16 MULTIPLE INTEGRALS ET 15

6 MULTIPLE INTEGRALS ET 5 6. Double Integrals over Rectangles ET 5.. (a) The subrectangles are shown in the figure. The surface is the graph of f(x, y) xy and A, so we estimate V f(x i,y j ) A (b) V i j f(, ) A + f(, ) A + f(, ) A + f(, ) A + f(6, ) A + f(6, ) A () + 8() + 8() + 6() + () + () 88 i j f x i, y j A f(, ) A + f(, ) A + f(, ) A + f(, ) A + f(5, ) A + f(5, ) A ()+()+()+9()+5()+5(). (a) The subrectangles are shown in the figure. Since A π /, we estimate R sin(x + y) da f x ij,yij A i j (b) R sin(x + y) da f(, ) A + f, π A + f π, A + f π, π A π π + π + π + π.95 i j f(x i, y j ) A f π, π A + f π, π π + π + A + f π, π π +( ) π A + f π, π A 5. (a) Each subrectangle and its midpoint are shown in the figure. The area of each subrectangle is A,soweevaluatef at each midpoint and estimate R f(x, y) da f x i, y j A i j f(.5, ) A + f(.5, ) A + f(.5, ) A + f(.5, ) A ()+( 8)() + 5() + ( )() 6

CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5 (b) The subrectangles are shown in the figure. In each subrectangle, the sample point farthest from the origin is the upper right corner, and the area of each subrectangle is A.Thusweestimate R f(x, y) da i j f(x i,y j) A f(.5, ) A + f(.5, ) A + f(.5, ) A + f(.5, ) A + f(, ) A + f(, ) A + f(, ) A + f(, ) A + f(.5, ) A + f(.5, ) A + f(.5, ) A + f(.5, ) A + f(, ) A + f(, ) A + f(, ) A + f(, ) A +( ) +( 8) +( 6) + + +( 5) +( 8) +5 + +( ) +( ) +8 +6 + +.5 7. The values of f(x, y) 5 x y get smaller as we move farther from the origin, so on any of the subrectangles in the problem, the function will have its largest value at the lower left corner of the subrectangle and its smallest value at the upper right corner, and any other value will lie between these two. So using these subrectangles we have U<V<L. (Note that this is true no matter how R is divided into subrectangles.) 9. (a) With m n,wehave A. Using the contour map to estimate the value of f at the center of each subrectangle, we have R f(x, y) da f x i, y j A A[f(, ) + f(, ) + f(, ) + f(, )] (7 + + + 7) 8 i j (b) f ave f(x, y) da (8) 5.5 A(R) R 6. z >, so we can interpret the integral as the volume of the solid S that lies below the plane z and above the rectangle [, ] [, 6]. S is a rectangular solid, thus da 5 6. R. z f(x, y) y for y. Thus the integral represents the volume of that part of the rectangular solid [, ] [, ] [, ] which lies below the plane z y. So ( y) da ()()() + ()()() R

SECTION 6. ITERATED INTEGRALS ET SECTION 5. 5 5. To calculate the estimates using a programmable calculator, we can use an algorithm similartothatofexercise5..7[et 5..7]. In Maple, we can define the function f(x, y) +xe y (calling it f), load the student package, and then use the command middlesum(middlesum(f,x..,m), y..,m); to get the estimate with n m squares of equal size. Mathematica has no special Riemann sum command, but we can define f andthenusenestedsum commands to calculate the estimates. n estimate.66.9 6.55 6.67 56.67.6 7. If we divide R into mn subrectangles, R kda m But f m x ij,yij k always and i points, m n i j R kda lim i j n f x ij,yij A for any choice of sample points x ij,yij. n A area of R (b a)(d c). Thus, no matter how we choose the sample j f m x ij,yij A k i m n m,n i j n A k(b a)(d c) and so j f x ij,yij A lim k m m,n i n A j lim m,n k(b a)(d c) k(b a)(d c). 6. Iterated Integrals ET 5... 5. 7. 9. x5 5 x y dx x y x y x5 x (5) y () y 5y, x x y dy x y y x y y y x () x () x y ( + xy) dx dy x +x y x dy ( + y) dy y + y (+9) ( + ) x π/ x sin ydydx xdx π/ sin ydy [as in Example 5] (x + (x + y) 9 y)8 dx dy 9 8 x y + y dy dx x x x [( + y) 9 ( + y) 9 ] dy 8 x π/ cos y ( )( + ) dy [substitute u x + y dx du] ( + y) 8 [( ) ( )],6,58 8 6,6 5 x ln y + x y y y dx y 8ln+ ln ln 5 ln + ln / ln x ln + dx x x ln + ln x

6 CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5.. (u v)5 du dv (u v)6 u dv 6 u 6 ( v) 6 ( v) 6 dv 6 ( v) 6 v 6 dv 6 ( 7 v)7 v7 7 [( + ) ( + )] π r sin θdθdr rdr π sin θdθ [as in Example 5] rdr π r θ sin θ π [(π ) ( )] π ( cos θ) dθ ( ) π sin π sin 5. 7. 9.. R (6x y 5y ) da R xy x + da (6x y 5y ) dy dx x y y 5 y dx y x dx x x 7 π/6 π/ x sin(x + y) dy dx π/6 xy dy dx x + (ln ln ) (7 + 7) 9 ln x cos(x + y) y π/ x sin x sin x + π y π/6 π/6 π 6 cos x +cos x + π R xyexy da xyexy dx dy x x + dx y dy ln(x +) y dx π/6 x cos x x cos x + π dx sin x sin x + π π/6 [(e ) ( )] (e ) dx [by integrating by parts separately for each term] π + + π x ex y dy (ey ) dy e y y x. z f(x, y) x y for x and y. Sothesolid is the region in the first octant which lies below the plane z x y and above [, ] [, ]. 5. V ( x y) da ( x y) dx dy R x x xy x y dy y y 95 7. V x y dx dy 9 x y dx dy 9 x x 9 y x x dy x x dy y dy y y 8 66 9 7 5 7

SECTION 6. DOUBLE INTEGRALS OVER GENERAL REGIONS ET SECTION 5. 7 9. Here we need the volume of the solid lying under the surface z x sec y and above the rectangle R [, ] [,π/] in the xy-plane. V π/ x sec ydydx xdx π/ sec ydy x π/ tan y ( )(tan π tan ) ( ). The solid lies below the surface z +x +(y ) and above the plane z for x, y. Thevolume of the solid is the difference in volumes between the solid that lies under z +x +(y ) over the rectangle R [, ] [, ] andthesolidthatliesunderz over R. V [ + x +(y ) ] dx dy () dx dy ( + +(y ) ) ( (y ) ) dy [x] [y] +(y ) dy [ ( )][ ] y + (y ) 56 + 6 6 8 88 6 8. In Maple, we can calculate the integral by defining the integrand as f and then using the command int(int(f,x..),y..);. In Mathematica, we can use the command Integrate[f,{x,,},{y,,}] We find that R x5 y e xy da e 57.89. Wecanuseplotd (in Maple) or PlotD (in Mathematica) to graph the function. 5. R is the rectangle [, ] [, 5]. Thus, A(R) 5and x + x + x(y ) x x dy dx dy ()() f ave A(R) f(x, y) da 5 R x ydxdy 5 x y x dy 5 x ydy y 5 5. 6 7. Let f(x, y) x y (x + y).thenacasgives f(x, y) dy dx and f(x, y) dx dy. To explain the seeming violation of Fubini s Theorem, note that f has an infinite discontinuity at (, ) and thus does not satisfy the conditions of Fubini s Theorem. In fact, both iterated integrals involve improper integrals which diverge at their lower limits of integration. 6. Double Integrals over General Regions ET 5... 5. y xy dx dy x y x y dy x y [( y ) ]dy y dy y (6 ) x ( + y)dy dx x y + y yx dx yx x + x x (x ) dx π/ cos θ e sin θ dr dθ π/ (x x )dx x 5 x5 5 + re sin θ rcos θ r dθ π/ (cos θ) e sin θ dθ e sin θ π/ e sin(π/) e e

8 CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5 7. 9.. D y da y y y dx dy xy xy dy x y y [y ( y )] dy (y +y )dy y + y + + D xda π sin x xdydx π [xy]ysin x y dx π x sin xdx x cos x +sinx π π cos π +sinπ + sin π D y e xy da y y e xy dx dy ye xy xy dy ye y y dy x ey y e6 8 + e6 7 integrate by parts with u x, dv sinxdx. x x cos ydydx y x x sin y dx x sin y x dx cos x ( cos ) 5. y y y dx dy xy xy dy x y (y y ) dy 5 y5 y 96 5 5 + 7 [(y ) ( y)] y dy 7. x (x y) dy dx x xy y y x x dx y x x x +x x + x dx x x dx x / [Or, note that x x is an odd function, so x x dx.] 9. V x (x +y) dy dx x xy + y yx dx yx (x x 5 x 8 ) dx x 6 x6 9 x9 6 9 7 8. V 7 y xy dx dy x y x 7 y dy x (8y y +9y ) dy y y + 9 y 8

. V x (6 x y) dy dx SECTION 6. DOUBLE INTEGRALS OVER GENERAL REGIONS ET SECTION 5. 9 6y xy y y x dx 6( y x) x( x) ( x) dx 9 x 9x +9 dx x 9 x +9x 6 6 5. V x x dy dx x y y yx dx (x x ) dx x x5 + 8 5 5 5 5 7. x y V ydydx x dx x x y x y dx 9. From the graph, it appears that the two curves intersect at x and at x.. Thus the desired integral is D xda. x x xdydx. x xy y x x y x. (x x x 5 ) dx x x x6. 6.7. The two bounding curves y x and y x intersect at (±, ) with x x on [, ]. Withinthis region, the plane z x +y +is above the plane z x y,so V x (x +y +)dy dx x x x x x (x +y + ( x y)) dy dx ( x y) dy dx x (x +y +8)dy dx y x xy + x y +8y dx yx x( x )+ ( x ) +8( x ) x(x ) (x ) 8(x ) dx ( 6x 6x +6x +6)dx x 6 x +x +6x 6 ++6+ 6 +6 6 dx

CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5. The solid lies below the plane z x y or x + y + z and above the region D {(x, y) x, y x} in the xy-plane. The solid is a tetrahedron. 5. The two bounding curves y x x and y x + x intersect at the origin and at x,withx + x>x x on (, ). Using a CAS, we find that the volume is V x + x x x zdydx x + x x x (x y + xy ) dy dx,98,75,66,59,55 7. The two surfaces intersect in the circle x + y, z and the region of integration is the disk D: x + y. Using a CAS, the volume is ( x y x ) da ( x D x y ) dy dx π. 9. Because the region of integration is D {(x, y) y x, x } (x, y) y x, y we have x f(x, y) dy dx f(x, y) da f(x, y) dx dy. D y. Because the region of integration is D (x, y) 9 y x 9 y, y (x, y) y 9 x, x f(x, y) dx dy we have 9 y f(x, y) da 9 y D 9 x f(x, y) dy dx. Because the region of integration is D {(x, y) y ln x, x } {(x, y) e y x, y ln } 5. we have ln x ln f(x, y) dy dx f(x, y) da D f(x, y) dx dy e y x/ yx/ e x dx dy e x dy dx e x y dx y y x e x dx 6 ex e9 6

SECTION 6. DOUBLE INTEGRALS OVER GENERAL REGIONS ET SECTION 5. 7. x y + dy dx y dx dy y + xy x dy y + x y y + dy ln y + (ln 9 ln ) ln 9 9. π/ arcsin y π/ cos x +cos xdxdy sin x cos x +cos xdydx π/ cos x +cos x y ysin x y π/ cos x +cos x sin xdx dx u +u du +u / 8 Let u cosx, du sin xdx, dx du/( sin x) 5. D {(x, y) x, x + y } {(x, y) x, x + y } D {(x, y) x, y x } {(x, y) x, y x }, all type I. x da x x x dy dx + x dy dx + x + x x dy dx + x x dy dx [by symmetry of the regions and because f(x, y) x ] x dy dx x dx 5. Here Q (x, y) x + y,x,y,and (x + y ) 6 (x + y ) so e /6 e (x +y ) e since e t is an increasing function. We have A(Q) π π, so by Property, 6 e /6 A(Q) Q e (x +y ) da A(Q) π 6 e /6 Q e (x +y ) da π or we can say 6.8 < Q e (x +y ) da <.96. (We have rounded the lower bound down and the upper bound up to preserve the inequalities.) 55. The average value of a function f of two variables defined on a rectangle R was defined in Section 6. [ET 5.] as f ave f(x, y)da. Extending A(R) R this definition to general regions D,wehavef ave f(x, y)da. A(D) Here D {(x, y) x, y x},soa(d) ()() and f ave x f(x, y)da xy dy dx A(D) D / xy yx dx y 9x dx x 57. Since m f(x, y) M, mda f(x, y) da MdAby (8) D D D m da f(x, y) da M da by (7) ma(d) f(x, y) da MA(D) by (). D D D D D

CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5 59. D (x tan x + y +)da D x tan xda+ D y da + da. D Butx tan x is an odd function of x and D is symmetric with respect to the y-axis, so D x tan xda. Similarly, y isanoddfunctionofy and D is symmetric with respect to the x-axis, so D y da.thus D (x tan x + y +)da da (areaofd) π D 8π 6. Since x y, we can interpret D x y da as the volume of the solid that lies below the graph of z x y and above the region D in the xy-plane. z x y is equivalent to x + y + z, z which meets the xy-plane in the circle x + y, the boundary of D. Thus, the solid is an upper hemisphere of radius which has volume π () π. 6. Double Integrals in Polar Coordinates ET 5.. The region R is more easily described by polar coordinates: R (r, θ) r, θ π. Thus f(x, y) da π/ f(r cos θ, r sin θ) rdrdθ. R. The region R is more easily described by rectangular coordinates: R (x, y) x, y x +. Thus f(x, y) da (x+)/ f(x, y) dy dx. R 5. The integral π 7 π rdrdθrepresents the area of the region R {(r, θ) r 7, π θ π},thelowerhalfofaring. π 7 rdrdθ π 7 dθ rdr π π θ π r 7 π π (9 6) π 7. The disk D can be described in polar coordinates as D {(r, θ) r, θ π}. Then D xy da π (r cos θ)(r sin θ) rdrdθ π sin θ cos θdθ r dr sin θ π r. 9. R cos(x + y ) da π cos(r ) rdrdθ π dθ r cos(r ) dr θ π sin(r ) π (sin 9 sin ) π sin 9. D e x y da π/ π/ π/ e r rdrdθ dθ π/ re r dr θ π/ π/ e r π (e e ) π ( e ). R is the region shown in the figure, and can be described by R {(r, θ) θ π/, r }. Thus R arctan(y/x) da π/ arctan(tan θ) rdrdθsince y/x tanθ. Also, arctan(tan θ) θ for θ π/, so the integral becomes π/ θrdrdθ π/ θdθ rdr θ π/ r π 6 π.

SECTION 6. DOUBLE INTEGRALS IN POLAR COORDINATES ET SECTION 5. 5. One loop is given by the region D {(r, θ) π/6 θ π/6, r cos θ },sotheareais π/6 cos θ π/6 rcos θ da rdrdθ D π/6 π/6 r dθ π/6 π/6 cos θdθ θ + 6 π/6 sin 6θ π/6 r +cos6θ dθ 7. By symmetry, π A π/ sin θ rdrdθ π/ r rsin θ dθ r π/ sin θdθ π/ ( cos θ) dθ θ sin θ π/ π sin π + sin (π ) 8 9. V x + y x + y da π r rdrdθ π dθ r dr θ π r π 8 6 π. The hyperboloid of two sheets x y + z intersects the plane z when x y +or x + y.sothe solid region lies above the surface z +x + y and below the plane z for x + y, and its volume is. By symmetry, V V x + y a θ π x + y +x + y da π ( +r ) rdrdθ π dθ (r r +r )dr θ π r ( + r ) / π 8 + π a x y da π a a (a r ) / (π) + a π a π a r rdrdθ dθ a r a r dr 5. The cone z x + y intersects the sphere x + y + z when x + y + x + y or x + y.so V x + y / dθ / π x y x + y da π / r r rdrdθ r r r dr θ π / ( r ) / r π π

CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5 7. The given solid is the region inside the cylinder x + y between the surfaces z 6 x y and z 6 x y.so V 6 x y 6 x y da 6 x y da x + y π x +y 6 r rdrdθ π dθ r 6 r dr θ π 8π ( / 6 / ) 8π 6 (6 r ) / 9. 9 x sin(x + y )dy dx π sin r rdrdθ π dθ r sin r dr [θ] π cos r π (cos 9 ) π ( cos 9). π/ (r cos θ + r sin θ) rdrdθ π/ (cos θ +sinθ) dθ r dr [sinθ cos θ] π/ r +. The surface of the water in the pool is a circular disk D with radius ft. If we place D on coordinate axes with the origin at 5. the center of D and define f(x, y) to be the depth of the water at (x, y), then the volume of water in the pool is the volume of the solid that lies above D (x, y) x + y and below the graph of f(x, y). We can associate north with the positive y-direction, so we are given that the depth is constant in the x-direction and the depth increases linearly in the y-direction from f(, ) to f(, ) 7. The trace in the yz-plane is a line segment from (,, ) to (,, 7). The slope of this line is 7, so an equation of the line is z 7 (y ) z y + 9 ( ) 8 8 8.Sincef(x, y) is independent of x, f(x, y) y + 9. Thus the volume is given by 8 f(x, y) da, which is most conveniently evaluated D using polar coordinates. Then D {(r, θ) r, θ π} and substituting x r cos θ, y r sin θ the integral becomes π r sin θ + 9 π 8 rdrdθ r sin θ + 9 r r r Thus the pool contains 8π 5655 ft of water. / x x xy dy dx + π/ 5 π/ x r cos θ sin θdrdθ cos θ + 9θ π 8π xy dy dx + π/ sin θ cos θdθ 5 sin θ x r cos θ sin θ π/ 5 6 xy dy dx r r dθ dθ π sin θ +9 dθ

SECTION 6.5 APPLICATIONS OF DOUBLE INTEGRALS ET SECTION 5.5 5 7. (a) We integrate by parts with u x and dv xe x dx. Thendu dx and v e x,so x e x t dx lim t x e x dx lim xe x t + t t e x dx lim te t + t e x dx + e x dx [by l Hospital s Rule] e x dx [since e x is an even function] π [by Exercise 6(c)] (b) Let u x.thenu x dx udu xe x t dx lim t xe x t dx lim ue u udu u e u du t π [by part(a)] π. 6.5 Applications of Double Integrals ET 5.5. Q D σ(x, y) da (xy + y ) dy dx xy + y y dx y x + 8 dx x + 8 x 6+6 6 C. m ρ(x, y) da D xy dy dx xdx y dy x y, x xρ(x, y) da m D x y dy dx x dx y dy x y 8, y yρ(x, y) da m D xy dy dx xdx y dy x y. Hence, (x, y),. 5. m x (x + y) dy dx x/ xy + y y x dx yx/ x x + ( x) x dx 8 9 8 x + 9 dx 9 x + 9 x 6, 8 M y x x/ (x + xy) dy dx x y + xy y x dx 9 x 9 x dx 9, yx/ 8 M x y (xy + x/ y ) dy dx xy + y y x dx yx/ 9 9 x dx 9. My Hence m 6, (x, y) m, M x m,. 7. m e x ydydx y ye x dx y ex dx ex (e ), M y e x xy dy dx xex dx xex ex 8 (e +), M x e x y dy dx y ye x dx y ex dx ex 9 (e ). Hence m (e 8 ), (x, y) (e +) (e ), 9 (e ) (e ) e + (e ), (e ). 9(e )

6 CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5 9. Note that sin(πx/l) for x L. m L sin(πx/l) ydydx L sin (πx/l) dx M y L sin(πx/l) x ydydx L x L sin(πx/l) L L, π x sin (πx/l) dx integrate by parts with u x, dv sin (πx/l) dx x x L sin(πx/l) L L x L sin(πx/l) dx π π L L x + L cos(πx/l) π L L + L L π π 8 L, M x L sin(πx/l) y ydydx L sin (πx/l) dx L cos (πx/l) sin(πx/l) dx [substitute u cos(πx/l)] du π sin(πx/l)] L L π cos(πx/l) cos (πx/l) L L π + + L. 9π Hence m L L, (x, y) /8 L/, L/(9π) L L/, 6. 9π. ρ(x, y) ky kr sin θ, m π/ kr sin θdrdθ k π/ sin θdθ k cos θ π/ k, M y π/ kr sin θ cos θdrdθ k π/ sin θ cos θdθ k 8 cos θ π/ k, 8 M x π/ kr sin θdrdθ k π/ sin θdθ k θ +sinθ π/ π k. 8 6 Hence (x, y) 8, π 6.. ρ(x, y) k x + y kr, m ρ(x, y)da π kr rdrdθ D k π dθ r dr k(π) r 7 πk, M y xρ(x, y)da π (r cos θ)(kr) rdrdθ k π cos θdθ D r dr k sin θ π r k() 5 [this is to be expected as the region and density function are symmetric about the y-axis] Hence (x, y) M x yρ(x, y)da π (r sin θ)(kr) rdrdθ k π sin θdθ D k cos θ π r k( + ) 5 5 k., 5k/ 7πk/, 5 π. 5. Placing the vertex opposite the hypotenuse at (, ), ρ(x, y) k(x + y ).Then m a a x By symmetry, k x + y dy dx k a Hence (x, y) 5 a, 5 a. r dr ax x + (a x) dx k ax x (a x) a 6 ka. M y M x a a x ky(x + y ) dy dx k a (a x) x + (a x) dx k 6 a x ax + x5 (a x)5 a 5 ka5

SECTION 6.5 APPLICATIONS OF DOUBLE INTEGRALS ET SECTION 5.5 7 7. I x D y ρ(x, y)da e x y ydydx y ye x dx y ex dx ex 6 (e ), I y D x ρ(x, y) da e x x ydydx x y ye x dx y x e x dx x x + e x [integrate by parts twice] 8 (e ), and I I x + I y 6 (e ) + 8 (e ) 6 (e +e ). 9. As in Exercise 5, we place the vertex opposite the hypotenuse at (, ) and the equal sides along the positive axes. I x a a x y k(x + y ) dy dx k a a x (x y + y ) dy dx k a x y + y5 ya x 5 dx y k a x (a x) + (a x)5 dx k 5 a x a x + 5 ax5 x6 (a x)6 a 7 6 8 ka6, I y a a x k a x k(x + y ) dy dx k a a x (x + x y ) dy dx k a x y + x y ya x dx y x (a x)+ x (a x) dx k 5 ax5 6 x6 + and I I x + I y 7 9 ka6. a x a x + 5 ax5 6 x6 a 7 8 ka6,. Using a CAS, we find m ρ(x, y) da π sin x xy dy dx π D 8.Then x xρ(x, y) da 8 π sin x x ydydx π m D π π and y yρ(x, y) da 8 π sin x xy dy dx 6 π,so(x, y) m π 9π π, 6. 9π D The moments of inertia are I x D y ρ(x, y) da π sin x xy dy dx π 6, I y D x ρ(x, y) da π sin x x ydydx π 6 (π ),andi I x + I y π 6 (π 9).. I x D y ρ(x, y)da h b ρy dx dy ρ b dx h y dy ρ x b y h ρb h ρbh, I y D x ρ(x, y)da h b ρx dx dy ρ b x dx h dy ρ x b [y]h ρb h, and m ρ (area of rectangle) ρbh since the lamina is homogeneous. Hence x Iy m ρb h ρbh b x b and y Ix m ρbh ρbh h y h. 5. In polar coordinates, the region is D (r, θ) r a, θ π,so I x D y ρda π/ a ρ(r sin θ) rdrdθ ρ π/ sin dθ a ρ θ sin θ π/ r a ρ π a 6 ρa π, r dr I y D x ρda π/ a ρ(r cos θ) rdrdθ ρ π/ cos dθ a ρ θ + sin θ π/ r a ρ π a 6 ρa π, r dr and m ρ A(D) ρ πa since the lamina is homogeneous. Hence x y 6 ρa π ρa π a x y a.

8 CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5 7. (a) f(x, y) is a joint density function, so we know R f(x, y) da.sincef(x, y) outside the rectangle [, ] [, ], we can say Then C C. (b) P (X,Y ) R f(x, y) da f(x, y) dy dx Cx( + y) dy dx C x y + y y y dx C xdx C x C f(x, y) dy dx x y + y y dx x y x( + y) dy dx dx x or.75 8 (c) P (X + Y ) P ((X, Y ) D) where D is the triangular region shown in the figure. Thus P (X + Y ) f(x, y) da D x x( + y) dy dx x y + y y x dx x y x x + dx x x +x dx x x +x 5 8. 9. (a) f(x, y),sof is a joint density function if R f(x, y) da. Here, f(x, y) outside the first quadrant, so R f(x, y) da.e (.5x +.y) dy dx.. lim t t e.5x dx lim t t e.y dy. lim t e.5x t lim t e.5x e.y dy dx. e.5x dx e.y dy 5e.y t. lim t (e.5t ) lim t 5(e.t ) (.) ( )( ) ( 5)( ) Thus f(x, y) is a joint density function. (b) (i) No restriction is placed on X,so P (Y ) (ii) P (X,Y ) f(x, y) dy dx. e.5x dx e.y dy. lim. lim t e.5x t lim t (c) The expected value of X is given by.e (.5x+.y) dy dx t t t e.5x dx lim t e.y dy 5e.y t. lim t (.) ( )( ) ( 5)( e. )e..887 f(x, y) dy dx.e (.5x+.y) dy dx. e.5x dx e.y dy. e.5x (.) ( )(e ) ( 5)(e.8 ) (e.5t ) lim t 5(e.t e. ) 5e.y (e )(e.8 ) + e.8 e.8 e.8 μ R xf(x, y) da x.e (.5x+.y) dy dx. xe.5x dx e.y dy. lim t t xe.5x dx lim t t e.y dy To evaluate the first integral, we integrate by parts with u x and dv e.5x dx (or we can use Formula 96

SECTION 6.5 APPLICATIONS OF DOUBLE INTEGRALS ET SECTION 5.5 9 in the Table of Integrals): xe.5x dx xe.5x e.5x dx xe.5x e.5x (x +)e.5x. Thus μ. lim (x +)e.5x t lim t 5e.y t t. lim( ) (t +)e.5t lim ( 5) e.t t t.( ) lim t t + e.5t ( 5)( ) [by l Hospital s Rule] The expected value of Y is given by μ yf(x, y) da y.e (.5+.y) dy dx R. e.5x dx ye.y dy. lim t t e.5x dx lim t t ye.y dy To evaluate the second integral, we integrate by parts with u y and dv e.y dy (oragainwecanuseformula96in the Table of Integrals) which gives ye.y dy 5ye.y + 5e.y dy 5(y +5)e.y.Then μ. lim e.5x t lim t 5(y +5)e.y t t. lim (e.5t ) lim 5 (t +5)e.t 5 t t t +5.( )( ) ( 5) lim t e 5 5 [by l Hospital s Rule].t. (a) The random variables X and Y are normally distributed with μ 5, μ, σ.5,andσ.. The individual density functions for X and Y, then, are f (x).5 /.5 π e (x 5) and f (y). /. π e (y ).SinceX and Y are independent, the joint density function is the product f(x, y) f (x)f (y) Then P ( X 5, Y 5) 5.5 /.5 π e (x 5). /. π e (y ) π e (x 5) 5(y ). 5 f(x, y) dy dx π 5 5 e (x 5) 5(y ) dy dx. Using a CAS or calculator to evaluate the integral, we get P ( X 5, Y 5).5. (b) P ((X 5) +(Y ) ) D π e (x 5) 5(y ) da,whered is the region enclosed by the ellipse (x 5) + (y ). Solving for y gives y ± (x 5), the upper and lower halves of the ellipse, and these two halves meet where y [since the ellipse is centered at (5, )] (x 5) x 5±.Thus D 5+/ 5(y ) π e (x 5) da π 5 / + (x 5) e (x 5) 5(y ) dy dx. (x 5) Using a CAS or calculator to evaluate the integral, we get P ((X 5) +(Y ) ).6.. (a) If f(p, A) is the probability that an individual at A will be infected by an individual at P,andkdAis the number of infected individuals in an element of area da, thenf(p, A)kdAis the number of infections that should result from exposure of the individual at A to infected people in the element of area da. IntegrationoverD gives the number of infections of the person at A due to all the infected people in D. In rectangular coordinates (with the origin at the city s

CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5 center), the exposure of a person at A is E kf(p, A) da k D (b) If A (, ),then E k k D π D x + y dx dy πk 5 5 r d(p, A) r rdrdθπk r 6 πk 9k da k D (x x) +(y y ) dx dy For A at the edge of the city, it is convenient to use a polar coordinate system centered at A. Then the polar equation for the circular boundary of the city becomes r cosθ instead of r, and the distance from A to a point P in the city is again r (see the figure). So E k π/ cos θ π/ r π/ r r cos θ rdrdθ k π/ r dθ 6 r + cos θ k π/ π/ cos θ cos θ dθ k π/ π/ sin θ cos θ dθ k θ + sin θ sin θ + sin θ π/ k π + + + π + + π/ 9 9 k π 8 9 6k Therefore the risk of infection is much lower at the edge of the city than in the middle, so it is better to live at the edge. 6.6 Triple Integrals ET 5.6.. 5. B xyz dv z x+z 6xz dy dx dz z xyz dy dz dx xy z y dz dx y xz z dx 7 7 xdx x 7 z yx+z 6xyz z ze y dx dz dy xz dz dx dx dz z 6xz(x + z) dx dz y x z +x z xz dz x (z +z ) dz 5z dz z 5 xze y x z dz dy zey z x dz dy z ( z ) / e y dy z ey dy ey (e ) 7. π/ y x cos(x + y + z)dz dx dy π/ y zx sin(x + y + z) dx dy z π/ π/ π/ y [sin(x + y) sin(x + y)] dx dy cos(x + y)+cos(x + y) xy x dy cos y +cosy + cos y cos y dy 6 sin y + sin y sin y π/ 6 9. E xdv y x y x y xdzdxdy y y dy ( x y )ydy y y zy xz dx dy y xy dx dy z

SECTION 6.6 TRIPLE INTEGRALS ET SECTION 5.6. Here E {(x, y, z) x, y x, z +x + y},so E 6xy dv x +x+y 6xy dz dy dx x 6xyz z+x+y z dy dx x xy +x y +xy y x y dx (x +x +x 5/ ) dx 6xy( + x + y) dy dx x + x + 7 x7/ 65 8. E is the region below the parabolic cylinder z y and above the square [, ] [, ] inthe xy-plane. E x e y dv y x e y dz dy dx x e y ( y ) dy dx x dx (ey y e y ) dy x e y (y y +)e y integrate by parts twice ()[e e e +5e ] 8 e 5. Here T {(x, y, z) x, y x, z x y},so T x dv x x y x dz dy dx x x (x x x y)dy dx x ( x) x ( x) x ( x) dx x x + x dx x5 x + 6 x + 6 6 x ( x y)dy dx x y x y x y y x dx y 7. The projection E on the yz-plane is the disk y + z. Using polar 9. The plane x + y + z intersects the xy-plane when x + y + y x,so E {(x, y, z) x, y x, z x y} and V x x y dz dy dx x ( x y) dy dx y xy y y x dx y ( x) x( x) ( x) dx (x 8x +8)dx x x +8x 6 coordinates y r cos θ and z r sin θ,weget xdv xdx da E D y +z D (y +z ) da 8 π ( r ) rdrdθ8 π dθ (r r5 ) dr 8(π) r r6 6π 6

CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5. V 9 x 9 x 5 y 9 x dz dy dx (5 y ) dy dx y 9 x y y 9 x y 9 x dx 8 9 x dx 8 x 9 x + 9 sin x 8 9 sin () 9 sin ( ) 6 π π 6π Alternatively, use polar coordinates to evaluate the double integral: 9 x π ( y) dy dx ( r sin θ) rdrdθ 9 x π r r sin θ r dθ r π (8 9sinθ) dθ 8θ +9cosθ π 6π. (a) The wedge can be described as the region D (x, y, z) y + z, x, y x (x, y, z) x, y x, z y So the integral expressing the volume of the wedge is D dv x y dz dy dx. (b) A CAS gives x y dz dy dx π. (Or use Formulas and 87 from the Table of Integrals.) 5. Here f(x, y, z) using trigonometric substitution or Formula in the Table of Integrals and V 6, so the Midpoint Rule gives ln( + x + y + z) B f(x, y, z) dv l m n i j k f x i, y j, z k V 6[f(,, ) + f(,, ) + f(, 6, ) + f(, 6, ) + f(,, ) + f(,, ) + f(, 6, ) + f(, 6, )] 6 ln 5 + ln 7 + ln 9 + ln + ln 7 + ln 9 + ln + ln 6.5 7. E {(x, y, z) x, z x, y z}, the solid bounded by the three coordinate planes and the planes z x, y z.

SECTION 6.6 TRIPLE INTEGRALS ET SECTION 5.6 9. If D, D, D are the projections of E on the xy-, yz-, and xz-planes, then Therefore Then D (x, y) x, y x (x, y) y, y x y D (y, z) y, y z y (y, z) z, y z D (x, z) x +z E (x, y, z) x, y x, x y z x y (x, y, z) y, y x y, x y z x y (x, y, z) z, y z, y z x y z (x, y, z) y, y z y, y z x y z (x, y, z) x, x z x, y x z (x, y, z) z, z x z, y x z E f(x, y, z) dv x z x y/ f(x, y, z) dz dy dx x y/ x / y z f(x, y, z) dx dy dz y z x / x z f(x, y, z) dy dz dx y y y/ y/ z x y/ x y/ y z y z x z z f(x, y, z) dz dx dy f(x, y, z) dx dz dy f(x, y, z) dy dx dz

CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5. If D, D,andD are the projections of E on the xy-, yz-, and xz-planes, then D D (x, y) x,x y (y, z) y, z y (x, y) y, y x y, (y, z) z, y z,and D (x, z) x, z x (x, z) z, z x z Therefore E (x, y, z) x, x y, z y (x, y, z) y, y x y, z y (x, y, z) y, z y, y x y (x, y, z) z, y z, y x y (x, y, z) x, z x, x y z (x, y, z) z, z x z, x y z Then E f(x, y, z) dv y/ f(x, y, z) dz dy dx y y/ x f(x, y, z) dz dx dy y y/ x / y f(x, y, z) dx dz dy y z z f(x, y, z) dy dz dx z x z y f(x, y, z) dx dy dz y z f(x, y, z) dy dx dz x

SECTION 6.6 TRIPLE INTEGRALS ET SECTION 5.6 5. The diagrams show the projections of E on the xy-, yz-, and xz-planes. Therefore x y f(x, y, z) dz dy dx y y y ( z) y f(x, y, z) dz dx dy f(x, y, z) dx dz dy z x f(x, y, z) dy dx dz z y x z x f(x, y, z) dx dy dz f(x, y, z) dy dz dx 5. y y f(x, y, z) dz dx dy f(x, y, z) dv where E {(x, y, z) z y, y x, y }. E If D, D,andD are the projections of E on the xy-, yz-andxz-planes then Thus we also have Then D {(x, y) y, y x } {(x, y) x, y x}, D {(y, z) y, z y} {(y, z) z, z y },and D {(x, z) x, z x} {(x, z) z, z x }. E {(x, y, z) x, y x, z y} {(x, y, z) y, z y, y x } {(x, y, z) z, z y, y x } {(x, y, z) x, z x, z y x} {(x, y, z) z, z x, z y x}. y f(x, y, z) dz dx dy x y z z y f(x, y, z) dz dy dx y f(x, y, z) dx dz dy y f(x, y, z) dx dy dz x x f(x, y, z) dy dz dx y z x z f(x, y, z) dy dx dz

6 CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5 7. m E ρ(x, y, z) dv y +xy + y y x dx y x +x+y dz dy dx x x +x / + x ( + x + y) dy dx dx x/ + 5 x5/ + x 79 M yz xρ(x, y, z) dv x +x+y xdzdydx x x( + x + y) dy dx E xy +x y + xy y x dx y (x/ +x 5/ + x ) dx 5 x5/ + 7 x7/ + x 79 5 M xz E yρ(x, y, z) dv y + xy + y y x y M xy E zρ(x, y, z) dv x +x+y ydzdydx x dx x + x + x/ dx x +x+y zdzdydx y( + x + y) dy dx x + x + 5 x5/ x z z+x+y x ( + x +y +xy + x + y ) dy dx y +xy + y + xy + x y + y y x x + 7 x/ + x + x + x 5/ dx z x/ + 5 x5/ + x + x + 7 x7/ 57 Thus the mass is 79 and the center of mass is (x, y, z) Myz m, Mxz m, Mxy 58 m 55, 79, 57. 55 dy dx x ( + x + y) dy dx y dx 9. m a a a M yz a a a a a (x + y + z ) dx dy dz a a x + xy + xz xa dy dz a a x a + ay + az dy dz a y + ay + ayz ya dz a y a + a z dz a z + a z a a5 + a5 a 5 x + x(y + z ) dx dy dz a a a + a (y + z ) dy dz a5 + 6 a5 + a z dz a6 + a6 7 a6 M xz M xy by symmetry of E and ρ(x, y, z) Hence (x, y, z) 7 a, 7 a, 7 a.. I x L L L k(y + z ) dz dy dx k L L By symmetry, I x I y I z kl5. Ly + L dy dx k L L dx kl5.. I z E (x + y ) ρ(x, y, z) dv h k(x + y ) dz da k(x + y )hda x +y a x +y a kh π a (r ) rdrdθ kh π dθ a r dr kh(π) r a πkh a πkha 5. (a) m 9 x 5 y 9 x x + y dz dy dx Myz (b) (x, y, z) m, Mxz m, Mxy where m M yz M xy (c) I z 9 x 9 x 9 x 9 x 5 y 9 x 5 y 9 x 5 y x x + y dz dy dx, M xz z x + y dz dy dx. (x + y ) x + y dz dy dx 9 x 9 x 9 x 5 y 9 x 5 y y x + y dz dy dx, and (x + y ) / dz dy dx

SECTION 6.6 TRIPLE INTEGRALS ET SECTION 5.6 7 7. (a) m x y (b) (x, y, z) m x (c) I z π ( + x + y + z) dz dy dx + m y x x( + x + y + z) dz dy dx, y y( + x + y + z) dz dy dx, m x y z( + x + y + z) dz dy dx 8 π +8 5π +8,, 9π + 5π + 5π + 66 x y (x + y )( + x + y + z) dz dy dx 68 + 5π 9. (a) f(x, y, z) is a joint density function, so we know R f(x, y, z) dv.herewehave R f(x, y, z) dv Then we must have 8C C 8. (b) P (X,Y,Z ) 8 f(x, y, z) dz dy dx C xdx ydy zdz C x f(x, y, z) dz dy dx xdx ydy zdz 8 x y y xyz dz dy dx 8 Cxyz dz dy dx z 8C z 8 6 (c) P (X + Y + Z ) P ((X, Y, Z) E) where E is the solid region in the first octant bounded by the coordinate planes and the plane x + y + z. The plane x + y + z meets the xy-plane in the line x + y,sowehave x x y P (X + Y + Z ) f(x, y, z) dv E x 8 xy z z x y dy dx z 6 5. V (E) L f ave L L x L x 6 6 8 xyzdzdydx x xy( x y) dy dx [(x x + x)y +(x x)y + xy ] dy dx (x x + x) y +(x x) y + x y y x dx y (x 9 x +6x x + x 5 ) dx 9 576 L L L y xyz dx dy dz L xdx L L z L L L L L L L 8 5. The triple integral will attain its maximum when the integrand x y z is positive in the region E and negative everywhere else. For if E contains some region F where the integrand is negative, the integral could be increased by excluding F from E,andifE fails to contain some part G of the region where the integrand is positive, the integral could be increased by including G in E. Sowerequirethatx +y +z. This describes the region bounded by the ellipsoid x +y +z. ydy L zdz

8 CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5 6.7 Triple Integrals in Cylindrical Coordinates ET 5.7. (a) (b) x cos π, y sin π, z, so the point is,, in rectangular coordinates. x cos π, y sin π, and z 5,sothepointis,, 5 in rectangular coordinates.. (a) r x + y +( ) so r ; tan θ y x and the point (, ) is in the fourth quadrant of the xy-plane, so θ 7π +nπ; z. Thus, one set of cylindrical coordinates is, 7π,. (b) r ( ) + so r ; tan θ and the point, is in the third quadrant of the xy-plane, so θ π +nπ; z. Thus, one set of cylindrical coordinates is, π,. 5. Since θ π but r and z may vary, the surface is a vertical half-plane including the z-axis and intersecting the xy-plane in the half-line y x, x. 7. z r (x + y ) or x y, so the surface is a circular paraboloid with vertex (,, ), axis the z-axis, and opening downward. 9. (a) x + y r, so the equation becomes z r. (b) Substituting x + y r and y r sin θ, the equation x + y y becomes r r sin θ or r sinθ.. r and z describe a solid circular cylinder with radius, axisthez-axis, and height, but π/ θ π/ restricts the solid to the first and fourth quadrants of the xy-plane, so we have a half-cylinder.. We can position the cylindrical shell vertically so that its axis coincides with the z-axisanditsbaseliesinthexy-plane. If we use centimeters as the unit of measurement, then cylindrical coordinates conveniently describe the shell as 6 r 7, θ π, z.

SECTION 6.7 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES ET SECTION 5.7 9 5. The region of integration is given in cylindrical coordinates by E {(r, θ, z) θ π, r, r z }. This represents the solid region bounded below by the cone z r and above by the horizontal plane z. π rdzdθdr π z r rz dθ dr π r( r) dθ dr zr (r r ) dr π dθ r r 6 (π) 6π 7. In cylindrical coordinates, E is given by {(r, θ, z) θ π, r, 5 z }. So x + y E dv π r rdzdrdθ π dθ 5 r dr θ π r z (π) 6 5 (9) 8π 5 dz θ π 9. In cylindrical coordinates E is bounded by the paraboloid z +r, the cylinder r 5or r 5,andthexy-plane, so E is given by (r, θ, z) θ π, r 5, z +r.thus E ez dv π 5 +r e z rdzdrdθ π 5 r e z z+r dr dθ π 5 z π dθ 5 re +r r dr π e+r 5 r π(e 6 e 5) r(e +r ) dr dθ. In cylindrical coordinates, E is bounded by the cylinder r, the plane z, and the cone z r. So E {(r, θ, z) θ π, r, z r} and E x dv π r r cos θ r dz dr dθ π r cos θz zr dr dθ π z r cos θdrdθ π 5 r5 cos θ r dθ π r 5 cos θdθ π +cosθ dθ θ + π 5 5 sin θ π 5. (a) The paraboloids intersect when x + y 6 x y x + y 9, so the region of integration is D (x, y) x + y 9. Then, in cylindrical coordinates, E (r, θ, z) r z 6 r, r, θ π and V π 6 r r rdzdrdθ π 6r r dr dθ π 8r r r dθ π 8 dθ 6π. r (b) For constant density K, m KV 6πK from part (a). Since the region is homogeneous and symmetric, M yz M xz and M xy π K π 6 r r (zk) rdzdrdθ K π r z z6 r dr dθ zr r((6 r ) r ) dr dθ K K (π) 8 6 r6 6 r + 96 r πk() πk Myz Thus (x, y, z) m, M xz m, M xy m π dθ (8r5 6r +96r) dr,, πk 6πK (,, 5).

5 CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5 5. The paraboloid z x +y intersects the plane z a when a x +y or x + y a. So, in cylindrical coordinates, E (r, θ, z) r a, θ π, r z a.thus m π a/ a K π r Kr dz dr dθ K ar r r a/ r Since the region is homogeneous and symmetric, M yz M xz and M xy Hence (x, y, z),, a. π a/ a K π r Krzdzdrdθ K a r r6 r a/ dθ K r π a/ (ar r ) dr dθ π dθ K 6 a dθ 8 a πk π a/ π a r 8r 5 dr dθ a dθ a πk 7. The region of integration is the region above the cone z x + y,orz r, and below the plane z.also,wehave y with y x y which describes a circle of radius in the xy-plane centered at (, ). Thus, y y x +y xz dz dx dy π r (r cos θ) z r dz dr dθ π r (cos θ) z z dr dθ zr r r dr π cos θdθ π π [sin θ]π r r (cos θ) zdzdrdθ r (cos θ) r dr dθ r 5 r5 9. (a) The mountain comprises a solid conical region C. The work done in lifting a small volume of material V with density g(p ) to a height h(p ) above sea level is h(p )g(p ) V. Summing over the whole mountain we get W h(p )g(p ) dv. C (b) Here C is a solid right circular cone with radius R 6, ft, height H, ft, and density g(p )lb/ft at all points P in C. We use cylindrical coordinates: W π H π R( z/h) z rdrdzdθπ H z r rr( z/h) dz r H πr z z R z H z H + dz πr H z z H + z H H z H πr H H 5 πr H 5 π(6,) (,). 9 ft-lb + H dz r R H z H z H

SECTION 6.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES ET SECTION 5.8 5 6.8 Triple Integrals in Spherical Coordinates ET 5.8. (a) (b) x ρ sin φ cos θ ()sincos, y ρ sin φ sin θ ()sinsin,and z ρ cos φ ()cosso the point is (,, ) in rectangular coordinates. x sin π cos π, y sinπ sin π 6, z cos π so the point is, 6, in rectangular coordinates.. (a) ρ x + y + z ++, cos φ z ρ cos θ x ρ sin φ sin(π/6) θ π φ π 6,and [since y>]. Thus spherical coordinates are, π, π. 6 (b) ρ ++, cos φ φ π,andcos θ sin(π/) θ π [since y<]., π Thus spherical coordinates are, π. 5. Since φ π, the surface is the top half of the right circular cone with vertex at the origin and axis the positive z-axis. 7. ρ sinθ sin φ ρ ρ sin θ sin φ x + y + z y x + y y + + z x +(y ) + z. Therefore, the surface is a sphere of radius centered at,,. 9. (a) x ρ sin φ cos θ, y ρ sin φ sin θ, andz ρ cos φ, so the equation z x + y becomes (ρ cos φ) (ρ sin φ cos θ) +(ρ sin φ sin θ) or ρ cos φ ρ sin φ.ifρ 6, this becomes cos φ sin φ.(ρ corresponds to the origin which is included in the surface.) There are many equivalent equations in spherical coordinates, such as tan φ, cos φ, cos φ,orevenφ π, φ π. (b) x + z 9 (ρ sin φ cos θ) +(ρ cos φ) 9 ρ sin φ cos θ + ρ cos φ 9or ρ sin φ cos θ +cos φ 9.. ρ represents a sphere of radius, centered at the origin, so ρ is this sphere and its interior. φ π restricts the solid to that portion of the region that lies on or above the xy-plane, and θ π further restricts the solidtothefirst octant. Thus the solid is the portion in the first octant of the solid ball centered at the origin with radius.

5 CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5. ρ represents the solid sphere of radius centered at the origin. π φ π restricts the solid to that portion on or below the cone φ π. 5. z x + y because the solid lies above the cone. Squaring both sides of this inequality gives z x + y z x + y + z ρ z ρ cos φ ρ cos φ. The cone opens upward so that the inequality is cos φ, or equivalently φ π. In spherical coordinates the sphere z x + y + z is ρ cos φ ρ ρ cosφ. ρ cos φ because the solid lies below the sphere. The solid can therefore be described as the region in spherical coordinates satisfying ρ cos φ, φ π. 7. The region of integration is given in spherical coordinates by E {(ρ, θ, φ) ρ, θ π/, φ π/6}. This represents the solid region in the first octant bounded above by the sphere ρ and below by the cone φ π/6. π/6 π/ ρ sin φdρdθdφ π/6 sin φdφ π/ dθ ρ dρ cos φ π/6 π/ θ ρ π (9) 9π 9. The solid E is most conveniently described if we use cylindrical coordinates: E (r, θ, z) θ π, r, z.then E f(x, y, z) dv π/ f(r cos θ, r sin θ, z) rdzdrdθ.. In spherical coordinates, B is represented by {(ρ, θ, φ) ρ 5, θ π, φ π }. Thus B (x + y + z ) dv π π 5 (ρ ) ρ sin φdρdθdφ π sin φdφ π dθ 5 π θ cos φ π,5 7 π,9.7 ρ7 5 ()(π) 78,5 7 7. In spherical coordinates, E is represented by (ρ, θ, φ) ρ, θ π, φ π. Thus E zdv π/ π/ (ρ cos φ) ρ sin φdρdθdφ π/ cos φ sin φdφ π/ dθ ρ dρ sin φ π/ θ π/ ρ π 5 5π 6 ρ6 dρ

5. E x dv π π SECTION 6.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES ET SECTION 5.8 5 (ρ sin φ cos θ) ρ sin φdρdφdθ π cos θdθ π sin φdφ θ + sin θ π ( + sin φ)cosφ π ρ5 π + 5 ρ dρ 5 (5 5 ) 56 5 π 7. The solid region is given by E (ρ, θ, φ) ρ a, θ π, π 6 φ π and its volume is V E dv π/ π/6 [ cos φ] π/ π/6 [θ]π ρ a π a ρ sin φdρdθdφ π/ sin φdφ π dθ a π/6 + (π) a πa ρ dρ 9. (a) Since ρ cosφ implies ρ ρ cos φ, the equation is that of a sphere of radius with center at (,, ). Thus V π π/ cosφ ρ sin φdρdφdθ π π/ ρ ρ cos φ ρ π 6 cos φ φπ/ dθ π 6 dθ 5θ φ 6 (b) By the symmetry of the problem M yz M xz.then M xy π Hence (x, y, z) (,,.). π sin φdφdθ π π π/ 6 π/ cosφ ρ cos φ sin φdρdφdθ π π/ cos φ sin φ 6 cos φ dφ dθ π 6 6 cos6 φ φπ/ dθ π dθ π φ. By the symmetry of the region, M xy and M yz. Assuming constant density K, and m KV K π π E Kπ cos φ π ρ Kπ 7 7 M xz ykdv K π π E K cos θ π Myz Thus the centroid is (x, y, z) m, M xz m, M xy m ρ sin φdρdφdθ K π dθ π sin φdφ ρ dρ cos φ sin φdφdθ πk (ρ sin φ sin θ) ρ sin φdρdφdθ K π sin θdθ π sin φdφ φ sin φ π ρ K() π 75 (56 8) πk, 75πK/ 7πK/,, 55,. 96 ρ dρ. (a) The density function is ρ(x, y, z) K, a constant, and by the symmetry of the problem M xz M yz.then M xy π π/ a Kρ sin φ cos φdρdφdθ πka π/ sin φ cos φdφ 8 πka.butthemassisk(volume of the hemisphere) πka, so the centroid is,, 8 a. (b) Place the center of the base at (,, ); the density function is ρ(x, y, z) K. By symmetry, the moments of inertia about any two such diameters will be equal, so we just need to find I x: I x π K π π/ a (Kρ sin φ) ρ (sin φ sin θ +cos φ) dρ dφ dθ π/ (sin φ sin θ +sinφ cos φ) a5 dφ dθ 5 5 Ka5 π sin θ cos φ + cos φ + cos φ φπ/ dθ φ 5 Ka5 π 5 Ka5 θ sin θ + θ π 5 Ka5 (π ) + (π ) 5 Ka5 π sin θ + dθ

5 CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5 5. In spherical coordinates z x + y becomes cos φ sinφ or φ π.then V π M xy π π/ ρ sin φdρdφdθ π dθ π/ sin φdφ π/ Hence (x, y, z) ρ sin φ cos φdρdφdθ π cos φ π/,, 8. ρ dρ π, π and by symmetry Myz Mxz. 8 7. In cylindrical coordinates the paraboloid is given by z r and the plane by z rsin θ and they intersect in the circle r sinθ. Then zdv π sinθ r sin θ rz dz dr dθ 5π E r 6 [using a CAS]. 9. The region E of integration is the region above the cone z x + y and below the sphere x + y + z in the first octant. Because E is in the first octant we have θ π. The cone has equation φ π (as in Example ), so φ π, and ρ. So the integral becomes π/ π/ (ρ sin φ cos θ)(ρ sin φ sin θ) ρ sin φdρdθdφ π/ sin φdφ π/ sin θ cos θdθ ρ dρ cos φ cos φ π/ 5 5 π/. In cylindrical coordinates, the equation of the cylinder is r, z. The hemisphere is the upper part of the sphere radius, center (,, ), equation cos φ sin φdφ sin θ π/ 5 ρ5 5 5 5 r +(z ), z.inmaple,wecanusethecoordscylindrical option in a regular plotd command. In Mathematica, we can use ParametricPlotD.. If E is the solid enclosed by the surface ρ + sin 6θ sin 5φ, it can be described in spherical coordinates as 5 E (ρ, θ, φ) ρ + 5 sin 6θ sin 5φ, θ π, φ π. Its volume is given by V (E) E dv π π +(sin6θ sin 5φ)/5 ρ sin φdρdθdφ 6π [using a CAS]. 99 5. (a) From the diagram, z r cot φ to z a r, r to r a sin φ (or use a r r cot φ ). Thus V π a sin φ a r r cot φ rdzdrdθ π a sin φ r a r r cot φ dr a sin φ (a r ) / r cot φ π π a a sin φ / a sin φ cot φ + a πa cos φ +sin φ cos φ πa ( cos φ )

SECTION 6.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS ET SECTION 5.9 55 (b) The wedge in question is the shaded area rotated from θ θ to θ θ. Letting V ij volume of the region bounded by the sphere of radius ρ i and the cone with angle φ j (θ θ to θ ) and letting V be the volume of the wedge, we have V (V V ) (V V ) (θ θ ) ρ ( cos φ ) ρ ( cos φ ) ρ ( cos φ )+ρ ( cos φ ) (θ θ) ρ ρ ( cos φ ) ρ ρ ( cos φ ) (θ θ) ρ ρ (cos φ cos φ ) Or: Show that V θ ρ sin φ r cot φ θ ρ sin φ r cot φ rdzdrdθ. (c) By the Mean Value Theorem with f(ρ) ρ there exists some ρ with ρ ρ ρ such that f(ρ ) f(ρ )f ( ρ)(ρ ρ ) or ρ ρ ρ ρ. Similarly there exists φ with φ φ φ such that cos φ cos φ sin φ φ. Substituting into the result from (b) gives V ( ρ ρ)(θ θ )(sin φ) φ ρ sin φ ρ φ θ. 6.9 Change of Variables in Multiple Integrals ET 5.9. x 5u v, y u +v. (x, y) The Jacobian is (u, v) x/ u x/ v y/ u y/ v 5 5() ( )() 6.. x e r sin θ, y e r cos θ. (x, y) (r, θ) x/ r x/ θ y/ r y/ θ e r sin θ e r cos θ e r cos θ e r sin θ e r e r sin θ e r e r cos θ sin θ cos θ or cos θ 5. x u/v, y v/w, z w/u. x/ u x/ v x/ w (x, y, z) /v u/v (u, v, w) y/ u y/ v y/ w /w v/w z/ u z/ v z/ w w/u /u /w v/w v /u u v/w v w/u /u + /w w/u v uw + uv v + u w uvw uvw

56 CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5 7. The transformation maps the boundary of S to the boundary of the image R,sowefirstlookatsideS in the uv-plane. S is described by v [ u ], so x u +v uand y u v u. Eliminating u,wehavex y, x 6. S is the line segment u, v,sox 6+v and y v. Thenv y x 6+( y) 5 y, 6 x. S is the line segment v, u,sox u +6and y u,givingu y + x y +, 6 x. Finally, S is the segment u, v,sox v and y v x y, x 6. Theimageof set S is the region R shown in the xy-plane, a parallelogram bounded by these four segments. 9. S is the line segment u v, u,soy v u and x u y.since u, the image is the portion of the parabola x y, y. S is the segment v, u,thusy v and x u,so x. Theimageis thelinesegmenty, x. S is the segment u, v,sox u and y v y. The image is the segment x, y. Thus, the image of S is the region R in the first quadrant bounded by the parabola x y,they-axis, and the line y... (x, y) (u, v) and x y (u + v) (u +v) u 5v. Tofind the region S in the uv-plane that corresponds to R we first find the corresponding boundary under the given transformation. The line through (, ) and (, ) is y x which is the image of u +v (u + v) v ; the line through (, ) and (, ) is x + y which is the image of (u + v)+(u +v) u + v ; the line through (, ) and (, ) is y x which is the image of u +v (u + v) u. Thus S is the triangle v u, u in the uv-plane and R (x y) da u ( u 5v) dv du uv + 5 v v u (x, y) (u, v) v du u u + 5 ( u) du u u 5 6 ( u) + 5 6 6, x u and the planar ellipse 9x +y 6 is the image of the disk u + v. Thus R x da u +v (u )(6) du dv π x + sin x π (r cos θ) rdrdθ π cos θdθ r dr r (π) 6π

5. (x, y) (u, v) SECTION 6.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS ET SECTION 5.9 57 /v u/v v, xy u, y x is the image of the parabola v u, y xis the image of the parabola v u, and the hyperbolas xy, xy are the images of the lines u and u respectively. Thus R xy da u u u v dv du u ln u ln u du u ln du ln ln. 7. (a) a (x, y, z) (u, v, w) b abc and since u x a, v y b, w z the solid enclosed by the ellipsoid is the image of the c c ball u + v + w. So E dv u +v +w (b) If we approximate the surface of the earth by the ellipsoid abc du dv dw (abc)(volume of the ball) πabc x 678 + y 678 + z, then we can estimate 656 the volume of the earth by finding the volume of the solid E enclosed by the ellipsoid. From part (a), this is E dv π(678)(678)(656).8 km. 9. Letting u x y and v x y,wehavex (v u) and y y) (v u). Then (x, 5 5 (u, v) and R is the image of the rectangle enclosed by the lines u, u, v,andv 8. Thus R x y x y da 8 u v 5 dv du 5 udu 8 v dv u 5. Letting u y x, v y + x,wehavey (u + v), x y) (v u). Then (x, (u, v) image of the trapezoidal region with vertices (, ), (, ), (, ),and(, ). Thus cos y x v y + x da cos u v du dv v sin u u v dv v u v R v /5 /5 /5 /5 5 8 ln v 8 ln 8. 5 / / / / and R is the v sin() dv sin. Let u x + y and v x + y. Thenu + v y y (u + v) and u v x x (u v). (x, y) (u, v) / / / /.Now u x + y x + y u,and v x + y x + y v. R is the image of the square region with vertices (, ), (, ), (, ),and(, ). So R ex+y da eu du dv e u v e e.

58 CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5 6 Review ET 5. (a) A double Riemann sum of f is m i j n f x ij,yij A,where A is the area of each subrectangle and x ij,yij is a sample point in each subrectangle. If f(x, y), this sum represents an approximation to the volume of the solid that lies above the rectangle R and below the graph of f. (b) f(x, y) da lim R m m,n i j n f x ij,yij A (c) If f(x, y), f(x, y) da represents the volume of the solid that lies above the rectangle R and below the surface R z f(x, y). Iff takes on both positive and negative values, f(x, y) da is the difference of the volume above R but R below the surface z f(x, y) and the volume below R but above the surface z f(x, y). (d) We usually evaluate f(x, y) da as an iterated integral according to Fubini s Theorem (see Theorem 6.. R [ET 5..]). (e) The Midpoint Rule for Double Integrals says that we approximate the double integral f(x, y) da by the double R Riemann sum m i j n f x i, y j A where the sample points xi, y j are the centers of the subrectangles. (f ) f ave f(x, y) da where A (R) is the area of R. A (R) R. (a) See () and () and the accompanying discussion in Section 6. [ET 5.]. (b) See () and the accompanying discussion in Section 6. [ET 5.]. (c) See (5) and the preceding discussion in Section 6. [ET 5.]. (d) See (6) () in Section 6. [ET 5.].. We may want to change from rectangular to polar coordinates in a double integral if the region R of integration is more easily described in polar coordinates. To accomplish this, we use R f(x, y) da β α given by a r b, α θ β.. (a) m ρ(x, y) da D (b) M x yρ(x, y) da, M D y xρ(x, y) da D b a f(r cos θ, r sin θ) rdrdθwhere R is (c) The center of mass is (x, y) where x M y m and y M x m. (d) I x D y ρ(x, y) da, I y D x ρ(x, y) da, I D (x + y )ρ(x, y) da 5. (a) P (a X b, c Y d) b a (b) f(x, y) and R f(x, y) da. d f(x, y) dy dx c (c) The expected value of X is μ R xf(x, y) da; the expected value of Y is μ R yf(x, y) da.

CHAPTER 6 REVIEW ET CHAPTER 5 59 6. (a) f(x, y, z) dv lim B l,m,n l m n i j k f x ijk,y ijk,z ijk V (b) We usually evaluate f(x, y, z) dv as an iterated integral according to Fubini s Theorem for Triple Integrals B (see Theorem 6.6. [ET 5.6.]). (c) See the paragraph following Example 6.6. [ET 5.6.]. (d) See (5) and (6) and the accompanying discussion in Section 6.6 [ET 5.6]. (e) See () and the accompanying discussion in Section 6.6 [ET 5.6]. (f ) See () and the preceding discussion in Section 6.6 [ET 5.6]. 7. (a) m ρ(x, y, z) dv E (b) M yz xρ(x, y, z) dv, M E xz yρ(x, y, z) dv, M E xy zρ(x, y, z) dv. E (c) The center of mass is (x, y, z) where x Myz m, y Mxz Mxy,andz m m. (d) I x E (y + z )ρ(x, y, z) dv, I y E (x + z )ρ(x, y, z) dv, I z E (x + y )ρ(x, y, z) dv. 8. (a) See Formula 6.7. [ET 5.7.] and the accompanying discussion. (b) See Formula 6.8. [ET 5.8.] and the accompanying discussion. (c) We may want to change from rectangular to cylindrical or spherical coordinates in a triple integral if the region E of integration is more easily described in cylindrical or spherical coordinates or if the triple integral is easier to evaluate using cylindrical or spherical coordinates. (x, y) 9. (a) (u, v) x/ u x/ v y/ u y/ v x y u v x y v u (b) See (9) and the accompanying discussion in Section 6.9 [ET 5.9]. (c) See () and the accompanying discussion in Section 6.9 [ET 5.9].. This is true by Fubini s Theorem.. True by Equation 6..5 [ET 5..5]. 5. True: D x y da the volume under the surface x + y + z and above the xy-plane the volume of the sphere x + y + z π() 6 π 7. The volume enclosed by the cone z x + y and the plane z is, in cylindrical coordinates, V π rdzdrdθ6 π dz dr dθ,sotheassertionisfalse. r r

6 CHAPTER 6 MULTIPLE INTEGRALS ET CHAPTER 5. Asshowninthecontourmap,wedivideR into 9 equally sized subsquares, each with area A. Then we approximate f(x, y) da by a Riemann sum with m n and the sample points the upper right corners of each square, so. 5. R R f(x, y) da i j f(x i,y j) A A [f(, ) + f(, ) + f(, ) + f(, ) + f(, ) + f(, ) + f(, ) + f(, ) + f(, )] Using the contour lines to estimate the function values, we have f(x, y) da [.7+.7+8.+.7+6.7+.+6.7+8.6+.9] 6. R (y +xey ) dx dy xy + x e y x dy (y x +ey ) dy y +e y +e e e e + x cos(x ) dy dx cos(x )y yx dx x y cos(x ) dx sin(x ) sin 7. π y y sin xdzdydx π π z y (y sin x)z dy dx π y y z sin xdydx y ( y ) / sin x dx π sin xdx cos x π y 9. The region R is more easily described by polar coordinates: R {(r, θ) r, θ π}. Thus R f(x, y) da π f(r cos θ, r sin θ) rdrdθ.. The region whose area is given by π/ sin θ rdrdθis (r, θ) θ π, r sin θ, which is the region contained in the loop in the first quadrant of the four-leaved rose r sinθ.. x cos(y ) dy dx y cos(y ) dx dy cos(y ) x xy x dy y cos(y ) dy sin(y ) sin 5. R yexy da yexy dx dy e xy x dy x (ey ) dy ey y e6 e6 7

CHAPTER 6 REVIEW ET CHAPTER 5 6 7. D y +x da x y dy dx +x +x y y x +x dx ln( + x ) ln y x dx 9. D yda 8 y ydxdy y y x x8 y xy dy y(8 y y ) dy (8y y ) dy y y 8. D x + y / da π/ π/ π dθ (r ) / rdrdθ 5 5 8π 5 r dr θ π/ 5 r5. 5. 7. E xy dv x x+y xy dz dy dx x xy z zx+y z 5 6 E y z dv x (x y + xy ) dy dx x dx 6 x5 8.5 π π E yz dv 9. V 6 5 y y z y x y + xy yx dx y y z dx dz dy (r cos θ)(r sin θ)( r ) rdrdθ π dy dx x xy(x + y) dy dx x + x dx y y y z ( y z ) dz dy sin θ(r 5 r 7 ) dr dθ ( cos θ) 8 6 r6 r8 r dθ 8 r 9 θ sin θ π π π 9 96 x π sin θdθ 6 5 (x +y ) dy dx y yz dz dy dx x y dy dx π cos θ + cos θ π 6 5 x y + y y y dx (x +8)dx 76 r (sin θ) rdrdθ. V y ( y)/ dz dx dy y y dx dy y y dy