PART 3: Integration. Calculus in Higher Dimensions. Compiled by M Frick. Assisted by CA Bohlman G Davie J Singleton K J Swanepoel

Transcription

1 Department of Mathematical Sciences PAT 3: Integration Study guide 3 for MAT615 alculus in Higher Dimensions ompiled by M Frick Assisted by A Bohlman G Davie J Singleton K J Swanepoel evised by J Singleton University of South Africa Pretoria

2 13 University of South Africa All rights reserved Printed and published by the University of South Africa Muckleneuk, Pretoria Page-layout by the Department MAT615/3/

3 iii MAT615/3. ontents Unit 1 Single Integrals 1 Page 1.1 Introduction 1 1. Definition of a Single Integral Geometric Interpretation of Single Integrals The Fundamental Theorem of alculus ules of Integration for Single Integrals Integration by Substitution 5 Unit 13 Double Integrals Introduction Definition of a Double Integral Geometric Interpretation of Double Integrals ules of Integration Evaluation of Double Integrals Transformation of oordinates alculation of Volumes by means of Double Integrals alculation of Areas by means of Double Integrals 4 Unit 14 Triple Integrals Introduction Definition of a Triple Integral Geometric Interpretation of Triple Integrals ules of Integration for Triple Integrals Evaluation of Triple Integrals oordinate Transformations alculating Volumes by using Triple Integrals 64

4 iv Unit 15 Line Integrals of eal valued Functions Introduction Line Integrals in the Plane, with respect to Arc Length Line Integrals in the Plane, with respect to the oordinate Variables Line Integrals of n functions Line Integrals over Piecewise Smooth urves 85 Unit 16 Line Integrals of Vector Fields Introduction Definition of a Line Integral of a Vector Field Physical Interpretation of F dr Evaluating F dr The Fundamental Theorem of Line Integrals Path independent Vector Fields Line Integrals over losed urves 99 Unit 17 Surface Integrals Introduction Definition of a Surface Integral Geometric Interpretation Evaluation of Surface Integrals 16 Unit 18 Flux Integrals Introduction Orientable Surfaces Flux Evaluation of Flux Integrals 115 Unit 19 The Theorems of Green, Gauss and Stokes Green s Theorem Stokes Theorem Gauss Theorem Generalizations of the Fundamental Theorem of alculus 131 Answers and hints to exercises 133 Index 138

5 Unit 1 Single Integrals 1.1 INTODUTION An integral of the form f xdx where f is an function and I is an interval on the - I axis, is called a single integral. Since single integrals are studied in first level alculus, we shall assume that you have grasped how an integral represents the limit of a sum, that you understand the connection between single integrals and areas and that you have mastered the standard techniques of integration. In this unit we briefly revise some important aspects of single integrals. We begin by revising the definition of a single integral, since this forms the basis for the definitions of the various types of integrals that we shall study in later units. 1. DEFINITION OF A SINGLE INTEGAL Thestepswhichleadtothedefinition of the single integral b a [a b] may be summarized as follows: 1. onsider an function f defined on a closed interval [a b]. Form a partition P of the interval [a b]bydividingitinton subintervals [c k1 c k ]; k 1 n f xdx over a bounded interval where a c c 1 c c n b 3. hoose a point x k in the kth subinterval of the partition, for k 1 n x k a x 1 c1 x c c x c cn-1 x b k-1 k k n Fig

6 4. Let x k denote the length of the kth subinterval, i.e. x k c k c k1 and let P (the norm of the partition) be the length of a longest subinterval, i.e. P max x k k 1n f(x ) k y = f(x) A k a c c x c c b x 1 1 k-1 k k n-1 Fig. 1. The area of A k is f x k x k 5. Form the sum n f x k x k k1 and consider the limit of this sum as P tends towards zero. Definition 1..1 Let f be an function defined on a closed interval [a b]. The single integral of f over [a b] is given by b n f xdx f x k x k a lim P provided this limit exists. If this limit exists, we say f is integrable over [a b] k1 emark 1.. If f x foreveryx I then, for each k 1n, thevalueof f x k x k gives the area of the kth rectangle, A k, shown in in Fig 1.. Let be the region below the graph of f above the - axis and between the lines x a and x b. Then the sum n k1 f x kx k is an approximation of the area of. Generally, the smaller the norm of the partition, the better the approximation. If the limit in Definition 1..1 exists, then it gives the exact value of the area of

7 3 MAT615/3 1.3 GEOMETI INTEPETATION OF SINGLE INTEGALS Suppose f is a continuous function definedonaclosedinterval[ab] If is the region between the -axis, the graph of f,andthelinesx a and x b then the following hold: If f (x forallx [a b] then If f x forallx [a b], then b a b a f xdx area of. (1.1) f xdx area of. (1.) y = f(x) a b a b (a) b a f xdx area of (b) b a y = f(x) f xdx (area of ) y = f(x) a 1 b (c) b a f xdx (area of 1 area of Fig. 1.3

8 4 1.4 THE FUNDAMENTAL THEOEM OF ALULUS We remind you of the following important theorem. Theorem The Fundamental Theorem of alculus If f is a smooth function that is definedonaclosedinterval[ab] then b a f xdx f b f a We can interpret f x as the rate of increase in f at the point x while f b f a gives the total increase in f over the interval [a b] Thus the Fundamental Theorem of alculus tells us that the integral of the rate of increase in a smooth function over an interval gives the total increase in the function over that interval. Let us explain this briefly. If f is constant over [a b] then the equation of f throughout [a b]isoftheform f x mxc where m f x, and the total increase in f over [a b] equals the slope of f times the length of the interval [a b], which equals mb a mb c ma c f b f a If f is not constant over [a b] then the total increase in f is given by the sum of the increases over the sub-intervals on which f x is constant. If f is a smooth function, the total change is given by the integral b a f xdx If F and f are functions such that Fx f x for all x [a b] then f is called an antiderivative of F on [a b]. We know that if F is continuous on [a b] then it has an antiderivative on [a b] Theorem can therefore also be stated as follows: Theorem 1.4. The Fundamental Theorem of alculus (Alternative form) If F is an function that is defined and continuous over a closed interval [a b] and f is any antiderivative of F on [a b],then b Fxdx f b f a a Theorem 1.4. tells us that, if F is continuous over a closed interval [a b], we can evaluate the integral b a Fxdx,byfinding an antiderivative of F and evaluating it only at the two boundary points of the interval.

9 5 MAT615/3 1.5 ULES OF INTEGATION FO SINGLE INTEGALS We remind you of the following rules: Theorem ules of Integration If f and g are both functions that are integrable over an interval [a b], then: 1. b a f x gxdx b a. b a kfxdx k b a 3. If c [a b]then f xdx b a gxdx f xdx for all k b a f xdx c a b f xdx c f xdx. 1.6 INTEGATION B SUBSTITUTION ecall the hange of Variable Formula for evaluating a single integral by means of a substitution. Theorem hange of Variable Formula Suppose g is a smooth, invertible function of u that maps the interval [c d] onto the interval [a b] If f is an function of x that is integrable over [a b],then b a f xdx g 1 b g 1 a f g u g u du The requirement that g be smooth and invertible ensures that g u forallu [c d] and that g is either monotone increasing (in which case g 1 a c and g 1 b d) ormonotone decreasing (in which case g 1 a d and g 1 b c). b The procedure for evaluating the integral f xdx by means of the substitution x gu involves the following four steps: a 1. eplace x with g u. eplace dx with g u du 3. Find the limits of integration for u (i.e. determine g 1 a and g 1 b). 4. Evaluate the new integral.

10 6 Example 1.6. Evaluate the integral Solution. Put 1 1 x 3 1 x dx x sin u for all u [ ] (Note that sin is a smooth function that is invertible on [ ] Now dx cos u du so we should replace dx with cos udu Now we determine the limits for u: x 1 u sin 1 1 and x 1 u sin 1 1 Hence 1 1 x 3 dx 1 x sin 3 u cos udu 1 sin u sin 3 udu since cos u foru sin u 1 cos u du 1 3 cos3 u cos u

11 7 MAT615/3 Exercises evision of first year work 1. Evaluate the following integrals. (a) (b) (c) x 1 x dx y 9 y dy x x 3 1 dx (d) 4 tan x sec xdx cos x (e) dx 4 3sinx (f) ln xe x dx. Use single integrals to calculate the area of the region enclosed by the curves y x 1andy 1 x 3. Use single integrals to calculate the area of the region enclosed by the curves y x andy x 4. Useasingleintegraltoderiveaformulafortheareaofacircle.

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13 Unit 13 Double Integrals 13.1 INTODUTION In this unit we consider the integral of an function over a bounded region in Such an integral is called a double integral. Double integrals, like single integrals, have important practical applications in various subjects. In order to model practical problems correctly in terms of integrals, it is crucial to have an intuitive understanding of the meaning of different types of integrals. In this unit we illustrate the meaning of double integrals and also provide techniques for evaluating such integrals. 13. DEFINITION OF A DOUBLE INTEGAL Just as the definition of a single integral is motivated by the problem of computing areas, the definition of a double integral is motivated by the problem of computing volumes. The following steps lead to the definition of the double integral f x yda,where f is an function and is a bounded region in (The concept of a bounded region is explained in Section.13.) 1. onsider an function f defined on a bounded region in the -plane.. Form a partition of into subregions by placing over a grid of lines parallel to the -and -axes. onsider only the rectangles of the grid that are completely contained in and number them 1 n. (In general, the finer the grid, the larger the part of that will be covered by these rectangles.) 9

14 1 (x k,y k ) 1 k n Fig hoose a point x k y k in k,fork 1 n 4. Let A k denote the area of the rectangle k ; k 1 n and let P be the length of the longest of the diagonals of the rectangles. Z z = f(x,y) f(x k,y k ) D k (x,y ) k k 1 Fig. 13. The volume of D k is f x k y k A k

15 11 MAT615/3 5. Form the sum n f x k y k A k k1 and consider the limit of this sum as P tends towards zero. Definition Let f be an function defined over a bounded region in the -plane. The double integral of f over is given by n f x yda f x k y k A k lim P provided this limit exists. If this limit exists, we say that f is integrable over k1 If f x y foreveryx y then, for each k 1n the value of f x k y k A k gives the volume of the rectangular prism D k with k as base and height f x k y k.nowletdbe the region below the graph of f that lies vertically above Then the sum n k1 f x k y k A k gives an approximation of the volume of D. Generally, the smaller the norm of the partition, the better the approximation. If the limit in Definition exists, then it gives the exact volume of D. We have seen that an function is integrable over a closed interval in if it is continuous over that interval. For functions we have a similar situation. Theorem 13.. If f is an function that is defined and continuous over a bounded region whose boundary consists of a finite number of smooth curves, then f is integrable over 13.3 GEOMETI INTEPETATION OF DOUBLE INTEGALS Interpretation as a volume The relationship between double integrals and volumes is similar to that between single integrals and areas. Let f be an function that is integrable over a bounded region in the -plane If D is the three-dimensional region below the graph of f that lies vertically above the region then the following holds:

16 1 If f x y forallx y then If f x y forallx y then f x yda volume of D. (13.1) f x ydavolume of D. (13.) Z z = f(x,y) D Z D z = f(x,y) (a) Volume of D f x yda (b)volume of D f x yda Fig Interpretation as an area. We know that the volume of a prism is obtained by multiplying its height with the area of its base. Thus if the height of a prism is 1 unit, then its volume has the same numerical value as the area of its base. If we integrate the function f x y 1 over a bounded region in the -plane, we therefore obtain the area of,i.e. 1 da area of (13.3)

17 13 MAT615/3 Z z = 1 1 Fig da area of 13.4 ULES OF INTEGATION The ules of Integration for double integrals are similar to those for single integrals. Theorem ules of Integration If f and g are both functions which are integrable over a bounded region in the -plane, then: 1. [ f x y gx y]da f x yda gx yda. kfx yda k f x yda 3. If 1,where 1 and are regions with no points other than boundary points in common, then f x yda f x yda f x yda 1 1 Fig Union of two regions with only boundary points in common

18 EVALUATION OF DOUBLE INTEGALS. In this section we shall show how to integrate a continuous function f over a bounded region in the -plane, by making use of the integration techniques for single integrals. We shall begin with rectangular regions and then proceed to bounded regions of a more general nature. Let T be the rectangle in the -plane defined by T x y a x b and c y d where a b, c and d are constants. d a T b c Fig The rectangle T x y a x b and c y d Let f be a continuous function defined on T For each x [a b]wenowdefine a function f x of y alone, by f x y f x y for all y [c d] Now f x is continuous and hence integrable over [c d] foreveryx [a b]. Thus, for every x [a b] we can consider the single integral d c f x ydy To evaluate this integral we regard x as a constant and integrate f x y with respect to y. We now define an function F of x by Fx d c f x ydy for every x [a b] Note that, after having substituted the limits of integration for y in the integral above, we are left with a function of x alone. It can be proved that F is integrable over [a b] so we may consider the single integral b a Fxdx.

19 15 MAT615/3 Now b a Fxdx b a d c f x ydy dx This integral is called a repeated integral (or an iterated integral) of f over T and we also write it as b d f x ydydx In a similar way we define the repeated integral d b c a a f x ydx dy c d c b a f y xdx dy The following theorem is named after the Italian mathematician Guido Fubini ( ), who proved a more general version of this result. Theorem Fubini s Theorem Let f be an function and let T be a rectangle in the domain of f defined by T x y a x b and c y d If f is continuous over T then the double integral f x yd A exists, and f x yda T b d a c T f x ydy dx d b c a f x ydx dy We shall not present the precise proof of Theorem here, but we illustrate the proof by means of sketches. In Fig the area of the shaded surface A k in (a) equals d c f xk ydy The volume of the shaded solid D k in (b) equals d f xk ydy x k and the volume of the solid D in (c) is given by lim P n d k1 c c f xk ydy x k b a d c f xk ydy dx d c Thus the repeated integral b a f x ydydx gives the volume of D But the double integral f x ydaalso gives the volume of D showing that T T f x yda b d a c f x ydydx

20 16 Z surface z = f(x,y) curve z = f (y) x k x b k a A k d y = d (a) Area of A k is d c f xk ydy Z x k b a (b) Volume of D k is d c D k x k f xk ydy x k d Z a D d b (c)volumeofdislim P n k1 d c f xk ydy x k b d a c f x k ydydx Fig. 13.7

21 17 MAT615/3 Theorem allows us to evaluate a double integral over a rectangle by successive evaluation of single integrals. Example Evaluate the double integral sketch below. T f x yda where f x y xy and T is the rectangle in the 3 T -1-1 Fig T x y 1 x 1 y 3 Solution.Since f is a continuous function, it follows from Theorem that T f x yda y 3 x 3 x xy dydx xdx 8 x dx dx

22 18 We now consider the case where the region of integration is not necessarily rectangular, but can be described in the form G x y a x b and g 1 x y g x where a and b are constants,andg 1 and g are smooth functions of x such that g 1 x g x for all x [a b] y = g (x) G a b y = g (x) 1 Fig.13.9AType1egion The lower and upper boundaries of the region G are formed by the smooth curves y g 1 x and y g x, respectively, while the left and the right boundaries of G areformedbythevertical lines x a and x b, respectively. We call such a region a Type 1 egion. For a Type 1 region the limits for x are constants,butthelimits for y may depend on x. The following generalization of Fubini s Theorem is useful for evaluating a double integral over a Type 1 egion. Theorem If f is an function that is defined and continuous on a Type 1 egion G, defined by G x y a x b and g 1 x y g x then the double integral f x yd A exists and G G f x yda b a g x g 1 x f x ydydx

23 19 MAT615/3 emarks The repeated integral in Theorem is calculated by first considering x as constant and integrating with respect to y. After substituting the integration limits g 1 x and g x for y, the integrand becomes a function of x alone and we then integrate normally with respect to x Since the limits for x are constants, our final answer will be independent of x and y.. The order of integration dydx indicates that we first integrate with respect to y,andthen with respect to x Example Evaluate G x yda where G is the region inside the circle x y, between the y-axis and the line x 1 Solution. Wefirst sketch the region of integration. y = - x G 1 y = - - x Fig The lower boundary for G is the smooth curve y x and the upper boundary is the smooth curve y x. The left boundary is the vertical line x andtheright boundary is the vertical line x 1. Thus G is a Type 1 egion, and can be described as G x y x 1and x y x

24 Thus, according to Theorem , G x yda 1 x x ydydx x 1 1 xy y x x x dx x dx 3 x Now we consider a region of integration of the form H x y c y d and h 1 y x h y wherec and d are constants, and h 1 and h are smooth functions of y such that h 1 y h y for all y [c d] d x = h (y) H 1 x = h (y) c Fig A Type egion

25 1 MAT615/3 The left and the right boundaries of the of the region H areformedbythesmooth curves x h 1 y and x h y, respectively, while the lower and upper boundaries are formed by the horizontal lines y c and y d, respectively. Such a region is called a Type egion. The following generalization of Fubini s Theorem is useful for evaluating a double integral over a Type egion. Theorem If f is an function that is defined and continuous on a Type egion H, where H x y c y d and h 1 y x h y then f is integrable over H and H f x yda d c h y h 1 y f x ydx dy emarks The repeated integral in Theorem is calculated by first considering y as constant and integrating with respect to x. After substituting the limits of integration for x,theintegrand becomes a function of y alone and we then integrate normally with respect to y. Sincethe limits for y are constants, our final answer will be independent of x and y.. The order of integration dx dy indicates that we first integrate with respect to x,andthen with respect to y Example Evaluate the integral H 4x yda where H is the region in the -plane that lies inside the circle x y 4 above the -axis and below the line y x Solution. The semicircle x 4 y and the line x y intersect where y 4 y i.e. y andx.

26 x = y H (, ) x = 4 - y Fig The region of integration, H,isaType egion,sincetheleft boundary is the smooth curve x y, itsright boundary is the smooth curve x 4 y,andthe lower and upper boundaries are the horizontal lines y andy, respectively. Thus H x y y andy x 4 y Hence, according to Theorem , 4x yda 4y 4x ydx dy H y x yx 4y y dy 8 3y y 4 y dy 8y y y

27 3 MAT615/3 eversing the Order of Integration Sometimes it is easier to evaluate a given repeated integral if we reverse the order of integration. Example Evaluate the repeated integral 4y y 1 dx dy. 4 x Solution. The region of integration of this integral is the region H in Fig Since integrating 1 4x with respect to x is laborious, we would prefer to integrate with respect to y first, i.e. we wish to use Theorem , with the order of integration dydx. However, H is not a Type 1 region, because its upper boundary is not smooth, sinceasudden change occurs where x. To the left of the vertical line x the upper boundary is formed by the smooth curve y x. To the right of the line x it is formed by the smooth curve y 4 x Thus H is not a Type 1 region, but it is the union of two Type 1 regions, H 1 and H where and H 1 x y x and y x H x y x and y 4 x y = x y = 4-x H 1 H y = Fig Since H H 1 H, ule 3 of Theorem implies 1 da 1 da 4 x 4 x H H 1 H 1 4 x da Thus

28 4 4y y 1 dx dy 4 x x 1 dydx 4 x y 4 x x dx x dx 1 dx 4 x 4 x 4 [x] 4x 1 dydx 4 x 4x y dx 4 x emarks Hints on reversing the order of integration 1. If the region of integration is a rectangle, then the limits of integration are all constants. In this case the limits of the inner and outer integrals are simply interchanged when the order of integration is reversed. Thus, if f is defined and continuous on the rectangle T x y a x b and c y d,then b d a c f x ydydx d b c a f x ydx dy. b. When we want to reverse the order of integration of the integral g x a g 1 x f x ydydx,where the limits of integration for y are dependent on x, then we cannot simply interchange the limits of integration. (emember, the limits of the outer integral must be constants!) In this case we need to describe the region of integration as a Type region, or as a union of Type regions. To achieve this we need to sketch the region and rewrite the inequalities that describe the region in such a way that the limits for x are smooth functions of y, and the limits for y are constants. d h y 3. When we want to reverse the order of integration of the integral f x ydx dy,where c h 1 y the limits of integration for x are dependent on y,weneedtodescribetheregionofintegration as a Type 1 region, or as a union of Type 1 regions. To achieve this we need to sketch the region and rewrite the inequalities that describe theregioninsuchawaythatthelimits for y are smooth functions of x, and the limits for x are constants. 4. When we want to describe a Type 1 region it is useful to shade the region by means of vertical lines in order to note which curves form the lower and upper boundaries for y (in terms of x 5. When we want to describe a Type region we shade it by means of horizontal lines in order to note which curves form the left and right boundaries for x (in terms of y

29 5 MAT615/3 Example Write the repeated integral 1 x ydydx 1 as a sum of repeated integrals in which the order of integration is reversed. Solution. Wefirst sketch the region of integration. y = x -1 1 y = Fig (a) We can shade the whole region by drawing vertical lines from the line y uptothecurve y x, between the vertical lines x 1andx 1 When we want to reverse the order of integration, the limits of integration for x must be written in terms of y, and the limits for y must be constants. In order to determine the left and right boundaries, we shade the region of integration by means of horizontal lines. 1 x = 1 x = y x = y x = 1 H 1 H Fig (b) For the region H 1 the left boundary is the line x 1andtheright boundary is the curve x y. For the region H,theleft boundary is the curve x y,andtheright boundary is the line x 1. Both H 1 and H lie between the horizontal lines y andy 1. Thus 1 x 1 ydydx 1 y 1 ydxdy 1 1 y ydxdy

30 6 Exercises In each of the following, sketch the region and calculate the double integral f x yda using the order of integration that you find most convenient. (a) f x y x y x y 1 x 4and y 4 (b) f x y x y x y y and x 4 y (c) f x y x y x y x and y cos x (d) f x y sinx y x y x y andx y (e) f x y ye x x y y 1andy x y. Let be the region in the -plane above the curve y x below the line y x Evaluate the integral x yda 3. Express each of the following repeated integrals as a repeated integral in which the order of integration is reversed (or as a sum of such integrals, if necessary). heck your answers by evaluating the integrals both ways. (a) (b) (c) (d) (e) 1 x 1 x ydy dx x 4x 1 1x y 1 y 1dydx x xy dydx 1x xdy dx x ydx dy

31 7 MAT615/ TANSFOMATION OF OODINATES onsider the transformation x T 1 u, y T u u Q and the corresponding function T defined by T T 1 T We can regard T as transforming the region Q in the UV plane into a region in the -plane. If T is an invertible function, then the above transformation establishes a one-to-one correspondence between the points in and the points in Q. (Study Theorem , Study Guide 1) V Q U T Fig If x y are the rectangular coordinates of a point in the -plane, we can think of u as specifying the same point, since by knowing u we can determine x y from the equations x T 1 u and y T u Theevaluationofagivenintegralcanoftenbesimplified by a change of variables. For double integralswehaveachange of variables formula which is similar to the one for single integrals (see Theorem 1.6.1), with the absolute value of the Jacobian taking the place of the ordinary derivative. Theorem hange of Variables Formula Suppose each of and Q is a bounded, -dimensional region, whose boundary consists of a finite number of smooth curves, and T is a smooth, invertible function of u and that maps Q onto If f is an function of x and y that is integrable over then f x y da f T u det J T u du d Q

32 8 The factor det J T u in the hange of Variables Formula compensates for the change in area brought about by the transformation x y T u. Let us explain this briefly. In defining the double integral f x y da, a rectangular grid is placed over the region in the -plane (see Fig. 13.1). This grid consists of horizontal lines, called x-lines and vertical lines, called y-lines. On each x-line y is constant and on each y-line x is constant. We now replace the rectangular grid with a u-grid, consisting of u-curves (where is constant) and -curves (where u is constant). onsider the point P x y and the region with an area of A shown in Fig Let u and be the width and length of the corresponding rectangle in the UV-plane. u - curve P a A b v - curve Fig A a b x At the point P, the vector derivative u y is tangent to the u-curve through P and the vector u derivative x y is tangent to the -curve through P. Nowlet a x u y x u and b u y Then A is approximately equal to the area of the parallelogram determined by the vectors a and b. From the geometric interpretation of the cross product (see Section.1) we know that the area of this parallelogram equals a b (see Fig..18). Hence A a b Now a b x u y u x u y u x u y x y u

33 9 MAT615/3 But, by Definition.1.3, x u y x u y x u x y u y k det J T u k Thus a b det JT u k u since k 1 and u and are positive. Hence det JT uu k by N.3, Section.5 det JT u u A det JT u u It can be shown that the limit of the sum f T u det J T u u taken over the entire region gives the integral f T u det J T u du d Q emark det J T u is called the Jacobian of the transformation x y T u It is also denoted by x y u alculation of an integral by means of a coordinate transformation Evaluating an integral of the form f x y daby using the transformation x y T 1 u T involves the following four steps. 1. eplace x with T 1 u and y with T u. eplace dawith det JT u du d 3. Describe the region Q in terms of u and and find the limits of integration for u and 4. Evaluate the new integral.

34 3 Polar coordinates Evaluation of certain double integrals can often be simplified by changing to polar coordinates. Definition Polar coordinates represent a point p in by means of an ordered pair r where (1) r is the distance from p to the origin, () is the angle that the vector p makes with the positive -axis (measured anticlockwise). If x y are the rectangular coordinates of p,andrthe polar coordinates, then it is clear from Figure that x r cos y r sin (13.4) Moreover, r x y y (x,y) = (rcos, rsin r rsin rcos x Fig elationship between polar and rectangular coordinates

35 31 MAT615/3 emarks Note that r is never negative, since it represents a distance.. If c is a constant, then the equation r c represents a circle in the -plane. 3. If is a constant, then the equation defines a ray from the origin in the -plane. r = c c - c c - c Fig (a) The circle r c (b) The ray Thus a rectangular grid in the r -plane corresponds to a curved grid in the - plane, as shown in Fig The -curves are circles (since r remains constant on a circle) and the r-curves are rays drawn from the origin (since remains constant on such a ray). 4 4 r r=1 r= r=3 4 4 Fig A rectangular grid in the r plane and the corresponding grid in the -plane

36 3 Now consider the region in Fig This region can be described in terms of polar coordinates as r 1 and r 1 r r where 1 and are constants, with 1 and r 1 and r are both smooth functions of with r 1 r for all [ 1 ] r = r( r = r 1 ( Fig. 13. The region r 1 and r 1 r r emark Note that neither of the curves bounding the region in the sketch above is the graph of a function of x (because on each curve some x-value corresponds to more than one y-value) and hence it is not easy to describe this region by means of rectangular coordinates. However, r 1 and r are functions of, since each value of corresponds to only one value of r on each of these curves. onsider the polar coordinate transformation x y P r where with P r r cos r sin r and I where I is an interval of length at most Then P is a smooth function. The restriction on the length of I ensures that P is one-to-one, and hence invertible. (We usually take I to lie

37 33 MAT615/3 in the interval [ ] but if negative angles are involved, we can take it to lie in [] for example.) Now The Jacobian of this transformation is x r y r cos sin x y r sin r cos r cos r sin r Hence det J P r r (13.4) Let f be an function which is continuous on. From Theorem we obtain the Transformation ule for Polar oordinates: f x y da r 1 r 1 f P r rdrd (13.6) Examples Evaluate the integral Solution. 4y y 4x ydx dy Note that this is the integral of Example Since the region of integration is a circle sector, polar coordinates are very suitable for describing this region, so we proceed as follows: Step 1. We put The integrand then becomes x r cos and y r sin 4x y 4r cos r sin Step. Since det J P rr, we replace dawith rdrd. The integral then becomes 4r cos r sin rdrd r 4cos sin dr d where denotes the region of integration.

38 34 Step 3. In order to describe the region by means of polar coordinates, we shade the region by drawing rays from the origin, starting on the -axis (i.e. where and ending on the line y x i.e. where r sin r cos i.e. tan 1 i.e. 4. The length of each of theses rays is units (since every point on the semicircle x 4 y lies at a distance of units from the origin). r = r = Fig Thus the region can be described by r 4 and r Step 4. 4y 4x y dx dy 4 r 4cos sin dr d y 4 r 3 3 4cos sin d 4 8 4cos sin d [4sin cos ]

39 35 MAT615/3. Evaluate the integral x y 3 da where is the region in the -plane below the graph of y x, above the line y 1 and between the lines y x and y 3x Solution. The integrand, x y 3 is laborious to integrate with respect to x as well as with respect to y, but it becomes simpler if we set x r cos and y r sin since then x y r. For that reason it seems worthwhile to switch to polar coordinates. Step 1. We set x r cos y r sin Then the integrand becomes r cos r sin 3 r 3 Step. Since detj F r r the integral becomes r 3 rdrd r dr d Step 3. Let us sketch the region y = - x y = 1 y = 3 x y = x Fig.13.(a)

40 36 We now shade the region by means of rays emanating form the origin. = 3 r = = 4 y = 3x 1 r = cosec y = x O Fig. 13.(b) lies between the ray y x and the ray y 3x. Hencethelower limit for is given by y x i.e. r sin r cos i.e. tan 1 i.e. 4 The upper limit for is given by y 3x i.e. r sin 3r cos i.e. tan 3 i.e. 3. This agrees with our sketch, since the lines y x and y 3x make angles of 4 and, respectively, with the positive -axis. 3 Since rays from the origin enter the region on the line y 1, the lower limit for r is given by y 1 i.e. r sin 1 i.e. r cosec

41 37 MAT615/3 The upper limit for r is given by the curve x y i.e. r i.e. r The limits for r may also be determined geometrically. onsider a ray that makes an angle of units with the positive -axis, where 4 Let B be the point where this rays 3 enters the region,and the point where it exits. cosec By considering the triangle OAB,weseethat Fig. 13.(c) sin OA OB 1 OB hence OB cosec Since lies on a circle with radius, centred at the origin, O Thus we can describe the region in terms of polar coordinates as r cosec r and 4 3 emark. All points on a circle centred at the origin lie at the same distance from the origin, but that is not the case for points on a straight line. That is why, for the given region, the lower limit for r depends on, while the upper limit is a constant.

42 38 Step 4. x y 3 da 3 r dr d 4 cosec 3 r 1 cosec d sin d cos Exercises Describe the following regions by means of polar coordinates. (a) (b) (c) (d) Describe the following regions by means of polar coordinates. (a) The region x y1 y andy x y (b) The region enclosed by the graph of x y and the semi-circle x 1 y

43 39 MAT615/3 3. Use polar coordinates to calculate the following double integrals: (a) (b) (c) (d) 4y 3 1 1y x y 3 dx dy sinx y dydx 9x e xy dx dy 1y 1 1 1y xy x y dx dy 13.7 ALULATION OF VOLUMES B MEANS OF DOUBLE INTEGALS The following examples illustrate how we can calculate the volume of a given 3-dimensional region by using the formulas (13.1) and (13.), given in Section Examples Let f be the function defined by f x y 4 x y and let be the triangle in the -plane with vertices 1 and 1 Determine the volume of the region D under the graph of f that lies vertically above. Solution. Let us sketch the region D Z z = 4 - x - y D 1 1 Fig. 13.3

44 4 is a Type 1 region and can be described as x y x 1and y 1 x By the formula (13.1), volume of D 1 1x 1 1 f x yda 4 x y dydx 4y x y 13 1x y3 dx 41 x x 1 x 13 1 x3 dx 4x x x3 3 x x Let D be the region in 3 inside the cylinder x y that lies above the surface z x y and below the surface z 4 x y. Find the volume of D by using double integrals. Solution. Let us sketch the region D 1 Z z = 4 - x - y D D 1 - z = x + y - Fig Let be the region in the -plane bounded by the circle x y. Let D 1 be the region above the paraboloid z x y thatliesbelow and let D be the region below the hemisphere z 4 x y that lies vertically above Then volume of D volume of D 1 volume of D

45 41 MAT615/3 By the formula 131 volume of D 1 x y da and by the formula 13 volume of D 4 x y da Since the area of integration, is bounded by a circle, polar coordinates will be most suitable for evaluating these integrals. Thus we set x r cos and y r sin Then and x y r 4 x y 4 r As shown in Section 13.6, the Jacobian of the transformation to polar coordinates is det J P r r We can describe as r and r Thus volume of D x y da r rdrd 4 x y da 4 r rdrd r 3 r dr d r 4 r dr d r 4 r 1 d d 8 d r 3 d

46 4 Exercises In each of the following cases, use double integrals to calculate the volume of the region below the graph of f that lies above the region. (a) f x y e x cosy and x y x ln 3 and y (b) f x y x 4and is the region in enclosed by the line y 3x and the curve y 4 x (c) f x y x y and is the region in between the -axis and the semicircle x 1 y. alculate the volume of the region inside the cylinder x y 4, above the -plane and below the paraboloid z x y 3. alculate the volume of the region enclosed by the functions f and g given by f x y 8 x y and gx y x y 4. Use a double integral and polar coordinates to deduce a formula for the volume of (a) a cylinder with radius a and height h (b) a sphere with radius a ALULATION OF AEAS B MEANS OF DOUBLE INTEGALS The following examples illustrate how we can calculate the area of a given region in,byusing the formula Area of 1 da which was given in Section Examples alculate the area of the region in enclosed by the curves y x andy x Solution. Let us sketch the region First, we note that the curves that bound the region intersect where x x i.e. where x 1or

47 43 MAT615/3 y = x -1 - y = x - Fig By shading the region by means of vertical lines, we note that is a Type 1 region and can be described as x y 1 x andx y x By (13.3), area of 1dA x 1 x 1 dydx y x x dx x x dx x 1 3 x3 x 1

48 44. Use a double integral to derive a formula for the area of a circle. Solution. Let be the circle in the -plane with radius a, centred at a Fig We can describe by means of polar coordinates as r and r a By (13.3) area of a 1dA rdrd 1 r a 1 a d 1 a a d Hence the area of a circle with radius a is a

49 45 MAT615/3 Exercises In each of the following cases, use a double integral to calculate the area of the region (a) x y x, y x (b) x y x, 4 x y 4 x (c) r 3 3 and r (d) is the region in enclosed by the curves y 1 x and y 1 x. (e) is the region in enclosed by the curves y x x 1 x 1andy 1 (f) is the region in enclosed by the lines y x y x x 1andx 1.

50 46

51 Unit 14 Triple Integrals 14.1 INTODUTION The definition of double integrals of functions and their calculation by means of repeated integrals can easily be generalized to triple integrals of 3 functions, and even to n-fold integrals of n functions, for all natural numbers n Since the analogy with double integrals is so close, the treatment of triple integrals will be brief. 14. DEFINITION OF A TIPLE INTEGAL The following steps lead to the definition of a triple integral: 1. onsider an 3 function f defined on a bounded region D in 3. Z (x k,y k,z k ) D k D Fig

52 48. Form a partition P of D into subregions by means of a 3-dimensional grid of planes parallel to the -plane, the Z-plane and the Z-plane, respectively. onsider only those blocks that lie completely inside D, and number them D 1 D D n 3. hoose a point x k y k z k in D k,fork 1 n 4. Let V k represent the volume of the block D k,fork 1 n, andletp be the length of the longest diagonal of the D k. 5. Form the sum n f x k y k z k V k k1 and consider the limit of this sum as P tends towards zero. Definition Let f be an 3 function that is defined on a bounded region D in 3.The triple integral of f over D is given by n f x y zdv f x k y k z k V k provided this limit exists. D lim P If the limit exists we say that f is integrable over D k1 The graph of a smooth function f is called a smooth surface. For triple integrals we have the following analogue of Theorem Theorem 14.. If f is an 3 function that is defined and continuous on a bounded region D, the boundary of which consists of a finite number of smooth surfaces, then f is integrable over D 14.3 GEOMETI INTEPETATION OF TIPLE INTEGALS In the same manner that a single integral of an function represents an area and a double integral of an function represents a volume, a triple integral of an 3 function can be regarded as representing a 4-dimensional volume (which, of course, has no immediate geometric meaning). As shown in Section 13.3, the double integral 1 da gives the area of a bounded region in. Similarly, if D is a bounded region in 3,then 1 dv volume of D (14.1) D

53 49 MAT615/ ULES OF INTEGATION FO TIPLE INTEGALS The rules of integration for triple integrals are similar to those for double integrals. Their formulation is left as an exercise for the reader EVALUATION OF TIPLE INTEGALS onsider a Type 1 region G in the -plane, given by G x y a x b and g 1 x y g x Suppose that h 1 and h are smooth functions of x and y such that h 1 x y h x y for all x y G Let D be the region in 3 defined by D x y z a x b g 1 x y g x and h 1 x y z h x y We call such a region a Type I egion. It is the 3-dimensional analogue of a -dimensional Type 1 egion. Z z = h (x,y) D z = h (x,y) 1 a y = g (x) G y = g (x) 1 b Fig. 14. A Type I egion.

54 5 Theorem If f is an 3 function that is continuous over a Type I egion D, then D f x y z dv b g x h xy a g 1 x h 1 xy f x y z dz dydx. emark In the formula above the limits of integration for z may depend on both x and y; those for y may depend only on x, and the limits for x must be constants. The limits of integration for a particular variable cannot depend on a variable with respect to which integration has already taken place. Example Evaluate xdzdydx D where D is the tetrahedron enclosed by the plane x y z and the three coordinate planes. Solution. Z z = - x - y D y = - x z = Fig The tetrahedron D is given by D x y z x y x z x y

55 51 MAT615/3 Thus xdzdydx x xy xdzdydx D x x [z] xy dydx x x x xydydx xy x y 1 xy x x x 1 x3 dx x 3 x3 1 8 x4 dx 3. Exercises Formulate the analogue of Theorem for the case where the order of integration is dy dx dz. (First describe an appropriate region D. Evaluate the following triple integrals. (a) (b) 1 x xy 1 x xy z z x y z dzdydx 1dydx dz 3. Evaluate the triple integral where f x y z dv D f x y z x y and D x y z x y z andx y z 3

56 OODINATE TANSFOMATIONS The change of variables formula for a triple integral is similar to the one for a double integral. Theorem Suppose each of D and B is a bounded, 3-dimensional region, whose boundary consists of a finite number of smooth surfaces, and T is a smooth, invertible 3 3 function of u and that maps B onto D If f is an 3 function of x y and z that is integrable over D then D f x y z dv B f T u det JT u du d d We shall apply this formula to two special cases, namely to change from rectangular to cylindrical coordinates and to change from rectangular to spherical coordinates. These two coordinate systems are especially useful for describing a 3-dimensional region that can be obtained by rotating a -dimensional region about the Z-axis. (Many of the graphs considered in Section 3. can be obtained in this manner.) ylindrical oordinates ylindrical coordinates are a natural mixture of polar and rectangular coordinates. Definition ylindrical coordinates represent a point p in 3 by means of an ordered triple rz where 1. rare the polar coordinates of the vertical projection of p onto the -plane,. z is the vertical rectangular coordinate.

57 53 MAT615/3 Z z (x, y, z) = (rcosrsin z) z r y x Fig elationship between rectangular and cylindrical coordinates. If x y z are the rectangular coordinates of p and rz its cylindrical coordinates, then x r cos, u r sin and z remains z. (14.) emarks The term cylindrical coordinates arises from the fact that if c is a constant then, in 3 the equation r c represents a cylinder with radius c. If is a constant then, in 3 the equation defines a vertical plane through the origin. Z Z c Fig. 14.5(a) The cylinder r c. (b) The plane.

58 54 Now consider the cylindrical coordinate transformation x y z rz where with rz r cos r sin z r I and z where I is an interval of length at most learly, is a one-to-one 3 3 function. The Jacobian of this transformation is given by x x x r z detj rz y y y r z z z z r z cos r sin sin r cos 1 r cos r sin r Hence detj rz r (14.3) Now let be the -dimensional region defined by r 1, r 1 r r where 1 and are constants, with 1 and r 1 and r are smooth functions of,with r 1 r r Let z 1 and z be smooth functions of r and,with z 1 r z rfor all r and let D be the region defined by D rz 1 r 1 r r z 1 r z z r

59 55 MAT615/3 Note that is the -projection of D. From Theorem we obtain the Transformation ule for ylindrical oordinates: If f is an 3 function that is continuous on D,then D f x y z dv r z r 1 r 1 z 1 r f rz rdzdrd (14.4) Example Evaluate the integral I x y z dv where D D x y z x y z Solution. Step 1. Put x r cos, y r sin, z z The integrand then becomes x y z r cos r sin z r z Step. Since det J rz r, the integral becomes I r z rdzdrd D Step 3. In order to find the limits of integration, we sketch the region D We note that it is bounded below by the cone z x y and bounded above by the plane z Z z = D z = x + y = r Fig. 14.6(a)

60 56 The cone z x y intersects the plane z where x y i.e. where x y 4 Thus the projection of the region D onto the -plane is the region inside the circle in the -plane with equation x y 4 Fig. 14.6(b) We can describe by means of polar coordinates as r and r ThewholeregionD is obtained by rotating the region D in Fig, 14.6(c) about the Z-axis through an angle of Z z = D * z = r Fig. 14.6(c) By shading the region D by means of vertical lines (lines parallel to the Z-axis), we note that the lower limit for z is given by z r and the upper limit for z is z (Thisisalsoclearfrom Fig. 14.6(a). Thus we can describe the region D in terms of cylindrical coordinates by D rz and r andr z

61 57 MAT615/3 Step 4. I r r r z rdzdr d r 3 rz dzdr d r 3 z 1 3 rz3 r dr d r 3 83 r 43 r 4 dr d 4 5 r 4 4r 3 4r 5 15 d 4 5 [] d Spherical oordinates Spherical coordinates locate points in 3 by means of two angles and a distance. Definition Spherical coordinates represent a point p in 3 by means of an ordered triple where (1) is the distance between p and the origin, () is the angle that the vector p makes with the positive Z-axis, (3) is the angle from cylindrical coordinates.

62 58 onsider a point in 3 with rectangular coordinates x y z cylindrical coordinates rz and spherical coordinates Z z (x,y,z) = ( sin cos, sin sin, cos) z = cos y x y = rsin r = sin x = rcos Fig elationship between spherical, cylindrical and rectangular coordinates. Using trigonometry, we see that z cos and r sin Since x r cos and y r sin we now have x sin cos y sin sin z cos (14.5) Furthermore, x y z emark cannot be negative, since it represents a distance.. The term spherical coordinates arises from the fact that, if c is a constant, the equation c defines a sphere in 3 (with radius c). 3. If is a constant, the equation defines a cone in 3 This cone is obtained by taking a ray that makes an angle of magnitude with the positive Z-axis and rotating it about the Z-axis through an angle of (i.e. by letting vary from to ).

63 59 MAT615/3 Z Z c Fig (a) The sphere c. (b)thecone. Now consider the spherical coordinate transformation x y z S where S sin cos sin sin cos with, I and [] where I is an interval of length at most S is clearly a one-to-one 3 3 function. The Jacobian for this transformation is x x x detj S y y y z z z sin cos sin sin cos cos sin sin sin cos cos sin cos sin sin The restriction ensures that sin Hence det J S sin (14.6)

64 6 Now let D be the region in 3 described by D 1 1 and 1 where 1 and are constants, with 1 1 and are smooth functions of with 1 and 1 and are smooth functions of and, with 1 If f is an 3 function that is continuous on D, then from Theorem we have the Transformation ule for Spherical oordinates D f x y z dv f S sin d d d (14.7) Example Use spherical coordinates to evaluate the integral of Example Solution. Step 1. We put x sin cos y sin sin z cos The integrand then becomes x y z sin cos sin sin cos sin cos sin cos sin cos Step. Since det J S sin, the integral now becomes I D 4 sin d d d

65 61 MAT615/3 Step 3. As noted in Example , the region D can be obtained by rotating the region D in Fig. 14.6(c) about the Z-axis through an angle of Geometrically it is clear that, in the region D the value of varies between and 4.(emember that is measured downward from the positive Z-axis.) We can check the upper limit for by noting that, if z x y then cos sin cos sin sin i.e. cos sin i.e. tan 1 i.e. 4. In order to find the limits for we note that the region D can be shaded by drawing, for each between and a ray from the origin up to the line z 4 Z = sec D * = Fig. 14.9(a) Thus the lower limit for is, and the upper limit is given by z i.e. cos i.e. sec We can also see this geometrically, by noting that the length of a ray in D that makes an angle of magnitude with the positive Z-axis, has length sec Z sec Fig. 14.9(b)

66 6 Thus the region D can be described in terms of spherical coordinates by D 4 and sec Step 4. I x y z dx dydz D 4 sec 4 sin d d d sec 5 5 sin d d sin sec 5 d d sec 4 tan d d sec 3 sec tan d d 1 4 sec4 [4 1] d 4 d 48 5.

67 63 MAT615/3 Exercises In each of the following cases, sketch the region D and describe it in terms of cylindrical coordinates. (a) D is the region in 3 inside the cylinder x y above the plane z 1and below the plane z 3 (b) D is the region inside the sphere x y z 4 (c) D is the region in 3 enclosed by the surfaces z 3 x y and z x y 3. In each of the following cases, sketch the region D and describe it in terms of spherical coordinates. (a) D is the region inside the sphere x y z 4 (b) D is the region inside the sphere x y z 4 above the cone z x y 3. Let D be the region in 3 enclosed by the surfaces z x y and z 9 x y Evaluate the integral xzdv using (a) cylindrical coordinates. (b) spherical coordinates. D 4. Let D be the region in 3 inside the cone z x y below the hemisphere z 4 x y and above the plane z 1 Evaluate the integral D x y dv using the coordinate system that you find most convenient. 5. Let D be the region in 3 inside the cylinder x y 1 above the cone z x y below the cone z x y Evaluate the integral 1 x y dv using the coordinate system that you find most convenient. D 6. Evaluate the integral x y z dv where D D x y z x y z 4

68 ALULATING VOLUMES B USING TIPLE INTEGALS The following examples illustrate how we can calculate the volume of a given 3-dimensional region by using the formula volume of D 1 dv which was given in Section D Examples Let D be the region in 3 above the plane z and below the paraboloid z 3 x y. alculate the volume of D Solution. The plane z and the paraboloid z 3 x y intersect where 3 x y i.e. x y 1 Thus the -projection of D is the region in the -plane inside the circle x y 1 Z 3 z = 3 - x - y D z = Z 3 z = 3 - x - y 1 D * z = 1 Fig