 0 ˆ ˆ ˆ 5 4 (4 2 ) C d t dt t dt t t dt ⎡ ⎤ × = − − + + ⎣ ⎦ ∫ ∫ A r i j k 1 6 6 4 5 0 ˆ ˆ ˆ 5 4 4 2 6 6 4 5 t t t t ⎛ ⎞ = − − + + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ i j k 5 2 7 ˆ ˆ ˆ 6 3 5 = − − + i j k. (c) Here ˆ ˆ ˆ ( cos ) ( sin ) ( ) x x e y yz xz e y xy z = + + − + + F i j k A force field F is conservative if and only if , f = ∇ F where f is some differentiable scalar function. i.e., if ˆ ˆ ˆ f f f x y z ∂ ∂ ∂ = + + ∂ ∂ ∂ F i j k i.e., if cos , x f = e y yz x ∂ + ∂ sin , x f = xz e y y ∂ − ∂ f = xy + z z ∂ ∂ i.e., if f f f df dy dz x y z ∂ ∂ ∂ = + + ∂ ∂ ∂ ( cos ) ( sin ) ( ) x x e y yz dx xz e y dy xy z dz = + + − + + 2 1 cos , 2 x d e y xyz z ⎛ ⎞ = + + ⎜ ⎟ ⎝ ⎠ which is true. Here F is conservative and , f = ∇ F where 2 cos . 2 x z f e y xyz = + + SAQ 3 (i) For 1 2 2 2 0 0 (1 ) , x x y dy dx + + ∫ ∫ the region of integration is 0, 2 , 0, y y x x = = = 1 x = which is shaded portion in the Figure Figure 192 Engineering Mathematics Now, 1 2 2 2 0 0 (1 ) x x y dy dx + + ∫ ∫ 1 3 3 0 8 2 2 3 x x x dx ⎛ ⎞ = + + ⎜ ⎟ ⎝ ⎠ ∫ 1 3 3 0 8 (2 2 ) 3 x x x dx = + + ∫ 1 2 4 0 14 2 2 3 4 x x = ⋅ + ⋅ 7 13 1 6 6 = + = (ii) For 2 1 0 (1 ) , x x y x xy dy dx = = − ∫ ∫ the region of integration is 0, 1, x x = = parabola 2 y x = and line y x = which is shaded portion in Figure. Figure Now 2 1 0 (1 ) , x x y x xy dy dx = = − ∫ ∫ 2 1 2 0 2 x x y y x dx = − ∫ 1 3 5 2 0 2 2 x x x x dx ⎛ ⎞ = − − + ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ∫ 1 2 3 4 6 0 1 1 2 3 2 4 2 6 x x x x = − − + ⋅ 1 1 1 1 12 8 3 2 1 2 3 8 12 24 8 − − + = − − + = = SAQ 4 (a) The base triangle in xy-plane is bounded by 2 3 , 0 x y y = = and 3. x = 193 Line and Surface Integrals Figure The required volume is under the plane 2 6 x y + + = ⇒ 6 z x y = − − . ∴ Volume 2 3 3 0 0 (6 ) x x y x y dy dx = = = − − ∫ ∫ 2 3 2 3 0 0 6 2 x y y xy dx = − − ∫ 3 2 2 0 2 2 4 3 9 x x x dx ⎛ ⎞ = − − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ∫ 3 2 3 3 0 2 2 4 2 3 3 9 3 x x x = − ⋅ − 2 2 2.9 27 27 10 9 27 ⎛ ⎞ = − ⋅ − ⋅ = ⎜ ⎟ ⎝ ⎠ (b) Here density of plate 2 2 ( ). x y = + K The plate lies between 3 y x = and 2 . x y = For the plate, range of integration is 0 x = to 1 x = and y x = to 2 . y x = Hence, mass of the plate 3 1 2 3 0 ( ) x x y x x y dy dx = = = + ∫ ∫ K 3 1 3 2 0 3 x x x y x y dx = = + ∫ K 1 5 9 5/ 2 3/ 2 0 1 1 3 3 x x x x dx ⎛ ⎞ = + − − ⎜ ⎟ ⎝ ⎠ ∫ K 1 6 10 7 / 2 5/ 2 0 1 1 6 3 10 7 / 2 3 5/ 2 x x x x = + ⋅ − − ⋅ K 194 Engineering Mathematics 1 1 2 2 6 30 7 15 = + − − K 35 7 60 28 210 + − − = K 46 23 210 105 = = K K SAQ 5 The required area is in the positive quadrant, bounded by the curves 2 2 2 2 4 , 4 , , . y ax y bx xy c xy d = = = = Let this area be transformed to uv-plane, using the transformation 2 , , 4 y u v xy x = = so that the transformed area is the rectangle A having sides parallel to u and v axes given by 2 , , u a u b v c = = = and 2 . v d = 3 1/ 3 4 , (4 ) y uv y uv = = and 2/ 3 1/ 3 1/ 3 (4 ) (4 ) v v v x y uv u = = = ∴ 2/ 3 1/ 3 1/ 3 1/ 3 1/ 3 4/ 3 1/ 3 1/ 3 2/ 3 1/ 3 1/ 3 2/ 3 1 1 2 (4) ( , ) 3 3 (4) ( , ) 1 1 (4) (4) 3 3 v v u x y u J u v v u u v − − − − − ⋅ ⋅ ∂ = = ∂ 1 1 2 1 1 9 9 3 u u u = − ⋅ = − ∴ Required Area 2 2 1 3 a b u a v c du dv u = = = − ∫ ∫ 2 2 1 log 3 d b a C u dv = − ∫ 2 2 1 log 3 d C b dv a ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ ∫ 2 2 1 log ( ) 3 b d c a ⎡ ⎤ ⎛ ⎞ = − − ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ 2 2 1 ( ) log 3 b d c a ⎛ ⎞ = − ⋅ ⎜ ⎟ ⎝ ⎠ SAQ 6 (a) Curve C is boundary of the square ABCD in Figure. Using Green’s theorem, 195 Line and Surface Integrals Figure 2 2 2 ( ) ( ) C x xy dx x y dy ⎡ ⎤ + + + ∫ ⎣ ⎦ 1 1 2 2 2 1 1 ( ) ( ) x y x y x xy dx dy x y + + =− =− ⎡ ⎤ ∂ ∂ = + − + ⎢ ⎥ ∂ ∂ ⎣ ⎦ ∫ ∫ 1 1 1 1 (2 ) x y x x dx dy + + =− =− = − ∫ ∫ 1 1 1 1 x xy dx − =− = ∫ 1 1 2 1 1 [ ( )] 0 x x dx x − − = − − = = ∫ (b) The boundary C of the region 2 8 , 2 y x x = = is the curve OABO and area S enclosed is shown as shaded in Figure. Figure Using Green’s theorem 2 2 ( 2 ) ( 3) C x xy dx x y dy ⎡ ⎤ − + + ∫ ⎣ ⎦ 196 Engineering Mathematics 3 2 ( 3) ( 2 ) S x y x xy dx dy x y ⎡ ⎤ ∂ ∂ = + − + ⎢ ⎥ ∂ ∂ ⎣ ⎦ ∫ ∫ 2 2 2 0 2 2 (2 2 ) x x y x xy x dy dx + = =− ⎡ ⎤ = + ⎢ ⎥ ⎣ ⎦ ∫ ∫ 2 2 2 2 0 2 2 2 x x x xy xy dx = − = + ∫ 2 0 (2 2 2 2 2 2 ) x x x x x dx = = ⋅ + ⋅ ∫ 2 5/ 2 0 8 2 5/ 2 x = 16 2 128 4 2 5 5 = ⋅ = SAQ 7 (a) The surface 2 2 2 2 : , 0 S x y z a z + + = ≥ can be represented by ˆ ˆ ˆ sin cos sin sin cos a u v a u v a u = + + r i j k ∴ ˆ ˆ ˆ cos cos cos sin sin a u v a u v a u = + + u r i j k ˆ ˆ sin sin sin cos a u v a u v = − + v r i j ∴ 2 2 2 2 2 2 2 2 2 cos cos cos sin sin a u v a u v a u a = ⋅ = + + = u u F r r 2 2 cos sin cos sin cos cos sin sin 0 a u u v v a u v v u = ⋅ = − + = u v F r r 2 2 2 2 2 2 2 2 sin sin sin cos sin a u v a u v a u = ⋅ = + = u v G r r If area A is the image of surface S in uv-plane, then u v dA du dv = × r r 2 EG F du dv = − 2 2 2 sin 0 a a u du dv = ⋅ − 2 sin a u du dv = Further, 2 2 2 2 2 2 (2 ) ( sin cos ) ( sin sin ) ( cos ) x y a a u v a u v a u a + + + = + + + 2 2 2 2 2 2 2 2 2 2 sin cos sin sin cos 2 cos a u v a u v a u a a u = + + + + 2 2 2 2 2 2 cos 4 cos 2 u a a u a = + = ∴ 2 2 ( ) S U dA x y z a σ = + + + ∫ ∫ 2 sin 2 cos / 2 S a u du dv a u σ σ = ⋅ ∫ ∫ 197 Line and Surface Integrals 2sinh/ 2cos / 2 2 cos / 2 S u a du dv a u σ = ∫ ∫ π / 2 2π 0 0 sin 2 u v u a du dv σ = = = ∫ ∫ (∵S is a hemi-sphere). π / 2 0 sin 2π 2 u u a σ = = ⋅ ∫ π / 2 0 cos / 2 2π 1/ 2 u aσ = ⋅ − 2π (2 2). aσ = − (b) Surface area S dS = ∫ ∫ 2 2 1 , x y S g g dx dy = + + ∫ ∫ where 2 2 ( , ) 2 z g x y x y = = − − is the surface of cylinder and R is its projection on xy-plane taken 2 2 1. x y + ≤ 2 2 2 2 1 2 2 x y dx dy x y + ≤ = − − ∫ ∫ Let cos θ, sin θ, x r y r = = then 2 2 1 x y + ≤ ⇒ 0 1 r ≤ ≤ and 0 θ 2π ≤ ≤ ∴ Required surface area 2π 1 2 θ 0 0 2 θ 2 r r dr d r = = = − ∫ ∫ 1 2π 2 1/ 2 θ 0 0 2 (2 ) θ 2 1/ 2 r d = − = − ∫ 2π 0 ( 2 2 2) θ d = − + ⋅ ∫ 2π (2 2). = − SAQ 8 (a) Here 2 ˆ ˆ x xy = − F i j The curve C is the boundary of the square in the plane 0 z = and bounded by the lines 0, 0, , . x y x a y a = = = = Here ˆ ˆ dr dx dy = + i j ∴ 2 x dx xy dy ⋅ = − F dr Now 2 ( ) C C x dx xy dy ⋅ = − ∫ ∫ F dr 198 Engineering Mathematics 2 2 0 , 0 ( ) ( ) y a x a x at x a y x dx xy dy x dx xy dy = = = = = = − + − ∫ ∫ 0 0 2 2 0 , 0 ( ) ( ) x y a at y a at x a y x dx xy dy x dx xy dy = = = = = + − + − ∫ ∫ 0 0 0 2 2 0 0 (0 0 ) a x y x a y a x dx a y dy x dx dx dy = = = = = − + + − ∫ ∫ ∫ ∫ 0 3 3 3 0 0 0 3 2 3 a a a x y x = − + + 2 2 3 3 . 3 2 3 2 a a a a a = − ⋅ − = − Figure For surface integral, 2 ˆ ˆ ˆ ˆ ˆ Curl ( 0) 0 y y x y z x xy ∂ ∂ ∂ = = − − = − ∂ ∂ ∂ − i j k F k k Now positive unit normal to the plane 0 z = is ˆ ˆ n = k . Also element of area . dS dx dy = Thus 0 0 ˆ ˆ Curl a a S x y dS y dx dy = = ⋅ = − ∫ ∫ ∫ ∫ F n 2 0 0 2 a a x y dx = = − ∫ 2 0 2 a dx = − ∫ a 2 3 0 . 2 2 a a a x = − ⋅ = − Hence 3 ˆ Curl , 2 C S a dr ds ⋅ = − = ⋅ ∫ ∫ ∫ F F n which verifies Stoke’s Theorem. 199 Line and Surface Integrals (b) Here ˆ ˆ ρ ρ ( ) V w x y = = − F j i and ˆ ˆ ˆ ˆ ˆ Curl (ρ ρ ) 2ρ ρ ρ 0 w w w y z wy wx ∂ ∂ ∂ = = + = ∂ ∂ ∂ − i j k F k k x The work done by a force equal to F is . C dr ⋅ ∫ F If C lies in a plane parallel to the xy-plane, then Stoke’s theorem gives ˆ ˆ ˆ Curl 2ρ C S S dr dS w dx dy ⋅ = ⋅ = ⋅ ∫ ∫ ∫ ∫ ∫ F F n k k 2ρ S w dx dy = ∫ ∫ Here S is the area enclosed by C, which is a circle of radius a ∴ 2 (2ρ ) (2ρ ) (π ) C S dr w dx dy w a ⋅ = = ∫ ∫ ∫ F Thus 2 1 ˆ (Curl ) 2ρ circulation density π C w dr= . a ⋅ = = ⋅ ∫ F k F Thus component of Curl F in the direction normal to the plane of V is equal to circulation density. (c) The elliptic shell S is 2 2 2 4 9 36 36, 0 x y z z + + = ≥ One parametric representative of the ellips C at the base of the shell is 3cos , 2sin , 0 2 . x t y t t π = = ≤ ≤ Here 2 2 2 3/ 2 ˆ ˆ ˆ ( ) sin ( ) xyz y x x y e = + + + F i j k By Stoke’s theorem ˆ Curl C S ds dr ⋅ = ⋅ ∫ ∫ ∫ F n F 2 2 xyz 2 2 2 3/ 2 ellipse 4 9 36 ˆ ˆ ˆ ˆ ˆ ˆ ( ) sin ( ) ( ) x y y x x y e dx dy dz + ≤ ⎡ ⎤ = + + + ⋅ + + ⎢ ⎥ ⎣ ⎦ ∫ i j k i j k xyz 2 2 2 3/ 2 ellipse ( ) sin ( ) y dx x dy x y e dz ⎡ ⎤ = + + + ⎢ ⎥ ⎣ ⎦ ∫ 2π 2 0 2sin ( 3sin ) 9cos (2cos ) t t t dt r t dt = = − + ∫ 2π 2 3 0 ( 6sin 18cos ) t t t dt = = − + ∫ 2π 3 0 18 0 3(cos 2 1) (cos 3cos ) 4 t t t t dt = ⎡ ⎤ = − + + ⎢ ⎥ ⎣ ⎦ ∫ 2π 0 sin 2 9 sin 3 3 3sin 2 2 3 t t t t = − + + + 2π = − 200 Engineering Mathematics (d) Here 2 2 ˆ ˆ ˆ 2 x x x = + + F i j k 2 2 ˆ ˆ ˆ ˆ Curl 2 2 y z x x z ∂ ∂ ∂ = = ∂ ∂ ∂ i j k F k x For region S bounded by ellips 2 2 : 4 4 C x y + = in plane ˆ ˆ 1, z = = n k . ∴ By Stoke’s Theorem ˆ Curl C S dr dS ⋅ = ⋅ ∫ ∫ ∫ F F n ( , , ) (0, 0, 0) x y z d Φ = ⋅ ∫ F r 2 S dS = ∫ ∫ 2 2π = ⋅ (∵ area of given ellipse π 1.2 2π = ⋅ = ) 4π = SAQ 9 (a) The given volume is symmetrical about the plane XOZ (y = 0). Therefore 0. y = To find , x we may consider only half the volume standing over the semicircle enclosed by 2 0, 2 , 0, 2 . y y ax x x x a = = − = = 2 2 2 2 0 0 2 2 0 0 ρ ρ a ax x nx x y z mx a ax x nx x y z mx x dz dy dz x dz dy dz − = = = − = = = = ∫ ∫ ∫ ∫ ∫ ∫ 2 2 2 2 0 0 2 2 0 0 ρ ( ) ρ( ) a ax x x y a ax x x y x nx mx dy dx nx mx dy dx − = = − = = − = − ∫ ∫ ∫ ∫ 2 2 2 0 2 2 0 ρ( ) 2 ρ( ) 2 a x a x n m x ax x dx n m x ax x dx = = − − = − − ∫ ∫ π / 2 4 6 2 0 π / 2 3 4 2 0 2(2 ) sin θcos θ θ 2(2 ) sin θcos θ θ a d a d = ∫ ∫ , on putting 2 2 sin θ x a = 5 4 2 . 8 4 a a = ⋅ = 2 2 2 2 0 0 2 2 0 0 ρ ρ a ax x nx x y z mx a ax x nx x y z mx z dz dy dx z dz dy dz − = = = − = = = = ∫ ∫ ∫ ∫ ∫ ∫ 201 Line and Surface Integrals 2 2 2 2 2 2 2 2 0 0 2 2 0 0 1 ρ( ) 2 ρ( ) a ax x x y a ax x x y n x m x dy dz nx mx dy dx − = = − = = − = − ∫ ∫ ∫ ∫ 1 ( ) expressin of 2 n m x = + × 1 5 5 ( ) ( ). 2 4 8 n m a a n m = + ⋅ = + (b) Region is the volume enclosed by the planes 0, , 0, , 0 x x a y y x z = = = = = and . z x y = + Here 0 0 0 a x x y x y z x y x e dz dy dx + + + = = = ∫ ∫ ∫ 0 0 0 a x x y x y z x y z e e e dz dy dx + = = = = ⋅ ⋅ ∫ ∫ ∫ 0 0 0 ( ) a x x y x v x y e e e e dy dx + = = = ⋅ ⋅ − ∫ ∫ 2 2 0 0 2 x x a x x y x e dx e e e = = ⋅ − ∫ 2 2 2 0 1 2 2 x a x x x x x x e dx e e e e e = ⎛ ⎞ = ⋅ − ⋅ − + ⎜ ⎟ ⎝ ⎠ ∫ 0 4 2 0 1 3 2 4 2 2 x x x e e e = − + 4 2 1 3 3 8 4 8 a a a e e e = − + − ⋅ (c) For the positive octant of the sphere 2 2 2 2 , x y z a + + = the limits of integration are 0 x = to , 0 x a y = = to , 0 y a z = = to . z a = ∴ Required mass 0 0 0 a a a x y z k xyz dz dy dx = = = = ∫ ∫ ∫ 2 0 0 2 a a x y a k xy dy dx = = = ⋅ ∫ ∫ 2 2 2 6 . 2 2 2 8 a a a a k k = ⋅ ⋅ ⋅ = SAQ 10 (a) Here ˆ ˆ ˆ x y z = + + F i j k ∴ div 3. x y z F x y z ∂ ∂ ∂ = + + = ∂ ∂ ∂ Also volume of sphere 2 2 2 2 x y z a + + = is 3 4π . 3 a ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 202 Engineering Mathematics ∴ 3 3 4π div 3 3 4π 3 R R a dV dV a ⎛ ⎞ = = ⋅ = ⎜ ⎟ ⎝ ⎠ ∫ ∫ ∫ ∫ ∫ ∫ F . . . (1) Here 2 2 2 2 ( , , ) f x y z x y z a = + + − ∴ Outward unit normal to S, calculated from gradient of ( , , ) f x y z is 2 2 2 ˆ ˆ ˆ ˆ ˆ ˆ 2( ) ˆ 4( ) x y z x y z x y z + + + + = = + + i j k i j k n a and ˆ ˆ ˆ ˆ ˆ ˆ ˆ ( ) x y z dS x y z dS a + + ⋅ = + + ⋅ i j k F n i j k 2 2 2 x y z dS a + + = 2 , a dS a = since on the surface of 2 2 2 2 : S x y z a + + = adS = ∴ 2 3 ˆ (4π ) 4π S S dS a dS a a a ⋅ = = = ∫ ∫ ∫ ∫ F n . . . (2) From (1) and (2), divergence theorem is verified. (b) By divergence theorem, ˆ div S R dS dV ⋅ = ∫ ∫ ∫ ∫ ∫ F n F Here 2 2 ˆ ˆ ˆ 4 2 x y z = − + F i j k ∴ div 2 2 (4 ) ( 2 ) ( ) x y z x y z ∂ ∂ ∂ = + − + ∂ ∂ ∂ F 4 4 2 y z = − + Here region of integration is 2 2 4, 0 x y z + = = and ⇒ 2 x = − to 2 2, 4 x y x = = − − to 2 4 y x = + − and 0 z = to 3 z = Now ˆ div S V dS dV ⋅ = ∫ ∫ ∫ ∫ ∫ F n F 2 2 2 4 3 2 4 0 (4 4 2 ) x x y x z y z dz dy dx − =− =− − = = − + ∫ ∫ ∫ 2 2 2 4 3 2 0 2 4 4 4 x x y x dy dx z yz z − =− =− − = − + ∫ ∫ 2 2 2 4 2 4 (21 12 ) x x y x y dy dx − =− =− − = − ∫ ∫ 2 2 2 2 21 4 21 4 x x z sz =− ⎡ ⎤ = − + − ⎢ ⎥ ⎣ ⎦ ∫ 2 2 1 2 2 2 42 84π 2sin 4 2 2 x x x − − − ⎡ ⎤ = = + − ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 203 Line and Surface Integrals SAQ 11 (a) ˆ ˆ ˆ N P = + + F M i j k ∴ ˆ ˆ ˆ Curl x y z M N P ∂ ∂ ∂ = ∂ ∂ ∂ i j k F ˆ ˆ ˆ P N N M M P y z x y z x ∂ ∂ ∂ ∂ ⎛ ⎞ ⎛ ⎞ ∂ ∂ ⎛ ⎞ − − = + + − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ∂ ∂ ∂ ∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ i j k and div (Curl F) P N N M M P y z x y x y x z x ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⎛ ⎞ ⎛ ⎞ ∂ ∂ ⎛ ⎞ − − = + + − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2 2 2 2 2 2 0 P N M P N M x y x y y z x y x z y z ∂ ∂ ∂ ∂ ∂ ∂ = − + − + − = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ . . . (1) BY divergence theorem ˆ (Curl ) div(Curl ) 0 S R dS dV ⋅ = = ∫ ∫ ∫ ∫ ∫ F n F (using result (1) above) Hence the result. (b) Flux of F outward through ˆ S S dS = ⋅ ∫ ∫ F n By divergence theorem, ˆ div S S dS dV = ∫ ∫ ∫ ∫ ∫ ⋅ F n F Here 2 ˆ ˆ ˆ xz yz y = − + F i j k ∴ 2 div ( ) ( ) ( ) xz yz y x y z ∂ ∂ ∂ = + − + ∂ ∂ ∂ F 0 z z = − + = 0. Hence Flux = div 0 0. R R F dV dV = = ∫ ∫ ∫ ∫ ∫ ∫ (c) Consider a volume V contained in a simply connected surface S. Let ρ( , , ) x y z be the fluid density at any point ( , , ) P x y z of the fluid in V at time t. Let ˆ n be unit outward normal at element δ S of S at P. Let ˆ δ (δ ) S S = n Let q be the fluid velocity at element δ S at P. The equation of continuity represents the conservation of mass, i.e., mass of fluid flowing out of S per unit time equals the decrease of mass within S per unit time in the absence of sources and sinks within V. 204 Engineering Mathematics Now mass of fluid flowing out of S per unit time. ρ S dS = ⋅ ∫ ∫ q and decrease of mass within S per unit time ρ ρ V dV dV t t ∂ ∂ = = − ∂ ∂ ∫ ∫ ∫ ∫ ∫ ∫ ∴ Continuity equation yields ρ ρ = S R dV t ∂ ⋅ − ∂ ∫ ∫ ∫ ∫ ∫ q dS ⇒ ρ div(ρ ) V V dV dV t ∂ = − ∂ ∫ ∫ ∫ ∫ ∫ ∫ q (using divergence theorem) ⇒ ρ 0 div(ρ ) V dV t ∂ ⎛ ⎞ = + ⎜ ⎟ ∂ ⎝ ⎠ ∫ ∫ ∫ q This result is true for all volumes V. Thus at any point of the fluid, we have ρ div(ρ ) 0. t ∂ + = ∂ q SAQ 12 (a) Here ˆ ˆ ˆ (sin cos ) (cos sin ) ( cos sin ) y z x y z y z x = + + + + + F i j k ∴ Curl ˆ ˆ ˆ (sin cos ) ( cos sin ) ( cos sin ) x y z y z x x y z y z x ∂ ∂ ∂ = ∂ ∂ ∂ + + + i j k F ˆ ˆ ˆ (cos cos ) (cos cos ) (cos cos) z z x x y = − + − + − = 0 i j k Hence F is an irrotatinal vector point function. To find Φ such that = ∇Φ F , we may take fixed point as O (0, 0, 0) and general point P as ( , , ). x y z Then ( , , ) (0, 0, 0) x y z d Φ = ⋅ ∫ F r ( , 0, 0) ( , , 0) (0, 0, 0) ( , 0, 0) (sin cos ) ( cos sin ) x x y x y z x dx x y z dy = + + + ∫ ∫ ( , , ) ( , , 0) ( cos sin ) x y z x y y z x dz + + ∫ ( , 0, 0) ( , , 0) (0, 0, 0) ( , 0, 0) sin sin sin sin x x y x x y z x x y y z = + + + ( , , ) ( , , 0) sin sin x y z x y y z z x + + (0 0) ( sin 0) ( sin sin 0) x y y z z x = − + − + + − sin sin sin . x y y z z x = + + Hence ( sin sin sin ) x y y z z x = ∇ + + F 205 Line and Surface Integrals (b) (i) Here ˆ ˆ ˆ ( ) ( ) ( ) x y z y z x z x y = − + − + − F i j k ∴ div [ ( )] [ ( )] [ ( )] x y z y z x z x y x y z ∂ ∂ ∂ = − + − + − ∂ ∂ ∂ F ( ) ( ) ( ) 0 y z z x x y = − + − + − = Hence vector function F is solenoidal. If Curl , f = F let 1 2 3 ˆ ˆ ˆ f f f f = + + i j k Suppose that 1 0, f = then using expression for curl, we get 0 0 2 3 3 2 , ( , ), x x x x f F dx f F dx y z = = − + Φ ∫ ∫ where 0 1 0 1 ( , ) ( , ) y y y z F x y z dy Φ = ∫ Let us take 0 0 0, 0 x y = = Now 2 2 2 0 0 1 ( ) . 2 2 x x x f z x y dx z zx xy yx ⎛ ⎞ − = = − − ⎜ ⎟ ⎝ ⎠ ∫ 2 3 0 1 ( ) ( , ) ( , ) 2 x f y z x dx y z xy x y y z = − + Φ = − + Φ ∫ and 0 ( , ) 0 0. y y z dy Φ = = ∫ Hence 1 2 3 ˆ ˆ ˆ f f f f = + + i j k 2 2 1 1 ˆ ˆ 2 2 zx xy x y xyz ⎛ ⎞ ⎛ ⎞ = + − − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ j k The general form of f, where F = curl f, is 2 2 1 1 ˆ ˆ , 2 2 f g zx xy x y xyz ⎛ ⎞ ⎛ ⎞ = + + ∇ − − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ j k where g is any scalar point function. (ii) Answer : 2 2 2 1 1 ˆ ˆ ( ) . 2 2 f x y y z x z g = + − + ∇ j k 206 Engineering Mathematics APPENDIX–I : PROOF OF GREEN’S THEOREM We first prove the theorem when C is a simple closed curve in the xy-plane such that a line parallel to either axis cuts C in at most two points. The special R can be represented in the form of , a x b ≤ ≤ ( ) ( ) u x y v x ≤ ≤ or , e y b ≤ ≤ ( ) ( ) p y x q y ≤ ≤ Thus curve C consists of lower portions ( ) y u x = and upper portion ( ) y v x = and both the portions extend from x a = to x b = (Figure 1(a)). (a) (b) Figure 1 : Special Region for Green’s Theorem Hence, (using relation 17 of section 7.3.2), we get ( ) ( ) v x b R x a y u x M M dx dy dy dx y y = = ⎡ ⎤ ∂ ∂ ⎢ ⎥ = ∂ ∂ ⎢ ⎥ ⎣ ⎦ ∫ ∫ ∫ ∫ . . . (1) Now integrating the inner integral, we get ( ) ( ) ( ) ( ) ( , ) [ , ( )] [ , ( )] y v x v x y u x y u x M dy M x y M x v x M x u x y = = = ∂ = = − ∂ ∫ Substituting this into Eq. (1), we get [ , ( ) [ , ( ) b b R a a M dx dy M x v x dx M x u x dx y ∂ = − ∂ ∫ ∫ ∫ ∫ Since ( ) ( ) b a a b f x dx f x dx = − ∫ ∫ [ , ( ) [ , ( ) a b b a M x v x dx M x u x dx = − − ∫ ∫ . . . (2) Since ( ) y u x = represents the oriented curve 1 C from x a = to x b = and ( ) y v x = represents the oriented curve 2 C from x b = to , x a = thus the integrals on the right may be written as line integrals over 1 C and 2 C and therefore, 1 [ , ( )] ( , ) b a C M x u x dx M x y dx = ∫ ∫ . . . (3) 207 Line and Surface Integrals and 2 [ , ( )] ( , ) b a C M x v x dx M x y dx = ∫ ∫ . . . (4) From Eqs. (2), (3) and (4), we get 1 2 1 2 ( , ) ( , ) ( , ) ( , ) R C C C C C M dx dy M x y dx M x y dx M x y dx M x y dx y + ∂ = − − − − ∂ ∫ ∫ ∫ ∫ ∫ ∫ . . . (5) Similarly by taking strips parallel to x-axis, we obtain ( ) ( ) d q y y c x p y R N N dx dy dx dy x x = = ∂ ∂ ⎡ ⎤ = ⎢ ⎥ ∂ ∂ ⎣ ⎦ ∫ ∫ ∫ ∫ { } [ ( ), ] [ ( ), ] d C N q y y y N p y y dy = − ∫ [ ( ), ] [ ( ), ] d d C d N q y y dy N p y y dy = + ∫ ∫ ( , ) C N x y dy = ∫ From Eqs. (5) and (6), we obtain ( ) R C N M dx dy M dx N dy x y ⎛ ⎞ ∂ ∂ − = + ⎜ ⎟ ∂ ∂ ⎝ ⎠ ∫ ∫ ∫ which complete the proof of Green’s Theorem in the plane form special regions which are enclosed by simple smooth curves. Next we consider that the region R itself is not a special region, but can be subdivided into finitely many special regions (see Figure 2). Figure 2 : Multiply-connected Region R In this case, we apply the theorem to each sub-region and then add the results. the left- hand members add up to the integral over R, which the right-hand members add up to the line integral over C (which now consists of outer boundary 1 C and inner boundary 2 C ) plus integrals over the curves introduced for sub-dividing R. Each of these later integrals occurs twice, one taken in each direction. Hence these two integrals along curves introduced cancel each other and we are left with the line integral over C. Hence the theorem, we must approximate R by a region of the type just considered and then use a limiting process. 208 Engineering Mathematics APPENDIX-II : PROOF OF STOKE’S THEOREM We shall first prove that in formula given by Eq. (7.38), the integrals over the terms involving 1 F are equal , i.e., 1 1 1 cosβ cos γ S F F dS F dx z y ∂ ∂ ⎡ ⎤ − = ∫ ⎢ ⎥ ∂ ∂ ⎣ ⎦ ∫ ∫ . . . (1) Let ( , ) z f x y = be the equation of the surface S and let the region R be the projection of S in the xy-plane. Then the projection of C in xy-plane is the curve τ bounding the region R (see Figure 3). Figure 3 : Surface S and its Projection on xy-plane Then we may integral over C as a line integral over τ as 1 1 ( , , ) [ , , ( , )] C F x y z dx F x y f x y dx τ = ∫ ∫ 1 , R F dx dy y ∂ = − ∂ ∫ ∫ applying Green’s theorem in plane to the function 1 [ , , ( , )] F x y f x y and O [instead of M and N in Section 7.4]. 1 1 R F F f dx dy y z y ∂ ∂ ⎛ ⎞ ∂ = − + ⎜ ⎟ ∂ ∂ ∂ ⎝ ⎠ ∫ ∫ . . . (2) In the integral on the right, 1 1 1 [ , , ( , )] ( , , ) ( , , ) as ( , ) F f x y f x y F x y z F x y z z f x y y y z y ∂ ∂ ∂ ∂ = + ⋅ = ∂ ∂ ∂ ∂ The direction consines of normal to the surface S, given by ( , ), z f x y = are cos α cosβ cos γ 1 f f x y = = ∂ ∂ − ∂ ∂ . . . (3) Also dx dy = projection of dS on xy-plane = dS cos γ 209 Line and Surface Integrals Thus 1 1 cosβ cos γ S F F dS z y ∂ ∂ ⎡ ⎤ − ⎢ ⎥ ∂ ∂ ⎣ ⎦ ∫ ∫ 1 1 cosβ cos γ cos γ R F F dx dy z y ∂ ∂ ⎡ ⎤ − ⎢ ⎥ ∂ ∂ ⎣ ⎦ ∫ ∫ 1 1 cosβ = cos γ cos γ R F F dx dy z y ∂ ∂ ⎡ ⎤ − ⎢ ⎥ ∂ ∂ ⎣ ⎦ ∫ ∫ 1 1 = , R F F F dx dy z y y ∂ ∂ ⎡ ⎤ ∂ − ⋅ − ⎢ ⎥ ∂ ∂ ∂ ⎣ ⎦ ∫ ∫ by virtue of Eq. (3) 1 = ( , , ) C F x y z dx ∫ by virtue of Eq. (2). Thus we have proved Eq. (1). Using the representation of surfaces S as = ( , ) y g x z and ( , ), x h y z − and reasoning exactly as above, we obtain 2 2 2 cos γ cos α = S C F F dS F dy z z ∂ ∂ ⎡ ⎤ − ⎢ ⎥ ∂ ∂ ⎣ ⎦ ∫ ∫ ∫ . . . (4) 3 3 3 cos γ cosβ = S C F F dS F dy y z ∂ ∂ ⎡ ⎤ − ⎢ ⎥ ∂ ∂ ⎣ ⎦ ∫ ∫ ∫ . . . (5) By adding Eqs. (1), (4) and (5), we obtain the formula given by Eq. (7.38). This proves Stoke’s Theorem for a surface S which can be represented simultaneously in explicit forms ( , ), ( , ) ( , ) z f x y y x z x h x y = = = We may immediately extend our results to a surface S which can be decomposed into finitely many pieces, each of which is of the type considered above. It should be noted that a person moving on C, with his head in the positive direction of the normal to surface S, should keep the surface on the left. 210 Engineering Mathematics APPENDIX-III : PROOF OF GAUSS DIVERGENCE THEOREM Let S be such a surface that a line parallel to z-axis meets it in two points only. Denote the lower and upper positions of S by S 1 and S 2 and let their equations be z = f 1 (x, y) and z = f 2 (x, y) respectively. Figure 4 We denote the projection of S on xy-plane by A. Then 3 , R u dx dy dz z ∂ ∂ ∫ ∫ ∫ 3 ( , ) 2 ( , ) 1 = A f x y f x y u dz dx dy z ∂ ⎡ ⎤ ⎢ ⎥ ∂ ⎣ ⎦ ∫ ∫ ∫ 2 1 3 = ( , , ) , f f A u x y z dx dy ∫ ∫ 3 2 3 1 = [ ( , , ) ( , , ) , A u x y f u x y f dx dy − ∫ ∫ . . . (1) Now for the upper position S 2 2 2 2 ˆ ˆ cos γ dx dy dS dS = = k . n and for the lower position S 1 1 2 2 ˆ ˆ cos γ dx dy dS dS = = − k . n a negative sign being taken as the normal to dS 1 makes an obtuse angle π − γ 1 with ˆ k . Therefore, 2 3 2 3 2 ˆ ˆ ( , , ) A S u x y f dx dy u dS = ∫ ∫ ∫ ∫ k . n and 1 3 1 3 1 ˆ ˆ ( , , ) A S u x y f dx dy u dS = − ∫ ∫ ∫ ∫ k . n Substituting in Eq. (1), we get 2 1 3 3 2 3 1 ˆ ˆ ˆ ˆ R S S u dx dy dz u dS u dS z ∂ = + ∂ ∫ ∫ ∫ ∫ ∫ ∫ ∫ k . n k . n 211 Line and Surface Integrals 3 ˆ ˆ S u dS = ∫ ∫ k . n This proves Eq. (7.44) of Section 7.7. Similarly, we can prove Eq. (7.42) and (7.43) of Section 7.7. This completes the proof of Eq. (7.50) and Gauss Divergence Theorem. FURTHER READING Ervin Kreyszig, Advanced Engineering Mathematic”, Wiley Eastern Ltd. Shanti Narain, Vector Algebra, S. Chand and Company. Chandrika Prasad, Mathematics for Engineers, Prasad Mudranalaya, Allahabad. G. B. Thomas and R. L. Finney, Calculus and Analytic Geometry, Narosa Publishing House. 212 Engineering Mathematics NOTATIONS AND SYMBOLS bold face letters, e.g. a : vector a. | a | : magnitude of vector a. ˆ ˆ ˆ , , i j k : unit vectors parallel to the axes of x, y, z respectively. 0 : null or zero vector. m a : multiplication of vector a by scalar m. a . b : scalar or dot or inner product of vector a and b. a × b : vector or cross product of vectors a and b. [a, b, c] or a . (b × c) : scalar triple product of vectors a, b and c. a × (b × c) : vector triple product of vectors a, b and c. f (p) : vector function. f (x) : vector field. 1 2 3 ˆ ˆ ˆ a a a = + + a i j k : a 1 , a 2 , a 3 are components of vector a w.r. to co-ordinate system. f s ∂ ∂ : directional derivative of f in the direction of s. grad f or ∇ f : gradient of f. div F or ∇ . F : divergence of F. curl F or ∇ × F : Curl of F. ˆ r : unit vector in the direction of r . or . or C C C w ds w ds w ds × ∫ ∫ ∫ : line integrals. ( , ) R f x y d A ∫ ∫ : double integral in plane region R. ( , , ) f x y z d ∑ ∑ ∫ ∫ : surface integral over surface Σ. ( , ) ( , ) x y x y u u J x y u v v v ∂ ∂ ∂ ∂ ∂ = = ∂ ∂ ∂ ∂ ∂ : Jacobian function. . C V d r ∫ : integration round a closed contour C. (Curl F) n : normal component of Curl F. ( , , ) V f x y z d V ∫ ∫ ∫ : triple or volume integral. 213 Line and Surface Integrals VECTOR CALCULUS The widespread application of computers to the engineering problems and the increasing use of linear analysis in handling large problems in systems analysis have affected the development of engineering mathematics during the past three decades. The usefulness of vector calculus in engineering mathematics results from the fact that many physical quantities – for example, forces and velocities – may be represented by vectors, and in several respects the rules of vector calculations are as simple as the rules governing the system of real numbers. However, vector analysis is a shorthand which simplifies many calculations considerably. Furthermore, it is a way of visualizing physical and geometrical quantities and relations between them. For all these reasons, extensive use is made of vector notation in modern engineering literature. This has motivated this block on vector calculus. This block is divided into three units, numbering 5 to 7. In Unit 5, we have presented the basic concepts : operation on vectors, vector products and methods of vector algebra and its applications to physical and geometrical problems. Unit 6 contains vector differential calculus, wherein scalar and vector fields have been defined along with limit, continuity and differentiability of vectors. The concepts of direction derivative, gradient, continuity and curl of vector fields and their applications have been developed in logical fashion in this unit. Unit 7 devoted to vector integral calculus. In particular, integration of a vector, double integrals, transformation of double integrals into line integrals (Green’s Theorem), surface integral and transformation of surface integrals into line integrals (Stoke’s Theorem) have been considered in Unit 7. Triple integrals, transformation of volume integrals into surface integrals (Divergence Theorem), irrotational and solenoidal vector fields form the subject matter of Unit 7.