M ULTIPLE I NTEGRALS. variables over regions in 3-space. Calculating such integrals will require some new techniques that will be a

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1 April, :4 g65-ch5 Sheet number Page number 5 can magenta ellow black M ULTIPLE I NTEALS n this chapter we will etend the concept of a definite integral to functions of two and three variables. Whereas functions of one variable are usuall integrated over intervals, functions of two variables are usuall integrated over regions in -space and functions of three variables over regions in 3-space. Calculating such integrals will require some new techniques that will be a central focus in this chapter. Once we have developed the basic methods for integrating functions of two and three variables, we will show how such integrals can be used to calculate surface areas and volumes of solids; and we will also show how the can be used to find masses and centers of gravit of flat plates and three-dimensional solids. In addition to our stud of integration, we will generalie the concept of a parametic curve in -space to a parametric surface in 3-space. This will allow us to work with a wider variet of surfaces than previousl possible and will provide a powerful tool for generating surfaces using computers and other graphing utilities.

2 April, :4 g65-ch5 Sheet number Page number 6 can magenta ellow black 6 Multiple Integrals 5. DOUBLE INTEALS The notion of a definite integral can be etended to functions of two or more variables. In this section we will discuss the double integral, which is the etension to functions of two variables. VOLUME ecall that the definite integral of a function of one variable b a f()d lim ma k k n f(k ) k lim n + k n f(k ) k () f(, ) arose from the problem of finding areas under curves. [In the rightmost epression in (), we use the limit as n + to encapsulate the process b which we increase the number of subintervals of [a,b] in such a wa that the lengths of the subintervals approach ero.] Integrals of functions of two variables arise from the problem of finding volumes under surfaces: 5.. THE VOLUME POBLEM. iven a function f of two variables that is continuous and nonnegative on a region in the -plane, find the volume of the solid enclosed between the surface f(,) and the region (Figure 5..). Figure 5.. Later, we will place more restrictions on the region, but for now we will just assume that the entire region can be enclosed within some suitabl large rectangle with sides parallel to the coordinate aes. This ensures that does not etend indefinitel in an direction. The procedure for finding the volume V of the solid in Figure 5.. will be similar to the limiting process used for finding areas, ecept that now the approimating elements will be rectangular parallelepipeds rather than rectangles. We proceed as follows: Using lines parallel to the coordinate aes, divide the rectangle enclosing the region into subrectangles, and eclude from consideration all those subrectangles that contain an points outside of. This leaves onl rectangles that are subsets of (Figure 5..). Assume that there are n such rectangles, and denote the area of the kth such rectangle b A k. Choose an arbitrar point in each subrectangle, and denote the point in the kth subrectangle b (k, k ). As shown in Figure 5..3, the product f( k, k ) A k is the volume of a rectangular parallelepiped with base area A k and height f(k, k ), so the sum n f(k, k ) A k k can be viewed as an approimation to the volume V of the entire solid. f(, ) Height f(* k, * k ) Area A k Figure 5.. ( k *, k *) Area A k Figure 5..3 ( k *, k *)

3 April, :4 g65-ch5 Sheet number 3 Page number 7 can magenta ellow black 5. Double Integrals 7 There are two sources of error in the approimation: first, the parallelepipeds have flat tops, whereas the surface f(,) ma be curved; second, the rectangles that form the bases of the parallelepipeds ma not completel cover the region.however,ifwe repeat the above process with more and more subdivisions in such a wa that both the lengths and the widths of the subrectangles approach ero, then it is plausible that the errors of both tpes approach ero, and the eact volume of the solid will be n V lim f(k, k ) A k n + k This suggests the following definition. 5.. DEFINITION (Volume Under a Surface). If f is a function of two variables that is continuous and nonnegative on a region in the -plane, then the volume of the solid enclosed between the surface f(,) and the region is defined b V lim n + k n f(k, k ) A k () Here, n + indicates the process of increasing the number of subrectangles of the rectangle enclosing in such a wa that both the lengths and the widths of the subrectangles approach ero. EMAK. Although this definition is satisfactor for our present purposes, there are various issues that would have to be resolved before it could be regarded as a rigorous mathematical definition. For eample, we would have to prove that the limit actuall eists and that its value does not depend on how the points (, ), (, ),...,( n, n ) are chosen. It can be proved that this is true if f is continuous on the region and this region is not too complicated. The details are beond the scope of this tet. It is assumed in Definition 5.. that f is nonnegative on the region.iff is continuous on and has both positive and negative values, then the limit lim n + k n f(k, k ) A k (3) no longer represents the volume between and the surface f(,); rather, it represents a difference of volumes the volume between and the portion of the surface that is above the -plane minus the volume between and the portion of the surface below the -plane. We call this the net signed volume between the region and the surface f(,). DEFINITION OF A DOUBLE INTEAL As in Definition 5.., the notation n + in (3) encapsulates a process in which the enclosing rectangle for is repeatedl subdivided in such a wa that both the lengths and the widths of the subrectangles approach ero. Note that subdividing so that the subrectangle lengths approach ero forces the mesh of the partition of the length of the enclosing rectangle for to approach ero. Similarl, subdividing so that the subrectangle widths approach ero forces the mesh of the partition of the width of the enclosing rectangle for to approach ero. Thus, we have etended the notion conveed b Formula () where the definite integral of a one-variable function is epressed as a limit of iemann sums. B etension, the sums in (3) are also called iemann sums, and the limit of the iemann sums is denoted b f(,)da lim n + k n f(k, k ) A k (4) which is called the double integral of f(,) over.

4 April, :4 g65-ch5 Sheet number 4 Page number 8 can magenta ellow black 8 Multiple Integrals If f is continuous and nonnegative on the region, then the volume formula in () can be epressed as V f(,)da (5) If f has both positive and negative values on, then a positive value for the double integral of f over means that there is more volume above than below, a negative value for the double integral means that there is more volume below than above, and a value of ero means that the volume above is the same as the volume below. POPETIES OF DOUBLE INTEALS f(, ) To distinguish between double integrals of functions of two variables and definite integrals of functions of one variable, we will refer to the latter as single integrals. Because double integrals, like single integrals, are defined as limits, the inherit man of the properties of limits. The following results, which we state without proof, are analogs of those in Theorem cf(,)da c [f(,) + g(,)] da f(,)da (c a constant) (6) f(,)da + g(,)da (7) [f(,) g(,)] da f(,)da g(,)da (8) The volume of the entire solid is the sum of the volumes of the solids above and. Figure 5..4 It is evident intuitivel that if f(,) is nonnegative on a region, then subdividing into two regions and has the effect of subdividing the solid between and f(,) into two solids, the sum of whose volumes is the volume of the entire solid (Figure 5..4). This suggests the following result, which holds even if f has negative values: f(,)da f(,)da + f(,)da (9) The proof of this result will be omitted. EVALUATIN DOUBLE INTEALS Ecept in the simplest cases, it is impractical to obtain the value of a double integral from the limit in (4). However, we will now show how to evaluate double integrals b calculating two successive single integrals. For the rest of this section, we will limit our discussion to the case where is a rectangle; in the net section we will consider double integrals over more complicated regions. The partial derivatives of a function f(,)are calculated b holding one of the variables fied and differentiating with respect to the other variable. Let us consider the reverse of this process, partial integration. The smbols b a f(,)d and d c f(,)d denote partial definite integrals; the first integral, called the partial definite integral with respect to, is evaluated b holding fied and integrating with respect to, and the second integral, called the partial definite integral with respect to, is evaluated b holding

5 April, :4 g65-ch5 Sheet number 5 Page number 9 can magenta ellow black 5. Double Integrals 9 fied and integrating with respect to. As the following eample shows, the partial definite integral with respect to is a function of, and the partial definite integral with respect to is a function of. Eample d d d ] ] d A partial definite integral with respect to is a function of and hence can be integrated with respect to ; similarl, a partial definite integral with respect to can be integrated with respect to. This two-stage integration process is called iterated (or repeated) integration. We introduce the following notation: d b d [ b ] f(,)d d f(,)d d () c a b d a c f(,)d d c b a a [ d These integrals are called iterated integrals. Eample (a) 3 Solution (a). 3 Solution (b). 3 Evaluate ( + 8) d d (b) ( + 8) d d ( + 8) d d c ] f(,)d d () 3 ( + 8) d d [ ] ( + 8) d d [ + 4 ] d [( + 6) ( + 4)] d ( + )d ( + 6 ) ] 3 57 [ 3 ] ( + 8) d d [ + 4 ] 3 d (3 + 36)d (3 + 8 ) ] 57 The following theorem shows that it is no accident that the two iterated integrals in the last eample have the same value.

6 April, :4 g65-ch5 Sheet number 6 Page number can magenta ellow black Multiple Integrals 5..3 THEOEM. Let be the rectangle defined b the inequalities a b, c d If f(,) is continuous on this rectangle, then d b f(,)da f(,)d d b d c a a c f(,)d d This important theorem allows us to evaluate a double integral over a rectangle b converting it to an iterated integral. This can be done in two was, both of which produce the value of the double integral. We will not formall prove this result; however, we will give a geometric motivation of the result for the case where f(,) is nonnegative on.in this case the double integral can be interpreted as the volume of the solid S bounded above b the surface f(,) and below b the region, soitsuffices to show that the two iterated integrals also represent this volume. For a fied value of, the function f(,) is a function of, and hence the integral A() b a f(,)d represents the area under the graph of this function of. This area, shown in ellow in Figure 5..5, is the cross-sectional area at of the solid S bounded above b f(,) and below b the region. Thus, b the method of slicing discussed in Section 6., the volume V of the solid S is d d [ b ] d b V A() d f(,)d d f(,)d d () c c a c a Similarl, the integral A() d c f(,)d represents the area of the cross section of S at (Figure 5..6), and the method of slicing again ields b b [ d ] b d V A() d f(,)d d f(,)d d (3) a a c This establishes the result in Theorem 5..3 for the case where f(,) is continuous and nonnegative on. a c f(, ) f(, ) a b Figure 5..5 c A() d b a Figure 5..6 c A() d Eample 3 Evaluate the double integral da over the rectangle {(, ) : 3, }.

7 April, :4 g65-ch5 Sheet number 7Page number can magenta ellow black 5. Double Integrals Solution. In view of Theorem 5..3, the value of the double integral ma be obtained from either of the iterated integrals 3 dd or 3 Using the first of these, we obtain da dd 3 d 3 6 ] dd (4) 3 [ ] 3 3 d You can check this result b evaluating the second integral in (4). 4 4 EMAK. We will often epress the rectangle {(, ) : a b, c d} as [a,b] [c, d] for simplicit. Eample 4 Use a double integral to find the volume of the solid that is bounded above b the plane 4 and below b the rectangle [, ] [, ] (Figure 5..7). Figure 5..7 (, ) Solution. V (4 )da ] [4 [ ] 7 5 d (4 )d d ( ) 7 d The volume can also be obtained b first integrating with respect to and then with respect to. FO THE EADE. Most computer algebra sstems have a built-in capabilit for computing iterated double integrals. If ou have a CAS, read the relevant documentation and use the CAS to check Eamples 3 and 4. EXECISE SET 5. C CAS In Eercises, evaluate the iterated integrals. d d d d ( + 3)dd. dd 4. 3 ( 4)d d ( + )dd 9.. ln π d d. ( + ) π/ e d d. 4 3 cos d d d d ( + ) 5. ln 3 ln e + d d 6. sin dd In Eercises 3 6, evaluate the double integral over the rectangular region.

8 April, :4 g65-ch5 Sheet number 8 Page number can magenta ellow black Multiple Integrals da; {(, ) :, } + + da; {(, ) :, } 5. da; {(, ) :, 3} 6. ( sin sin )da; {(, ) : π/, π/3} 7. (a) Let f(,) +, and as shown in the accompaning figure, let the rectangle [, ] [, ] be subdivided into 6 subrectangles. Take (k, k ) to be the center of the kth rectangle, and approimate the double integral of f over b the resulting iemann sum. (b) Compare the result in part (a) to the eact value of the integral. Figure E-7 8. (a) Let f(,), and as shown in Figure E-7, let the rectangle [, ] [, ] be subdivided into 6 subrectangles. Take (k, k ) to be the center of the kth rectangle, and approimate the double integral of f over b the resulting iemann sum. (b) Compare the result in part (a) to the eact value of the integral. In Eercises 9, use a double integral to find the volume. 9. The volume under the plane + and over the rectangle {(, ) :3 5, }.. The volume under the surface and over the rectangle {(, ) : 3, }.. The volume of the solid enclosed b the surface and the planes,, 3,, and.. The volume in the first octant bounded b the coordinate planes, the plane 4, and the plane (/3) + (/5). In Eercises 3 and 4, each iterated integral represents the volume of a solid. Make a sketch of the solid. (You do not have to find the volume.) 3. (a) 4. (a) 5 4 d d (b) ( )d d 3 4 (b) 5 d d ( + )dd 5. Evaluate the integral b choosing a convenient order of integration: cos() cos π da; [ ], [,π] 6. (a) Sketch the solid in the first octant that is enclosed b the planes,, 5,, and + 6. (b) Find the volume of the solid b breaking it into two parts. The average value or mean value of a continuous function f(,) over a rectangle [a,b] [c, d] isdefined as f ave f(,)da A() where A() (b a)(d c) is the area of the rectangle (compare to Definition 5.7.5). Use this definition in Eercises Find the average value of f(,) sin over the rectangle [, ] [, π/]. 8. Find the average value of f(,) ( + ) / over the interval [, ] [, 3]. 9. Suppose that the temperature in degrees Celsius at a point (, ) on a flat metal plate is T (,) 8, where and are in meters. Find the average temperature of the rectangular portion of the plate for which and. 3. Show that if f(,) is constant on the rectangle [a,b] [c, d], sa f(,) k, then f ave k over. C 3. C 3. Most computer algebra sstems have commands for approimating double integrals numericall. For Eercises 3 and 3, read the relevant documentation and use a CAS to find a numerical approimation of the double integral. sin d d e ( + ) d d 33. In this eercise, suppose that f(,) g()h() and {(, ) : a b, c d}. Show that [ b ][ d ] f(,)da g() d h() d 34. Use the result in Eercise 33 to evaluate the integral ln e + tan dd b inspection. Eplain our reasoning. a c

9 April, :4 g65-ch5 Sheet number 9 Page number 3 can magenta ellow black 5. Double Integrals Over Nonrectangular egions 3 C 35. Use a CAS to evaluate the iterated integrals d d and ( + ) 3 Does this violate Theorem 5..3? Eplain. d d ( + ) 3 C 36. Use a CAS to show that the volume V under the surface 3 sin over the rectangle shown in the accompaning figure is V 3/π. p Figure E DOUBLE INTEALS OVE NONECTANULA EIONS In this section we will show how to evaluate double integrals over regions other than rectangles. ITEATED INTEALS WITH NONCONSTANT LIMITS OF INTEATION Later in this section we will see that double integrals over nonrectangular regions can often be evaluated as iterated integrals of the following tpes: b g () b [ g () ] f(,)d d f(,)d d () a g () a g () d h () c h () f(,)d d d [ h () c h () ] f(,)d d () We begin with an eample that illustrates how to evaluate such integrals. DOUBLE INTEALS OVE NONECTANULA EIONS Eample (a) Solution (a). Solution (b). π cos Evaluate dd (b) dd sin dd [ ( 4 π cos ] d d π [ cos π sin dd ) [ 5 d [ ] ] ] sin d d cos sin d π d [ [ 6 cos3 sin ] π 3 ] cos Plane regions can be etremel comple, and the theor of double integrals over ver general regions is a topic for advanced courses in mathematics. We will limit our stud of double integrals to two basic tpes of regions, which we will call tpe I and tpe II; the are defined as follows: d

10 April, :4 g65-ch5 Sheet number Page number 4 can magenta ellow black 4 Multiple Integrals g () g () a b A tpe I region (a) d h () h () c A tpe II region (b) Figure DEFINITION. (a) A tpe I region is bounded on the left and right b vertical lines a and b and is bounded below and above b continuous curves g () and g (), where g () g () for a b (Figure 5..a). (b) A tpe II region is bounded below and above b horiontal lines c and d and is bounded on the left and right b continuous curves h () and h () satisfing h () h () for c d (Figure 5..b). The following theorem will enable us to evaluate double integrals over tpe I and tpe II regions using iterated integrals. 5.. THEOEM. (a) If is a tpe I region on which f(,) is continuous, then b g () f(,)da f(,)d d (3) a g () (b) If is a tpe II region on which f(,) is continuous, then d h () f(,)da f(,)d d (4) c h () We will not prove this theorem, but for the case where f(,) is nonnegative on the region, it can be made plausible b a geometric argument that is similar to that given for Theorem Since f(,) is nonnegative, the double integral can be interpreted as the volume of the solid S that is bounded above b the surface f(,) and below b the region, soitsuffices to show that the iterated integrals also represent this volume. Consider the iterated integral in (3), for eample. For a fied value of, the function f(,) is a function of, and hence the integral A() g () g () f(,)d represents the area under the graph of this function of between g () and g (). This area, shown in ellow in Figure 5.., is the cross-sectional area at of the solid S, and hence b the method of slicing, the volume V of the solid S is V b g () a g () f(,)d d which shows that in (3) the iterated integral is equal to the double integral. Similarl for (4). f(, ) Figure 5.. b a g () A() g () SETTIN UP LIMITS OF INTEATION FO EVALUATIN DOUBLE INTEALS To appl Theorem 5.., it is helpful to start with a two-dimensional sketch of the region. [It is not necessar to graph f(,).] For a tpe I region, the limits of integration in Formula (3) can be obtained as follows:

11 April, :4 g65-ch5 Sheet number Page number 5 can magenta ellow black 5. Double Integrals Over Nonrectangular egions 5 Step. Step. Since is held fied for the first integration, we draw a vertical line through the region at an arbitrar fied value (Figure 5..3). This line crosses the boundar of twice. The lower point of intersection is on the curve g () and the higher point is on the curve g (). These two intersections determine the lower and upper -limits of integration in Formula (3). Imagine moving the line drawn in Step first to the left and then to the right (Figure 5..3). The leftmost position where the line intersects the region is a and the rightmost position where the line intersects the region is b. This ields the limits for the -integration in Formula (3). g () g () a b Figure 5..3 Figure 5..4 h () 4 4 (4, ) h () Eample Evaluate da over the region enclosed between,,, and 4. Solution. We view as a tpe I region. The region and a vertical line corresponding to a fied are shown in Figure This line meets the region at the lower boundar and the upper boundar. These are the -limits of integration. Moving this line first left and then right ields the -limits of integration, and 4. Thus, 4 4 [ ] 4 ( ) da / [ d d ] 4 ( ) / d ( ) If is a tpe II region, then the limits of integration in Formula (4) can be obtained as follows: d d Step. Since is held fied for the first integration, we draw a horiontal line through the region at a fied value (Figure 5..5). This line crosses the boundar of twice. The leftmost point of intersection is on the curve h () and the rightmost point is on the curve h (). These intersections determine the -limits of integration in (4). c Figure 5..5 Step. Imagine moving the line drawn in Step first down and then up (Figure 5..5). The lowest position where the line intersects the region is c, and the highest position where the line intersects the region is d. This ields the -limits of integration in (4).

12 April, :4 g65-ch5 Sheet number Page number 6 can magenta ellow black 6 Multiple Integrals Eample 3 Evaluate ( )da over the triangular region enclosed between the lines +, +, and 3. Solution. We view as a tpe II region. The region and a horiontal line corresponding to a fied are shown in Figure This line meets the region at its left-hand boundar and its right-hand boundar. These are the -limits of integration. Moving this line first down and then up ields the -limits, and 3. Thus, 3 3 ( )da ( [ )dd ] d 3 3 [( + 3 ) ( + 3 )] d [ ( 3 3 ] 3 )d EMAK. To integrate over a tpe II region, the left- and right-hand boundaries must be epressed in the form h () and h (). This is wh we rewrote the boundar equations + and + as and in the last eample. In Eample 3 we could have treated as a tpe I region, but with an added complication: Viewed as a tpe I region, the upper boundar of is the line 3 (Figure 5..7) and the lower boundar consists of two parts, the line + to the left of the origin and the line + to the right of the origin. To carr out the integration it is necessar to decompose the region into two parts, and, as shown in Figure 5..7, and write ( )da ( )da+ ( )da 3 + ( )dd+ 3 + This will ield the same result that was obtained in Eample 3. ( )dd 4 (, 3) 3 3 (, 3) ( ) + ( ) Figure 5..8 Figure 5..6 Figure 5..7 Eample 4 Use a double integral to find the volume of the tetrahedron bounded b the coordinate planes and the plane 4 4. Solution. The tetrahedron in question is bounded above b the plane 4 4 (5) and below b the triangular region shown in Figure Thus, the volume is given b V (4 4 )da

13 April, :4 g65-ch5 Sheet number 3 Page number 7 can magenta ellow black 5. Double Integrals Over Nonrectangular egions 7 4 The region is bounded b the -ais, the -ais, and the line [set in (5)], so that treating as a tpe I region ields V (4 4 )da (4 4 )d d [ 4 4 ] d ( )d Eample 5 Find the volume of the solid bounded b the clinder + 4 and the planes + 4 and. - 4 Solution. The solid shown in Figure 5..9 is bounded above b the plane 4 and below b the region within the circle + 4. The volume is given b V (4 )da Treating as a tpe I region we obtain Figure V 4 4 (4 )d d [4 ] 4 d d 8(π) 6π See Formula (3) of Section 8.4. EVESIN THE ODE OF INTEATION (, ) (or ) Sometimes the evaluation of an iterated integral can be simplified b reversing the order of integration. The net eample illustrates how this is done. Eample 6 / Since there is no elementar antiderivative of e, the integral e d d cannot be evaluated b performing the -integration first. Evaluate this integral b epressing it as an equivalent iterated integral with the order of integration reversed. Figure 5.. Solution. For the inside integration, is fied and varies from the line / tothe line (Figure 5..). For the outside integration, varies from to, so the given iterated integral is equal to a double integral over the triangular region in Figure 5... To reverse the order of integration, we treat as a tpe I region, which enables us to write the given integral as [ e d d e da e d d e ] d / e d e ] e AEA CALCULATED AS A DOUBLE INTEAL Although double integrals arose in the contet of calculating volumes, the can also be used to calculate areas. To see wh this is so, recall that a right clinder is a solid that is generated when a plane region is translated along a line that is perpendicular to the region. In Formula () of Section 6. we stated that the volume V of a right clinder with cross-sectional area A and height h is V A h (6) Now suppose that we are interested in finding the area A of a region in the -plane. If we translate the region upward unit, then the resulting solid will be a right clinder that

14 April, :4 g65-ch5 Sheet number 4 Page number 8 can magenta ellow black 8 Multiple Integrals Clinder with base and height Figure 5.. has cross-sectional area A, base, and the plane as its top (Figure 5..). Thus, it follows from (6) that da (area of ) which we can rewrite as area of da da (7) EMAK. Formula (7) is sometimes confusing because it equates an area and a volume; the formula is intended to equate onl the numerical values of the area and volume and not the units, which must, of course, be different. Eample 7 Use a double integral to find the area of the region enclosed between the parabola and the line. Solution. The region ma be treated equall well as tpe I (Figure 5..a) or tpe II (Figure 5..b). Treating as tpe I ields 4 4 [ ] area of da d d / / d 4 ( ) ] 4 d [ Treating as tpe II ields 8 8 [ ] area of da d d / d 8 ( / ) d [ 3 3/ 4 ] (4, 8) 8 (4, 8) 4 4 (a) (b) Figure 5..

15 April, :4 g65-ch5 Sheet number 5 Page number 9 can magenta ellow black 5. Double Integrals Over Nonrectangular egions 9 EXECISE SET 5. raphing Utilit C CAS In Eercises, evaluate the iterated integral d d. 9 dd 4. π 3 π π π/ sin d d 6. cos d d 8. 3/ 3 /4 dd d d ( )d d e d d 9. d d. e / d d. In each part, find da over the shaded region. (a). In each part, find (a) (b) (, 3) (, ) (4, ) (5, 3) ( + )da over the shaded region. In Eercises 3 6, evaluate the double integral in two was using iterated integrals: (a) viewing as a tpe I region, and (b) viewing as a tpe II region. 3. (b) da; is the region bounded b 6/,, and da; is the region enclosed b,,, and. 5. (3 )da; is the region enclosed b the circle da; is the region in the first quadrant enclosed between the circle + 5 and the line + 5. In Eercises 7, evaluate the double integral. ( + ) / da; is the region in the first quadrant enclosed b, 4, and. 8. cos da; is the triangular region bounded b the lines,, and π. 9. da; is the region enclosed b, 6, and.. da; is the region enclosed b sin, /, and.. ( )da; is the region in the first quadrant enclosed between and 3.. da; is the region in the first quadrant enclosed b,, and. 3. (a) B hand or with the help of a graphing utilit, make a sketch of the region enclosed between the curves + and e. (b) Estimate the intersections of the curves in part (a). (c) Viewing as a tpe I region, estimate da. (d) Viewing as a tpe II region, estimate da. 4. (a) B hand or with the help of a graphing utilit, make a sketch of the region enclosed between the curves and (b) Find the intersections of the curves in part (a). (c) Find da. In Eercises 5 8, use double integration to find the area of the plane region enclosed b the given curves. 5. sin and cos, for π/4.

16 April, :4 g65-ch5 Sheet number 6 Page number 3 can magenta ellow black 3 Multiple Integrals 6. and and cosh, sinh,, and. In Eercises 9 and 3, use double integration to find the volume of the solid e π/ sin f(,)d d 4. f(,)d d 44. f(,)d d e ln f(,)d d f(,)d d f(,)d d In Eercises 3 38, use double integration to find the volume of each solid. 3. The solid bounded b the clinder + 9 and the planes and The solid in the first octant bounded above b the paraboloid + 3, below b the plane, and laterall b and. 33. The solid bounded above b the paraboloid 9 +, below b the plane, and laterall b the planes,, 3, and. 34. The solid enclosed b,, and The wedge cut from the clinder b the planes and The solid in the first octant bounded above b 9, below b, and laterall b The solid that is common to the clinders + 5 and The solid bounded above b the paraboloid +, bounded laterall b the circular clinder +( ), and bounded below b the -plane. C In Eercises 47 5, evaluate the integral b first reversing the order of integration e d d 48. / 3 ln cos( )dd 49. e 3 d d 5. dd 5. Evaluate sin( 3 )da, where is the region bounded b,, and. [Hint: Choose the order of integration carefull.] 5. Evaluate da, where is the region bounded b ln,, and e. 53. Tr to evaluate the integral with a CAS using the stated order of integration, and then b reversing the order of integration. (a) (b) 4 π/ sin π 3 d d sin sec (cos )d d 54. Use the appropriate Wallis formula (see Eercise Set 8.3) to find the volume of the solid enclosed between the circular paraboloid +, the right circular clinder + 4, and the -plane (see the accompaning figure for cut view). 55. Evaluate da over the region shown in the accompaning figure. In Eercises 39 and 4, use a double integral and a CAS to find the volume of the solid. C 39. The solid bounded above b the paraboloid and below b the -plane. C 4. The solid in the first octant that is bounded b the paraboloid +, the clinder + 4 and the coordinate planes. In Eercises 4 46, epress the integral as an equivalent integral with the order of integration reversed. Figure E-54 Figure E ive a geometric argument to show that d d π 6

17 April, :4 g65-ch5 Sheet number 7Page number 3 can magenta ellow black 5.3 Double Integrals in Polar Coordinates 3 The average value or mean value of a continuous function f(,) over a region in the -plane is defined as f ave f(,)da A() where A() is the area of the region (compare to the definition preceding Eercise 7 of Section 5.). Use this definition in Eercises 57 and Find the average value of /( + ) over the triangular region with vertices (, ), (, ), and (, ). 58. Find the average value of f(,) over the region enclosed b and 3. C 59. Suppose that the temperature in degrees Celsius at a point (, ) on a flat metal plate is T (,) 5 +, where and are in meters. Find the average temperature of the diamond-shaped portion of the plate for which + 4 and A circular lens of radius inches has thickness (r /4) inches at all points r inches from the center of the lens. Find the average thickness of the lens. 6. Use a CAS to approimate the intersections of the curves sin and /, and then approimate the volume of the solid in the first octant that is below the surface + + and above the region in the -plane that is enclosed b the curves. 5.3 DOUBLE INTEALS IN POLA COODINATES In this section we will stud double integrals in which the integrand and the region of integration are epressed in polar coordinates. Such integrals are important for two reasons: first, the arise naturall in man applications, and second, man double integrals in rectangular coordinates can be evaluated more easil if the are converted to polar coordinates. SIMPLE POLA EIONS Some double integrals are easier to evaluate if the region of integration is epressed in polar coordinates. This is usuall true if the region is bounded b a cardioid, a rose curve, a spiral, or, more generall, b an curve whose equation is simpler in polar coordinates than in rectangular coordinates. Moreover, double integrals whose integrands involve + also tend to be easier to evaluate in polar coordinates because this sum simplifies to r when the conversion formulas r cos θ and r sin θ are applied. Figure 5.3.a shows a region in a polar coordinate sstem that is enclosed between two ras, θ α and θ β, and two polar curves, r r (θ) and r r (θ). If, as shown in that figure, the functions r (θ) and r (θ) are continuous and their graphs do not cross, then the region is called a simple polar region. Ifr (θ) is identicall ero, then the boundar r r (θ) reduces to a point (the origin), and the region has the general shape shown in Figure 5.3.b. If, in addition, β α + π, then the ras coincide, and the region has the general shape shown in Figure 5.3.c. The following definition epresses these geometric ideas algebraicall DEFINITION. A simple polar region in a polar coordinate sstem is a region that is enclosed between two ras, θ α and θ β, and two continuous polar curves, r r (θ) and r r (θ), where the equations of the ras and the polar curves satisf the following conditions: (i) α β (ii) β α π (iii) r (θ) r (θ) EMAK. Conditions (i) and (ii) together impl that the ra θ β can be obtained b rotating the ra θ α counterclockwise through an angle that is at most π radians. This is consistent with Figure Condition (iii) implies that the boundar curves r r (θ) and r r (θ) can touch but cannot actuall cross over one another (wh?). Thus, in keeping with Figure 5.3., it is appropriate to describe r r (θ) as the inner boundar of the region and r r (θ) as the outer boundar.

18 April, :4 g65-ch5 Sheet number 8 Page number 3 can magenta ellow black 3 Multiple Integrals u b u b r r (u) r r (u) r r (u) r r (u) u a u a b a + p b a Simple polar regions Figure 5.3. (a) (b) (c) DOUBLE INTEALS IN POLA COODINATES f(r, u) Net, we will consider the polar version of Problem THE VOLUME POBLEM IN POLA COODINATES. iven a function f(r,θ) that is continuous and nonnegative on a simple polar region, find the volume of the solid that is enclosed between the region and the surface whose equation in clindrical coordinates is f(r,θ) (Figure 5.3.). r r (u) Figure 5.3. u a r r (u) u b To motivate a formula for the volume V of the solid in Figure 5.3., we will use a limit process similar to that used to obtain Formula () of Section 5., ecept that here we will use circular arcs and ras to subdivide the region into blocks, called polar rectangles. As shown in Figure 5.3.3, we will eclude from consideration all polar rectangles that contain an points outside of, leaving onl polar rectangles that are subsets of. Assume that there are n such polar rectangles, and denote the area of the kth polar rectangle b A k. Let (rk,θ k ) be an point in this polar rectangle. As shown in Figure 5.3.4, the product f(rk,θ k ) A k is the volume of a solid with base area A k and height f(rk,θ k ), so the sum n f(rk,θ k ) A k k can be viewed as an approimation to the volume V of the entire solid. (r k *, u k *) f(r, u) Area A k f(r k *, u k *) r r (u) u b u b r r (u) Figure u a Figure u a Area A k (r k *, u k *) If we now increase the number of subdivisions in such a wa that the dimensions of the polar rectangles approach ero, then it seems plausible that the errors in the approimations approach ero, and the eact volume of the solid is n V lim f(rk,θ k ) A k () n + k

19 April, :4 g65-ch5 Sheet number 9 Page number 33 can magenta ellow black 5.3 Double Integrals in Polar Coordinates 33 If f(r,θ) is continuous on and has both positive and negative values, then the limit lim n + k n f(rk,θ k ) A k () represents the net signed volume between the region and the surface f(r,θ)(as with double integrals in rectangular coordinates). The sums in () are called polar iemann sums, and the limit of the polar iemann sums is denoted b f(r,θ)da lim n + k n f(rk,θ k ) A k (3) which is called the polar double integral of f(r,θ) over. Iff(r,θ) is continuous and nonnegative on, then the volume formula () can be epressed as V f(r,θ)da (4) EMAK. Polar double integrals are also called double integrals in polar coordinates to distinguish them from double integrals over regions in the -plane, which are called double integrals in rectangular coordinates. Because double integrals in polar coordinates are defined as limits, the have the usual integral properties, such as those stated in Formulas (6), (7), and (8) of Section 5.. EVALUATIN POLA DOUBLE INTEALS r k * r k r k (r* k, u* k ) r k In Sections 5. and 5. we evaluated double integrals in rectangular coordinates b epressing them as iterated integrals. Polar double integrals are evaluated the same wa. To motivate the formula that epresses a double polar integral as an iterated integral, we will assume that f(r,θ) is nonnegative so that we can interpret (3) as a volume. However, the results that we will obtain will also be applicable if f has negative values. To begin, let us choose the arbitrar point (rk,θ k ) in (3) to be at the center of the kth polar rectangle as shown in Figure Suppose also that this polar rectangle has a central angle θ k and a radial thickness r k. Thus, the inner radius of this polar rectangle is rk r k and the outer radius is rk + r k. Treating the area A k of this polar rectangle as the difference in area of two sectors, we obtain Figure u b Figure u k r r (u) r r (u) u a u ( A k r k + r ) ( k θk r k r ) k θk which simplifies to A k rk r k θ k (5) Thus, from (3) and (4) n V f(r,θ)da lim f(rk,θ k )r k r k θ k n + k which suggests that the volume V can be epressed as the iterated integral β r (θ) V f(r,θ)da f(r,θ)rdrdθ (6) α r (θ) in which the limits of integration are chosen to cover the region ; that is, with θ fied between α and β, the value of r varies from r (θ) to r (θ) (Figure 5.3.6). Although we assumed f(r,θ)to be nonnegative in deriving Formula (6), it can be proved that the relationship between the polar double integral and the iterated integral in this formula

20 April, :4 g65-ch5 Sheet number Page number 34 can magenta ellow black 34 Multiple Integrals also holds if f has negative values. Accepting this to be so, we obtain the following theorem, which we state without formal proof THEOEM. If is a simple polar region whose boundaries are the ras θ α and θ β and the curves r r (θ) and r r (θ) shown in Figure 5.3.6, and if f(r,θ) is continuous on, then β r (θ) f(r,θ)da f(r,θ)rdrdθ (7) α r (θ) To appl this theorem, ou will need to be able to find the ras and the curves that form the boundar of the region, since these determine the limits of integration in the iterated integral. This can be done as follows: Step. Step. Since θ is held fied for the first integration, draw a radial line from the origin through the region at a fied angle θ (Figure 5.3.7a). This line crosses the boundar of at most twice. The innermost point of intersection is on the inner boundar curve r r (θ) and the outermost point is on the outer boundar curve r r (θ). These intersections determine the r-limits of integration in (7). Imagine rotating a ra along the polar -ais one revolution counterclockwise about the origin. The smallest angle at which this ra intersects the region is θ α and the largest angle is θ β (Figure 5.3.7b). This determines the θ-limits of integration. u b u b r r (u) u r r (u) u a u u a Figure (a) (b) r u 6 Figure r ( + cos u) u Eample Evaluate sin θda where is the region in the first quadrant that is outside the circle r and inside the cardioid r ( + cos θ). Solution. The region is sketched in Figure Following the two steps outlined above we obtain π/ (+cos θ) sin θda (sin θ)rdrdθ π/ r sin θ ] (+cos θ) r dθ

21 April, :4 g65-ch5 Sheet number Page number 35 can magenta ellow black 5.3 Double Integrals in Polar Coordinates 35 π/ [( + cos θ) sin θ sin θ] dθ [ 3 ( + cos θ)3 + cos θ [ 3 ( 5 )] ] π/ a r Eample The sphere of radius a centered at the origin is epressed in rectangular coordinates as + + a, and hence its equation in clindrical coordinates is r + a. Use this equation and a polar double integral to find the volume of the sphere. a r a Figure a a a Solution. In clindrical coordinates the upper hemisphere is given b the equation a r so the volume enclosed b the entire sphere is V a r da where is the circular region shown in Figure Thus, π a V a r da a r (r)dr dθ π [ 3 a3 θ [ 3 (a r ) 3/ ] a r dθ ] π π 3 a3 dθ 4 3 πa3 FINDIN AEAS USIN POLA DOUBLE INTEALS ecall from Formula (7) of Section 5. that the area of a region in the -plane can be epressed as area of da da (8) r sin 3u u 4 The argument used to derive this result can also be used to show that the formula applies to polar double integrals over regions in polar coordinates. Figure 5.3. u Eample 3 Use a polar double integral to find the area enclosed b the three-petaled rose r sin 3θ. Solution. The rose is sketched in Figure We will use Formula (8) to calculate the area of the petal in the first quadrant and multipl b three. π/3 sin 3θ A 3 da 3 rdrdθ π/3 [ θ sin 3θdθ 3 4 π/3 ( cos 6θ)dθ ] sin 6θ π/3 6 4 π

22 April, :4 g65-ch5 Sheet number Page number 36 can magenta ellow black 36 Multiple Integrals CONVETIN DOUBLE INTEALS FOM ECTANULA TO POLA COODINATES Sometimes a double integral that is difficult to evaluate in rectangular coordinates can be evaluated more easil in polar coordinates b making the substitution r cos θ, r sin θ and epressing the region of integration in polar form; that is, we rewrite the double integral in rectangular coordinates as f(,)da f(r cos θ,r sin θ)da f(r cos θ,r sin θ)rdrdθ (9) appropriate limits Eample 4 Use polar coordinates to evaluate ( + ) 3/ d d. Figure 5.3. Solution. In this problem we are starting with an iterated integral in rectangular coordinates rather than a double integral, so before we can make the conversion to polar coordinates we will have to identif the region of integration. To do this, we observe that for fied the -integration runs from to, which tells us that the lower boundar of the region is the -ais and the upper boundar is a semicircle of radius centered at the origin. From the -integration we see that varies from to, so we conclude that the region of integration is as shown in Figure In polar coordinates, this is the region swept out as r varies between and and θ varies between and π. Thus, ( + ) 3/ d d ( + ) 3/ da π (r 3 )r dr dθ π 5 dθ π 5 EMAK. The conversion to polar coordinates worked so nicel in this eample because the substitution r cos θ, r sin θ collapsed the sum + into the single term r, thereb simplifing the integrand. Whenever ou see an epression involving + in the integrand, ou should consider the possibilit of converting to polar coordinates. EXECISE SET 5.3 C CAS In Eercises 6, evaluate the iterated integral π/ sin θ π/ a sin θ π sin θ r cos θdrdθ. r dr dθ 4. r cos θdrdθ 6. π +cos θ π/6 cos 3θ π/ cos θ rdrdθ rdrdθ r 3 dr dθ In Eercises 7, use a double integral in polar coordinates to find the area of the region described.. The region inside the circle r 4 sin θ and outside the circle r.. The region inside the circle r and outside the cardioid r + cos θ. In Eercises 3 8, use a double integral in polar coordinates to find the volume of the solid that is described The region enclosed b the cardioid r cos θ. 8. The region enclosed b the rose r sin θ. 9. The region in the first quadrant bounded b r and r sin θ, with π/4 θ π/.. The region inside the circle + 4 and to the right of the line. Inside of Outside of + Below + Inside of + Above

23 April, :4 g65-ch5 Sheet number 3 Page number 37 can magenta ellow black 5.3 Double Integrals in Polar Coordinates Below Inside of + Above 6. Below ( + ) / Outside of + Inside of + 9 Above 7. The solid in the first octant bounded above b the surface r sin θ, below b the -plane, and laterall b the plane and the surface r 3 sin θ. 8. The solid inside of the surface r + 4 and outside of the surface r cos θ. In Eercises 9, use polar coordinates to evaluate the double integral. 9. e ( + ) da, where is the region enclosed b the circle da, where is the region in the first quadrant within the circle da, where is the sector in the first quadrant bounded b,, and da, where is the region in the first quadrant bounded above b the circle ( ) + and below b the line. In Eercises 3 3, evaluate the iterated integral b converting to polar coordinates d d + + d d 3. Use a double integral in polar coordinates to find the volume of a clinder of radius a and height h. 3. (a) Use a double integral in polar coordinates to find the volume of the oblate spheroid a + a + c ( <c<a) (b) Use the result in part (a) and the World eodetic Sstem of 984 (WS-84) discussed in Eercise 5 of Section.7 to find the volume of the Earth in cubic meters. 33. Use polar coordinates to find the volume of the solid that is inside of the ellipsoid a + a + c above the -plane, and inside of the clinder + a. 34. Find the area of the region enclosed b the lemniscate r a cos θ. 35. Find the area in the first quadrant that is inside of the circle r 4 sin θ and outside of the lemniscate r 8 cos θ. 36. Show that the shaded area in the accompaning figure is a φ a sin φ. a Figure E-36 r a sin u f ( + )dd 4 e ( +) d d a a + d d cos( + )dd d d ( + + ) 3/ (a > ) + d d The integral e d, which arises in probabilit theor, can be evaluated using the following method. Let the value of the integral be I. Thus, I + e d + e d since the letter used for the variable of integration in a definite integral does not matter. (a) ive a reasonable argument to show that I + + e ( + ) d d (b) Evaluate the iterated integral in part (a) b converting to polar coordinates. (c) Use the result in part (b) to show that I π/.

24 April, :4 g65-ch5 Sheet number 4 Page number 38 can magenta ellow black C 38 Multiple Integrals 38. (a) Use the numerical integration capabilit of a CAS to approimate the value of the double integral e ( + ) d d (b) Compare the approimation obtained in part (a) to the approimation that results if the integral is first converted to polar coordinates. 39. Suppose that a geser, centered at the origin of a polar coordinate sstem, spras water in a circular pattern in such a wa that the depth D of water that reaches a point at a distance of r feet from the origin in hour is D ke r. Find the total volume of water that the geser spras inside a circle of radius centered at the origin. 4. Evaluate da over the region shown in the accompaning figure. Figure E PAAMETIC SUFACES; SUFACE AEA In previous sections we considered parametric curves in -space and 3-space. In this section we will discuss parametric surfaces in 3-space. As we will see, parametric representations of surfaces are not onl important in computer graphics but also allow us to stud more general kinds of surfaces than those encountered so far. In Section 6.5 we showed how to find the surface area of a surface of revolution. Our work on parametric surfaces will enable us to derive area formulas for more general kinds of surfaces. PAAMETIC EPESENTATION OF SUFACES We have seen that curves in 3-space can be represented b three equations involving one parameter, sa (t), (t), (t) Surfaces in 3-space can be represented parametricall b three equations involving two parameters, sa (u, v), (u, v), (u, v) () To visualie wh such equations represent a surface, think of (u, v) as a point that varies over some region in a uv-plane. If u is held constant, then v is the onl varing parameter in (), and hence these equations represent a curve in 3-space. We call this a constant u-curve (Figure 5.4.). Similarl, if v is held constant, then u is the onl varing parameter in (), so again these equations represent a curve in 3-space. We call this a constant v-curve. B varing the constants we generate a famil of u-curves and a famil of v-curves that together form a surface. v u constant Constant u-curve v constant u Constant v-curve Figure 5.4.

25 April, :4 g65-ch5 Sheet number 5 Page number 39 can magenta ellow black 5.4 Parametric Surfaces; Surface Area 39 Eample Consider the paraboloid 4. One wa to parametrie this surface is to take u and v as the parameters, in which case the surface is represented b the parametric equations u, v, 4 u v () Figure 5.4.a shows a computer-generated graph of this surface. The constant u-curves correspond to constant -values and hence appear on the surface as traces parallel to the -plane. Similarl, the constant v-curves correspond to constant -values and hence appear on the surface as traces parallel to the -plane. u + v 4 r u o r 3 u o r u o (a) (b) (c) (d) Figure 5.4. Eample The paraboloid 4 that was considered in Eample can also be parametried b first epressing the equation in clindrical coordinates. For this purpose, we make the substitution r cos θ, r sin θ, which ields 4 r. Thus, the paraboloid can be represented parametricall in terms of r and θ as r cos θ, r sin θ, 4 r (3) Figure 5.4.b shows a computer-generated graph of this surface for r and θ π. The constant r-curves correspond to constant -values and hence appear on the surface as traces parallel to the -plane. The constant θ-curves appear on the surface as traces from vertical planes through the origin at varing angles with the -ais. Parts (c) and (d) of Figure 5.4. show the effect of restrictions on the parameters r and θ. FO THE EADE. If ou have a graphing utilit that can generate parametric surfaces, read the relevant documentation and then tr to make reasonable duplicates of the surfaces in Figure Eample 3 One wa to generate the sphere + + with a graphing utilit is to graph the upper and lower hemispheres and on the same screen. However, this usuall produces a fragmented sphere (Figure 5.4.3a) because roundoff error sporadicall produces negative values inside the radical when is near ero. A better graph can be generated b first epressing the sphere in spherical coordinates as ρ and then using the spherical-to-rectangular conversion formulas in Table.8. to obtain the parametric equations sin φ cos θ, sin φ sin θ, cos φ with parameters θ and φ. Figure 5.4.3b shows the graph of this parametric surface for