Solucao impares Calculo 6ed CAP.15

16

MULTIPLE INTEGRALS

ET 15

16.1

Double Integrals over Rectangles

ET 15.1

1. (a) The subrectangles are shown in the gure.

The surface is the graph ofi({> |) = {| and D = 4, so we estimate Y  S3

l = 1 2

S

m = 1i({l> |m) D

= i(2> 2) D + i(2> 4) D + i(4> 2) D + i(4> 4) D + i(6> 2) D + i(6> 4) D = 4(4) + 8(4) + 8(4) + 16(4) + 12(4) + 24(4) = 288 (b)Y  S3 l = 1 2 S m = 1i  {l> |m 

D = i(1> 1) D + i(1> 3) D + i(3> 1) D + i(3> 3) D + i(5> 1) D + i(5> 3) D = 1(4) + 3(4) + 3(4) + 9(4) + 5(4) + 15(4) = 144

3. (a) The subrectangles are shown in the gure. SinceD = 2@4, we estimate UU Usin({ + |) gD  2 S l=1 2 S m=1i  { lm> |lmD = i(0> 0) D + i0> 2  D + i 2> 0  D + i 2>2  D = 02 4  + 12 4  + 12 4  + 02 4  =2 2  4=935 (b) UU_Usin({ + |) gD  S2 l=1 2 S m=1i({l> |m) D = i 4>4  D + i 4>34  D + i3 4>4  D + i3 4 >34  D = 12 4  + 02 4  + 02 4  + (1)2 4  = 0

5. (a) Each subrectangle and its midpoint are shown in the gure. The area of each subrectangle isD = 2, so we evaluate i at each midpoint and estimate UU Ui({> |) gD  2 S l = 1 2 S m = 1i  {l> |m  D = i(1=5> 1) D + i(1=5> 3) D + i(2=5> 1) D + i(2=5> 3) D = 1(2) + (8)(2) + 5(2) + (1)(2) = 6 223

TX.10

(b) The subrectangles are shown in the gure. In each subrectangle, the sample point farthest from the origin is the upper right corner, and the area of each subrectangle isD =1₂. Thus we estimate UU Ui({> |) gD  4 S l = 1 4 S m = 1i({l> |m) D

= i(1=5> 1) D + i(1=5> 2) D + i(1=5> 3) D + i(1=5> 4) D + i(2> 1) D + i(2> 2) D + i(2> 3) D + i(2> 4) D

+ i(2=5> 1) D + i(2=5> 2) D + i(2=5> 3) D + i(2=5> 4) D + i(3> 1) D + i(3> 2) D + i(3> 3) D + i(3> 4) D = 11 2  + (4)1 2  + (8)1 2  + (6)1 2  + 31 2  + 01 2  + (5)1 2  + (8)1 2  + 51 2  + 31 2  + (1)1 2  + (4)1 2  + 81 2  + 61 2  + 31 2  + 01 2  = 3=5

7. The values ofi({> |) =s52  {2 |2get 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 haveX ? Y ? O. (Note that this is true no matter howU is divided into subrectangles.)

9. (a) Withp = q = 2, we have D = 4. Using the contour map to estimate the value of i at the center of each subrectangle, we have UU Ui({> |) gD  2 S l = 1 2 S m = 1i  {l> |m 

D = D[i(1> 1) + i(1> 3) + i(3> 1) + i(3> 3)]  4(27 + 4 + 14 + 17) = 248 (b)i_ave=_D(U)1 UU_Ui({> |) gD  ₁₆1(248) = 15=5

11. } = 3 A 0, so we can interpret the integral as the volume of the solid V that lies below the plane } = 3 and above the rectangle[2> 2] × [1> 6]. V is a rectangular solid, thusUU_U3 gD = 4 · 5 · 3 = 60.

13. } = i({> |) = 4  2|  0 for 0  |  1. Thus the integral represents the volume of that part of the rectangular solid[0> 1] × [0> 1] × [0> 4] which lies below the plane } = 4  2|. So

UU

SECTION 16.2 ITERATED INTEGRALS ET SECTION 15.2 ¤ 225 15.To calculate the estimates using a programmable calculator, we can use an algorithm

similar to that of Exercise 5.1.7 [ET 5.1.7]. In Maple, we can dene the function i({> |) =1 + {h|_{(calling it f), load the student package, and then use the}

command

middlesum(middlesum(f,x=0..1,m), y=0..1,m);

to get the estimate withq = p2squares of equal size. Mathematica has no special Riemann sum command, but we can dene f and then use nested Sum commands to calculate the estimates.

q estimate 1 1=141606 4 1=143191 16 1=143535 64 1=143617 256 1=143637 1024 1=143642

17.If we divideU into pq subrectangles,UU_Un gD  Sp

l = 1 q S m = 1i  {

lm> |lmD for any choice of sample points{lm> |lm.

Buti{_lm> |_lm= n always and Sp

l = 1 q

S

m = 1D = area of U = (e  d)(g  f). Thus, no matter how we choose the sample

points, Sp l = 1 q S m = 1i  { lm> |lmD = n p S l = 1 q S m = 1D = n(e  d)(g  f) and so UU Un gD = lim_p>q p S l = 1 q S m = 1i  { lm> |lmD = lim_p>qn p S l = 1 q S

m = 1D = limp>qn(e  d)(g  f) = n(e  d)(g  f).

16.2

Iterated Integrals

ET 15.2

1.U₀512{2|3g{ =  12 {₃3 |3{=5 {=0= 4{ 3_|3{=5 {=0= 4(5)3|3 4(0)3|3= 500|3, U1 0 12{2|3g| =  12{2|4 4 |=1 |=0= 3{ 2_|4|=1 |=0= 3{2(1)4 3{2(0)4= 3{2 3.U₁3U₀1(1 + 4{|) g{ g| =U₁3{ + 2{2|{=1 {=0g| = U3 1(1 + 2|) g| =  | + |23 1= (3 + 9)  (1 + 1) = 10

5.U₀2U₀@2{ sin | g| g{ =U₀2{ g{U₀@2sin | g| [as in Example 5] =  {2 2 2 0 k  cos |l@2 0 = (2  0)(0 + 1) = 2 7.U₀2U₀1(2{ + |)8g{ g| = ] 2 0  1 2(2{ + |) 9 9 {=1 {=0 g| [substitutex = 2{ + |  g{ =1₂gx] = 1₁₈] 2 0 [(2 + |) 9_{ (0 + |)}9_{] g| = 1} 18  (2 + |)10 10  | 10 10 2 0 = 1 180[(410 210)  (210 010)] =1,046,528180 = 261,63245 9. ] 4 1 ] 2 1  { | + |{  g| g{ =] 4 1  { ln ||| + 1_{· 1₂|2|=2 |=1g{ = ] 4 1  { ln 2 + 3_2{  g{ =1 2{2ln 2 +32ln |{| 4 1 = 8 ln 2 +3 2ln 4 12ln 2 = 152 ln 2 + 3 ln 41@2= 212 ln 2

TX.10

11. U₀1U₀1(x  y)5gx gy =U₀11₆(x  y)6x=1_x=0gy = 1₆U₀1(1  y)6 (0  y)6gy =1 6 U1 0  (1  y)6_{ y}6_{gy =} 1 6  1 7(1  y)717y7 1 0 = 1 42[(0 + 1)  (1 + 0)] = 0

13. U₀2U₀u sin2 g gu =U₀2u guU₀sin2 g [as in Example 5] =U₀2u guU₀1₂(1  cos 2) g =1 2u2 2 0·12   1 2sin 2  0 = (2  0) ·12   1 2sin 2  0 1 2sin 0  = 2 ·1 2[(  0)  (0  0)] =  15. UU_U(6{2|3 5|4) gD =U₀3U₀1(6{2|3 5|4) g| g{ =U₀33₂{2|4 |5_|=0|=1g{ =U₀33₂{2 1g{ =1 2{3 { 3 0=272  3 = 212 17. ]] U {|2 {2_{+ 1}gD = ] 1 0 ] 3 3 {|2 {2_{+ 1}g| g{ = ] 1 0 { {2_{+ 1}g{ ] 3 3| 2_{g| =}1 2ln({2+ 1) 1 0  1 3|3 3 3 =1 2(ln 2  ln 1) ·13(27 + 27) = 9 ln 2 19. U₀@6U₀@3{ sin({ + |) g| g{ =U@6 0  { cos({ + |)| = @3_{| = 0} g{ =U@6 0  { cos {  { cos{ + 3  g{ = {sin {  sin{ + 3 @6 0  U@6 0  sin {  sin{ + 3 

g{ [by integrating by parts separately for each term] = 6 ₁ 2  1   cos { + cos{ + 3 @6 0 = 12  k 3 2 + 0   1 +1 2 l = 31 2 12 21. UU_U{|h{2|gD=U₀2U₀1{|h{2|g{ g| =U₀2k1₂h{2|l{=1 {=0 g| = 1 2 U2 0(h| 1) g| =12  h|_{ |}2 0 =1 2[(h2 2)  (1  0)] =12(h2 3)

23. } = i({> |) = 4  {  2|  0 for 0  {  1 and 0  |  1. So the solid is the region in the rst octant which lies below the plane} = 4  {  2| and above[0> 1] × [0> 1]. 25. Y =UU_U(12  3{  2|) gD =U_23 U₀1(12  3{  2|) g{ g| =U_23 12{ 3₂{2 2{|{=1_{=0g| =U3 2 ₂₁ 2  2|  g| =21 2|  |2 3 2=952 27. Y =U_22 U_11 1 1₄{21₉|2g{ g| = 4U₀2U₀11 ₄1{21₉|2g{ g| = 4U₀2{  1 12{319|2{ { = 1 { = 0g| = 4 U2 0 ₁₁ 1219|2  g| = 411 12| 271|3 2 0= 4 · 8354= 16627

SECTION 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS ET SECTION 15.3 ¤ 227 29.Here we need the volume of the solid lying under the surface} = { sec2| and above the rectangle U = [0> 2] × [0> @4] in the

{|-plane. Y =U2 0 U@4 0 { sec2| g| g{ = U2 0 { g{ U@4 0 sec2| g| = ₁ 2{2 2 0  tan |@4₀ = (2  0)(tan 4  tan 0) = 2(1  0) = 2

31.The solid lies below the surface} = 2 + {2+ (|  2)2and above the plane} = 1 for 1  {  1, 0  |  4. The volume of the solid is the difference in volumes between the solid that lies under} = 2 + {2+ (|  2)2over the rectangle

U = [1> 1] × [0> 4] and the solid that lies under } = 1 over U. Y =U₀4U_11 [2 + {2_{+ (|  2)}2_{] g{ g| }U4 0 U1 1(1) g{ g| = U4 0  2{ +1 3{3+ {(|  2)2 { = 1 { = 1g|  U1 1g{ U4 0g| =U₀4(2 +1 3+ (|  2)2)  (2 13 (|  2)2)  g|  [{]1 1[|]40 =U₀414 3 + 2(|  2)2  g|  [1  (1)][4  0] =14 3| +23(|  2)3 4 0 (2)(4) =56 3 +163  0 16 3   8 =88 3  8 = 643

33.In Maple, we can calculate the integral by dening the integrand as f and then using the command int(int(f,x=0..1),y=0..1);. In Mathematica, we can use the command

Integrate[f,{x,0,1},{y,0,1}]

We nd thatUU_U{5|3h{|gD = 21h  57  0=0839. We can use plot3d (in Maple) or Plot3D (in Mathematica) to graph the function.

35.U is the rectangle [1> 1] × [0> 5]. Thus, D(U) = 2 · 5 = 10 and iave= 1 D(U) UU Ui({> |) gD =101 U5 0 U1 1{2| g{ g| = 101 U5 0 ₁ 3{3| { = 1 { = 1g| =101 U5 0 23| g| =101 ₁ 3|2 5 0= 56. 37.Leti({> |) = {  |

({ + |)3. Then a CAS gives

U1 0 U1 0 i({> |) g| g{ =12 and U1 0 U1 0 i({> |) g{ g| = 12.

To explain the seeming violation of Fubini’s Theorem, note thati has an innite discontinuity at (0> 0) 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.

16.3

Double Integrals over General Regions

ET 15.3

1.U₀4U₀|{|2g{ g| =U₀4₂1{2|2{=|_{=0 g| =U₀41₂|2[(s| )2 02]g| = 1₂U₀4|3g| = 1₂1₄|44₀=1₂(64  0) = 32 3.U₀1U_{{2(1 + 2|)g| g{ =U₀1| + |2|={_|={2g{ = U1 0  { + {2_{ {}2_{ ({}2₎2_g{ =U₀1({  {4_{)g{ =}1 2{215{5 1 0= 1215 0 + 0 = 103

5.U₀@2U₀cos hsin gu g =U₀@2uhsin u=cos _u=0 g =U₀@2(cos ) hsin g = hsin @2₀ = hsin(@2) h0= h  1 6.3.3 DOUDOU

7. UU_G|2gD =U_11 U_|2| |2g{ g| =U_11 {|2{=|_{=|2g| =U_11 |2[|  (|  2)] g| =U1 1(2|3+ 2|2)g| = ₁ 2|4+23|3 1 1= 12+2312+23 =43

9. UU_G{ gD =U₀U₀sin {{ g| g{ =U₀[{|]|=sin {_|=0 g{ =U₀{ sin { g{ 

integrate by parts withx = {> gy = sin { g{

 ={ cos { + sin {₀ =  cos  + sin  + 0  sin 0 = 

11. UU_G|2h{|gD =U₀4U₀||2h{|g{ g| =U₀4|h{|{=|_{=0g| =U₀4  |h|2_{ |}_g| =k1 2h| 2 1 2|2 l4 0= 1 2h16 8 12+ 0 = 12h16172 13. U₀1U₀{2{ cos | g| g{ =U₀1{ sin || = {2 | = 0 g{ = U1 0 { sin {2g{ = 12cos {2 1 0=12(1  cos 1) 15. ] 2 1 ] 2|1 2| | 3_{g{ g| =}] 2 1 k {|3l{=2|1 {=2| g| = ] 2 1 [(2|  1)  (2  |)] | 3_g| =U₁2(3|4_{ 3|}3_{) g| =}3 5|534|4 2 1 =96 5  12 35 +34 = 14720 17. ] 2 2 ] 4{2 4{2(2{  |) g| g{ =] 2 2 k 2{| 1 2|2 l|=4{2 |=4{2g{ =U_22 2{4  {2_1 2  4  {2_{+ 2{}_{4  {}2₊1 2  4  {2_g{ =U_22 4{4  {2_{g{ = }4 3  4  {23@2l2 2= 0

[Or, note that4{4  {2is an odd function, soU_22 4{4  {2g{ = 0.]

19. Y =U₀1U_{{4({ + 2|) g| g{ =U1 0  {| + |2|={ |={4 g{ = U₁ 0(2{2 {5 {8) g{ =2 3{316{619{9 1 0= 231619 =187 21. Y =U₁2U₁7  3|{| g{ g| =U₁21₂{2|{ = 7  3|_{{ = 1} g| =1 2 U2 1(48|  42|2+ 9|3) g| =1 2  24|2_{ 14|}3₊9 4|4 2 1= 318

SECTION 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS ET SECTION 15.3 ¤ 229 23. Y =U₀2U3 32{ 0 (6  3{  2|) g| g{ =U₀26|  3{|  |2| = 3 3 2{ | = 0 g{ =U2 0  6(3  3 2{)  3{(3  32{)  (3  32{)2  g{ =U₀29 4{2 9{ + 9  g{ =3 4{392{2+ 9{ 2 0= 6  0 = 6 25. Y =U_22 U_{42{2g| g{ =U_22 {2_||=4 |={2 g{ = U2 2(4{2 {4) g{ =4 3{315{5 2 2=323 325 +323 325 = 12815 27. Y =] 1 0 ] 1  {2 0 | g| g{ = ] 1 0  |2 2 | =1  {2 | = 0 g{ =] 1 0 1  {2 2 g{ =12  { 1 3{3 1 0= 13

29. From the graph, it appears that the two curves intersect at{ = 0 and

at{  1=213. Thus the desired integral is UU G{ gD  U1=213 0 U3{  {2 {4 { g| g{ = U1=213 0 k {|l| = 3{  {2 | = {4 g{ =U₀1=213(3{2_{ {}3_{ {}5_{) g{ =}_{3_1 4{416{6 1=213 0  0=713

31.The two bounding curves| = 1  {2and| = {2 1 intersect at (±1> 0) with 1  {2 {2 1 on [1> 1]. Within this region, the plane} = 2{ + 2| + 10 is above the plane } = 2  {  |, so

Y =U_11 U_{1{2_12(2{ + 2| + 10) g| g{  U1 1 U1{2 {2_1(2  {  |) g| g{ =U1 1 U1{2 {2_1(2{ + 2| + 10  (2  {  |)) g| g{ =U_11 U_{1{2_12(3{ + 3| + 8) g| g{ = U1 1 k 3{| +3 2|2+ 8| l|=1{2 |={2_1 g{ =U_11 3{(1  {2_{) +}3 2(1  {2)2+ 8(1  {2)  3{({2 1) 32({2 1)2 8({2 1)  g{ =U_11 (6{3_{ 16{}2_{+ 6{ + 16) g{ =}_3 2{4163{3+ 3{2+ 16{ 1 1 = 3 2163 + 3 + 16 +32163  3 + 16 = 643

TX.10

6.3.3 DOUDOU

33. The solid lies below the plane} = 1  {  | or{ + | + } = 1 and above the region G = {({> |) | 0  {  1> 0  |  1  {} in the{|-plane. The solid is a tetrahedron.

35. The two bounding curves| = {3 { and | = {2+ { intersect at the origin and at { = 2, with {2+ { A {3 { on (0> 2). Using a CAS, we nd that the volume is

Y = ] 2 0 ]{2_{+ {} {3_{ {} } g| g{ = ] 2 0 ]{2_{+ {} {3_{ {} ({ 3_|4_{+ {|}2_{) g| g{ = 13},984,735,616 14,549,535

37. The two surfaces intersect in the circle{2+ |2= 1, } = 0 and the region of integration is the disk G: {2+ |2 1. Using a CAS, the volume is

]] G(1  { 2_{ |}2_{) gD =}] 1 1 ] 1{2 1{2(1  { 2_{ |}2_{) g| g{ =} 2.

39. Because the region of integration is

G = {({> |) | 0  | {> 0  {  4} =({> |) | |2_{ {  4> 0  |  2}

we haveU₀4U₀{i({> |) g| g{ =UU_G i({> |) gD =U₀2U_|4₂ i({> |) g{ g|.

41. Because the region of integration is

G =q({> |) | s9  |2_{ { }s_{9  |}2_{> 0  |  3}r =({> |) | 0  | 9  {2_{> 3  {  3} we have ] 3 0 ] 9|2 9|2 i({> |) g{ g| = ]] Gi({> |) gD = ] 3 3 ] 9{2 0 i({> |) g| g{

43. Because the region of integration is

G = {({> |) | 0  |  ln {, 1  {  2} = {({> |) | h|_{ {  2, 0  |  ln 2}} we have ] 2 1 ] ln { 0 i({> |) g| g{ = ]] Gi({> |) gD = ]ln 2 0 ] 2 h|i({> |) g{ g| 45. ]1 0 ]3 3|h {2_{g{ g| =}] 3 0 ] {@3 0 h {2_{g| g{ =}] 3 0 k h{2_|l|={@3 |=0 g{ =] 3 0 { 3  h{2 g{ =1 6 h{ 2l3 0= h 9_{ 1} 6

SECTION 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS ET SECTION 15.3 ¤ 231 47. ] ₄ 0 ] 2  { 1 |3_{+ 1}g| g{ = ]2 0 ]|2 0 1 |3_{+ 1}g{ g| =]2 0 1 |3_{+ 1}  {{=|_{=02g| =] 2 0 |2 |3_{+ 1}g| = 1 3ln|3+ 1l 2 0= 1 3(ln 9  ln 1) =13ln 9 49. ] 1 0 ] @2 arcsin |cos { s 1 + cos2_{{ g{ g|}

=U₀@2U₀sin {cos {1 + cos2_{{ g| g{}

=U₀@2cos {1 + cos2_{_||=sin { |=0 g{

=U₀@2cos {1 + cos2_{{ sin { g{}



Letx = cos {, gx =  sin { g{, g{ = gx@( sin {)  =U₁0x1 + x2_{gx = }1 3  1 + x23@2l0 1 =1 3  8  1=1 3  22  1 51.G = {({> |) | 0  {  1,  { + 1  |  1}  {({> |) | 1  {  0, { + 1  |  1}  {({> |) | 0  {  1,  1  |  {  1}  {({> |) | 1  {  0,  1  |  {  1}, all type I. ]] G{ 2_{gD =}] 1 0 ] 1 1  {{ 2_{g| g{ +}] 0 1 ] 1 { + 1{ 2_{g| g{ +}] 1 0 ] {  1 1 { 2_{g| g{ +}] 0 1 ] {  1 1 { 2_{g| g{} = 4] 1 0 ] 1 1  {{

2_{g| g{ [by symmetry of the regions and becausei({> |) = {}2_{ 0]}

= 4U₀1{3_{g{ = 4}1 4{4 1 0= 1 53.HereT =({> |) | {2+ |2 1₄> {  0> |  0, and0  ({2+ |2)21₄2  ₁₆1  ({2+ |2)2 0 so h1@16_{ h}({2_+|2₎2

 h0_{= 1 since h}w_{is an increasing function. We haveD(T) =}1 4 ₁ 2 2₌  16, so by Property 11, h1@16_{D(T) }UU Th({ 2_+|2₎2 gD  1 · D(T)   16h1@16 UU Th({ 2_+|2₎2 gD   16or we can say 0=1844 ?UU_Th({2_+|2₎2

gD ? 0=1964. (We have rounded the lower bound down and the upper bound up to preserve the inequalities.)

55.The average value of a functioni of two variables dened on a rectangle U was dened in Section 16.1 [ET 15.1] asiave= _D(U)1

UU

Ui({> |)gD. Extending

this denition to general regionsG, we have iave=_D(G)1

UU Gi({> |)gD. HereG = {({> |) | 0  {  1> 0  |  3{}, so D(G) =1₂(1)(3) = 3₂ and iave= _D(G)1 UU Gi({> |)gD =3@21 U1 0 U3{ 0 {| g| g{ = 2 3 U1 0 ₁ 2{|2 |=3{ |=0 g{ =13 U1 0 9{3g{ = 34{4 1 0=34

57.Sincep  i({> |)  P,UU_Gp gD UU_Gi({> |) gD UU_GPgD by (8) 

pUU_G1 gD UU_Gi({> |) gD  PUU_G1 gD by (7)  pD(G) UU_Gi({> |) gD  PD(G) by (10). 6.3.3 DOUDOU

59. UU_G({2tan { + |3+ 4) gD =UU_G{2tan { gD +UU_G|3gD +UU_G4 gD. But {2tan { is an odd function of { and G is symmetric with respect to the|-axis, soUU_G{2tan { gD = 0. Similarly, |3is an odd function of| and G is symmetric with respect to the{-axis, soUU_G|3gD = 0. Thus

UU G({2tan { + |3+ 4) gD = 4 UU G gD = 4(area of G) = 4 ·   22 = 8

61. Sinces1  {2 |2 0, we can interpretUU_Gs1  {2 |2gD as the volume of the solid that lies below the graph of } =s1  {2_{ |}2_{and above the regionG in the {|-plane. } =}s_{1  {}2_{ |}2_{is equivalent to{}2_{+ |}2_{+ }}2_{= 1, }  0}

which meets the{|-plane in the circle {2+ |2= 1, the boundary of G. Thus, the solid is an upper hemisphere of radius 1 which has volume1₂4₃(1)3=2₃.

16.4

Double Integrals in Polar Coordinates

ET 15.4

1. The regionU is more easily described by polar coordinates: U =(u> ) | 0  u  4, 0    3₂ . ThusUU_Ui({> |) gD =U₀3@2U₀4i(u cos > u sin ) u gu g.

3. The regionU is more easily described by rectangular coordinates: U =({> |) | 1  {  1, 0  | 1₂{ +1₂. ThusUU_Ui({> |) gD =U_11 U₀({+1)@2i({> |) g| g{.

5. The integralU2U₄7 u gu g represents the area of the region U = {(u> ) | 4  u  7,     2}, the lower half of a ring.

U2  U7 4 u gu g = U2  g U7 4 u gu  =2 1 2u2 7 4=  ·12(49  16) =332

7. The diskG can be described in polar coordinates as G = {(u> ) | 0  u  3, 0    2}. Then UU

G {| gD =

U2 0

U3

0 (u cos )(u sin ) u gu g =

U2 0 sin  cos  g U3 0 u3gu  =1 2sin2 2 0 ₁ 4u4 3 0= 0.

9. UU_Ucos({2+ |2) gD =U₀U₀3cos(u2) u gu g =U₀gU₀3 u cos(u2) gu  =₀ 1

2sin(u2)

3

0=  ·12(sin 9  sin 0) =2sin 9

11. UU_Gh{2|2gD =U_@2@2 U₀2hu2u gu g =U_@2@2 gU₀2uhu2gu  =@2_@2k1 2hu 2l2 0=   1 2  (h4_{ h}0_{) =}  2(1  h4)

13. U is the region shown in the gure, and can be described byU = {(u> ) | 0    @4> 1  u  2}. Thus UU Uarctan(|@{) gD = U@4 0 U2

1 arctan(tan ) u gu g since |@{ = tan .

Also, arctan(tan ) =  for 0    @4, so the integral becomes U@4 0 U2 1  u gu g = U@4 0  g U2 1 u gu = ₁ 22 @4 0 ₁ 2u2 2 1= 2 32 ·32 =6432.

SECTION 16.4 DOUBLE INTEGRALS IN POLAR COORDINATES ET SECTION 15.4 ¤ 233 15.One loop is given by the region

G = {(u> ) |@6    @6, 0  u  cos 3 }, so the area is ]] GgD = ] @6 @6 ]cos 3 0 u gu g = ]@6 @6  1 2u2 u=cos 3 u=0 g =] @6 @6 1 2cos23 g = 2 ] @6 0 1 2  1 + cos 6 2  g = 1₂   + 1₆sin 6 @6 0 = 12 17.By symmetry, D = 2U₀@4U₀sin u gu g = 2U₀@41 2u2 u=sin  u=0 g =U₀@4sin2_{ g =}U@4 0 12(1  cos 2) g = 1 2   1 2sin 2 @4 0 = 1 2  4 12sin2  0 +12sin 0  =1 8(  2) 19.Y =UU_{{2+ |24}s{2+ |2gD =U₀2U₀2u2u gu g =U₀2gU₀2u2gu =2₀ 1₃u32₀= 28₃=16₃

21.The hyperboloid of two sheets{2 |2+ }2= 1 intersects the plane } = 2 when {2 |2+ 4 = 1 or {2+ |2= 3. So the solid region lies above the surface} =s1 + {2+ |2and below the plane} = 2 for {2+ |2 3, and its volume is

Y = ]] {2_{+ |}2_{ 3}  2 s1 + {2_{+ |}2_{gD =} ] 2 0 ]  3 0 (2  s 1 + u2_{) u gu g} =U₀2gU₀3(2u  u1 + u2_{)gu =}_2 0 k u2_1 3(1 + u2)3@2 l 3 0 = 23 8 3 0 +13  = 4 3 23.By symmetry, Y = 2 ]] {2_{+ |}2_{ d}2 s d2_{ {}2_{ |}2_{gD = 2} ] 2 0 ] d 0 s d2_{ u}2_{u gu g = 2} ] 2 0 g ]d 0 u s d2_{ u}2_gu = 22₀ k1 3(d2 u2)3@2 ld 0 = 2(2)  0 +1 3d3  = 4 3 d3

25.The cone} =s{2+ |2intersects the sphere{2+ |2+ }2= 1 when {2+ |2+s{2+ |22= 1 or {2+ |2=1₂. So

Y = ]] {2_{+ |}2_{ 1@2} s 1  {2_{ |}2_s_{2_{+ |}2_{gD =} ] 2 0 ] 1@2 0 s 1  u2_{ u}_{u gu g} =U₀2gU₀1@2u1  u2_{ u}2_{gu =}_2 0 k 1 3(1  u2)3@213u3 l1@2 0 = 2  1 3  ₁  2 1  =  3  2 2

TX.10

16.46.4 DODO

27. The given solid is the region inside the cylinder{2+ |2= 4 between the surfaces } =s64  4{2 4|2 and} = s64  4{2 4|2. So Y = ]] {2_{+ |}2_{ 4} ks 64  4{2_{ 4|}2__s_{64  4{}2_{ 4|}2l_{gD =} ]] {2_+|2_{ 4} 2s64  4{2_{ 4|}2_gD = 4U₀2U₀216  u2_{u gu g = 4}U2 0 g U2 0 u  16  u2_{gu = 4}_2 0 k 1 3(16  u2)3@2 l2 0 = 81 3  (123@2_{ 16}2@3_{) =}8 3  64  243 29. ] 3 3 ] 9{2 0 sin({ 2_{+ |}2_{)g| g{ =}]  0 ] 3 0 sin  u2_{u gu g} =U₀gU₀3u sinu2_{gu = []} 0  1 2cos  u23 0 = 1 2  (cos 9  1) =  2 (1  cos 9) 31. U₀@4U  2 0 (u cos  + u sin ) u gu g = U@4 0 (cos  + sin ) g U 2 0 u2gu = [sin   cos ]@4 0 ₁ 3u3  2 0 =k2 2   2 2  0 + 1 l ·1 3  22  0=22 3

33. The surface of the water in the pool is a circular diskG with radius 20 ft. If we place G on coordinate axes with the origin at the center ofG and dene i({> |) to be the depth of the water at ({> |), then the volume of water in the pool is the volume of the solid that lies aboveG =({> |) | {2+ |2 400and below the graph ofi({> |). We can associate north with the positive|-direction, so we are given that the depth is constant in the {-direction and the depth increases linearly in the |-direction from i(0> 20) = 2 to i(0> 20) = 7. The trace in the |}-plane is a line segment from (0> 20> 2) to (0> 20> 7). The slope of this line is_{20  (20)}7  2 =1₈, so an equation of the line is}  7 =₈1(|  20)  } = 1₈| +9₂. Sincei({> |) is independent of{, i({> |) = 1₈| +9₂. Thus the volume is given byUU_Gi({> |) gD, which is most conveniently evaluated using polar coordinates. ThenG = {(u> ) | 0  u  20, 0    2} and substituting { = u cos , | = u sin  the integral becomes U2 0 U20 0 ₁ 8u sin  +92  u gu g =U₀21 24u3sin  +94u2 u = 20 u = 0 g = U2 0 ₁₀₀₀ 3 sin  + 900  g =1000 3 cos  + 900 2 0 = 1800

Thus the pool contains1800  5655 ft3of water.

35. ] 1 1@2 ]{  1  {2 {| g| g{ + ]  2 1 ] { 0 {| g| g{ + ] 2  2 ] 4  {2 0 {| g| g{ = ] @4 0 ] 2 1 u 3_{cos  sin  gu g =}] @4 0  u4 4 cos  sin  u = 2 u = 1g = 15 4 ] @4 0 sin  cos  g = 154  sin2_ 2 @4 0 = 1516

SECTION 16.5 APPLICATIONS OF DOUBLE INTEGRALS ET SECTION 15.5 ¤ 235 37. (a) We integrate by parts withx = { and gy = {h{2g{. Then gx = g{ and y = 1₂h{2, so

U 0 {2h{ 2 g{ = lim w Uw 0{2h{ 2 g{ = lim w  1 2{h{ 2lw 0+ Uw 0 12h{ 2 g{  = lim_w1 2whw 2 +1 2 U 0 h{ 2 g{ = 0 +1 2 U 0 h{ 2

g{ [by l’Hospital’s Rule] =1

4

U_

h{

2

g{ [sinceh{2is an even function] =1 4   [by Exercise 36(c)] (b) Letx ={. Then x2= {  g{ = 2x gx  U 0  {h{_{g{ = lim} w Uw 0  { h{_{g{ = lim} w U w 0 xhx 2 2x gx = 2U₀x2_hx2 gx = 21 4   [by part(a)] =1₂.

16.5

Applications of Double Integrals

ET 15.5

1.T =UU_G({> |) gD =U₁3U₀2(2{| + |2) g| g{ =U₁3{|2+1₃|3|=2_|=0 g{ =U₁34{ +8 3  g{ =2{2₊8 3{ 3 1= 16 +163 =643 C 3.p =UU_G({> |) gD =U₀2U_11 {|2g| g{ =U₀2 { g{U_11 |2g| =1₂{22₀1₃|31_1= 2 · 2₃ =4₃, { = 1 p UU G {({> |) gD = 34 U2 0 U1 1{2|2g| g{ =34 U2 0 {2g{ U1 1|2g| =34 ₁ 3{3 2 0 ₁ 3|3 1 1=34·83 ·23 = 43, | = 1 p UU G |({> |) gD =34 U2 0 U1 1{|3g| g{ =34 U2 0 { g{ U1 1|3g| = 34 ₁ 2{2 2 0 ₁ 4|4 1 1= 34· 2 · 0 = 0. Hence,({> |) =4₃> 0. 5.p =U₀2U_{@23{({ + |) g| g{ =U₀2{| +1₂|2|=3{_|={@2g{ =U₀2{3 3₂{+1₂(3  {)21₈{2g{ =U₀29 8{2+92  g{ =9 8 ₁ 3{3  +9 2{ 2 0= 6, P|=U₀2U_{@23{({2+ {|) g| g{ =U₀2{2| +1₂{|2|=3{_|={@2 g{ =U₀2₂9{ 9₈{3g{ =9₂, P{=U₀2U_{@23|({| + |2) g| g{ =U₀21₂{|2+1₃|3|=3{_|={@2 g{ =U₀29 9₂{g{ = 9. Hencep = 6, ({> |) =  P| p > Pp{  =  3 4> 32  . 7.p =U₀1U₀h{| g| g{ =U₀11₂|2|=h_|=0{g{ =1₂U₀1h2{g{ = 1₄h2{1₀=1₄(h2 1), P|=U₀1Uh { 0 {| g| g{ =12 U1 0 {h2{g{ =12 ₁ 2{h2{14h2{ 1 0= 18(h2+ 1), P{=U₀1Uh { 0 |2g| g{ = U1 0 ₁ 3|3 |=h{ |=0 g{ =13 U1 0 h3{g{ =13 ₁ 3h3{ 1 0=19(h3 1). Hencep = 1₄(h2 1), ({> |) = #₁ 8(h2+ 1) 1 4(h2 1) > 19(h3 1) 1 4(h2 1) $ =  h2_{+ 1} 2(h2_{ 1)}> 4(h 3_{ 1)} 9(h2_{ 1)}  .

TX 10

CTION 16TION 16.

9. Note thatsin({@O)  0 for 0  {  O. p =U₀OU₀sin({@O)| g| g{ =U₀O1 2sin2({@O) g{ = 12 ₁ 2{ 4O sin(2{@O) O 0 =14O,

P|=U₀OU₀sin({@O){ · | g| g{ =1₂U₀O{ sin2({@O) g{



integrate by parts with x = {> gy = sin2_{({@O) g{}  = 1 2 · { ₁ 2{ 4O sin(2{@O) O 0 12 UO 0 ₁ 2{ 4O sin(2{@O)  g{ =1 4O212 k 1 4{2+ O 2 42cos(2{@O) lO 0 = 1 4O212  1 4O2+ O 2 42  O 2 42  = 1 8O2>

P{=U₀OU₀sin({@O)| · | g| g{ =U₀O₃1sin3({@O) g{ = 1₃U₀O1  cos2({@O)sin({@O) g{

[substitutex = cos ({@O)]  gx = _Osin({@O)] = 1 3  O   cos({@O) 1 3cos3({@O) O 0 = 3O  1 +1 3  1 +13  = 4 9O. Hencep = O 4,({> |) =  O2_@8

O@4> 4O@(9)O@4  =  O 2> 169  .

11. ({> |) = n| = nu sin , p =U₀@2U₀1nu2sin  gu g =1₃nU₀@2sin  g =1₃n cos @2₀ = 1₃n, P|=U₀@2U₀1nu3sin  cos  gu g =1₄nU₀@2sin  cos  g = 1₈n cos 2@2₀ = 1₈n,

P{=U₀@2U₀1nu3sin2 gu g =1₄nU₀@2sin2 g =1₈n + sin 2@2₀ = ₁₆n.

Hence({> |) =3₈>3₁₆. 13. ({> |) = ns{2+ |2= nu, p =UU_G({> |)gD =U₀U₁2nu · u gu g = nU₀gU₁2u2_{gu = n()}1 3u3 2 1=73n,

P|=UU_G{({> |)gD =U₀U₁2(u cos )(nu) u gu g = nU₀cos  gU₁2u3gu

= nsin ₀ 1 4u4 2 1= n(0) ₁₅ 4 

= 0 [this is to be expected as the region and density

function are symmetric about they-axis] P{=UU_G|({> |)gD =U₀U₁2(u sin )(nu) u gu g = nU₀sin  gU₁2u3gu

= n cos ₀ 1 4u4 2 1= n(1 + 1) ₁₅ 4  =15 2n= Hence({> |) =0>_7n@315n@2=0>₁₄45.

15. Placing the vertex opposite the hypotenuse at(0> 0), ({> |) = n({2+ |2). Then p =Ud 0 Ud  { 0 n  {2_{+ |}2_{g| g{ = n}Ud 0  d{2_{ {}3₊1 3(d  {)3  g{ = n1 3d{314{4121 (d  {)4 d 0 =16nd4. By symmetry, P|= P{ =U₀dU₀d  {n|({2+ |2) g| g{ = nU₀d1₂(d  {)2{2+1₄(d  {)4g{ = n1 6d2{314d{4+101{5201(d  {)5 d 0 =151nd5 Hence({> |) =2₅d>2₅d.

SECTION 16.5 APPLICATIONS OF DOUBLE INTEGRALS ET SECTION 15.5 ¤ 237 17.L_{ =UU_G|2({> |)gD =U₀1U₀h{|2· | g| g{ =U₀11₄|4_|=0|=h{g{ = 1₄U₀1h4{g{ =1₄1₄h4{1₀= ₁₆1(h4 1), L| =UU_G{2({> |) gD =U₀1Uh { 0 {2| g| g{ = U1 0 {2 ₁ 2|2 |=h{ |=0 g{ =12 U1 0 {2h2{g{ =1 2 ₁ 2{212{ +14  h2{1

0 [integrate by parts twice] = 18(h2 1),

andL₀= L_{+ L_|= ₁₆1(h4 1) +1₈(h2 1) =₁₆1(h4+ 2h2 3).

19.As in Exercise 15, we place the vertex opposite the hypotenuse at(0> 0) and the equal sides along the positive axes. L{=U₀dU₀d{|2n({2+ |2) g| g{ = nU₀dU₀d{({2|2+ |4) g| g{ = nU₀d1₃{2|3+1₅|5|=d{_|=0 g{ = nU₀d1 3{2(d  {)3+15(d  {)5  g{ = n1 3 ₁ 3d3{334d2{4+35d{516{6   1 30(d  {)6 d 0 =1807 nd6, L|=U₀dU₀d{{2n({2+ |2) g| g{ = nU₀dU₀d{({4+ {2|2) g| g{ = nU₀d{4| +1₃{2|3|=d{_|=0 g{ = nU₀d{4_{(d  {) +}1 3{2(d  {)3  g{ = n1 5d{516{6+13 ₁ 3d3{334d2{4+35d{516{6 d 0 =1807 nd6, andL0= L{+ L|= ₉₀7nd6.

21.Using a CAS, we ndp =UU_G({> |) gD =U₀U₀sin {{| g| g{ = 

2 8 . Then { = 1_p ]] G{({> |) gD = 82 ]  0 ] sin { 0 { 2_{| g| g{ = 2} 3  1 and | = 1_p ]] G|({> |) gD = 82 ] 0 ] sin { 0 {| 2_{g| g{ = 16} 9, so({> |) =  2 3  1> 169  .

The moments of inertia areL{ =UU_G|2({> |) gD =U₀U₀sin {{|3g| g{ = 3 2 64 , L|=UU_G{2({> |) gD =U₀U₀sin {{3| g| g{ =  2 16(2 3), and L0= L{+ L|=  2 64(42 9). 23.L{ =UU_G|2({> |)gD =U₀kU₀e|2g{ g| = U₀eg{U₀k|2g| = {e₀ 1₃|3k₀ = e1₃k3= 1₃ek3, L|=UU_G{2({> |)gD =U₀kU₀e{2g{ g| = U₀e{2g{U₀kg| = 1₃{3e₀ [|]k0 = 13e3k,

andp =  (area of rectangle) = ek since the lamina is homogeneous. Hence {2= L|

p = 1 3e3k ek = e 2 3  { = e3 and|2= L{ p = 1 3ek3 ek = k 2 3  | = k3.

25.In polar coordinates, the region isG =(u> ) | 0  u  d> 0   ₂, so L{=UU_G|2 gD =U₀@2U₀d(u sin )2u gu g = U₀@2sin2gU₀du3gu

= 1 2 14sin 2 @2 0 ₁ 4u4 d 0 =   4  ₁ 4d4  = 1 16d4,

L|=UU_G{2 gD =U₀@2U₀d(u cos )2u gu g = U₀@2cos2gU₀du3gu

= 1 2 +14sin 2 @2 0 ₁ 4u4 d 0 =   4  ₁ 4d4  = 1 16d4,

andp =  · D(G) =  ·1₄d2since the lamina is homogeneous. Hence{2= |2=

1 16d4 1 4d2 = d₄2  { = | = d₂.

TX 10

CTION 16TION 16.

27. (a)i({> |) is a joint density function, so we knowUU_R2i({> |) gD = 1. Since i({> |) = 0 outside the rectangle[0> 1] × [0> 2], we can say

UU R2i({> |) gD = U  U i({> |) g| g{ = U1 0 U2 0 F{(1 + |) g| g{ = FU₀1{| +1 2|2 |=2 |=0g{ = F U1 0 4{ g{ = F  2{21 0= 2F Then2F = 1  F = 1₂. (b)S ([  1> \  1) =U_1 U_1 i({> |) g| g{ =U₀1U₀11₂{(1 + |) g| g{ =U₀1 1 2{  | +1 2|2 | = 1 | = 0g{ = U1 0 12{ ₃ 2  g{ =3 4 ₁ 2{2 1 0= 38or0=375

(c)S ([ + \  1) = S (([> \ ) G) where G is the triangular region shown in the gure. Thus

S ([ + \  1) =UU_Gi({> |) gD =U1 0 U_{1  {} 0 12{(1 + |) g| g{ =U₀11 2{  | +1 2|2 | = 1{ | = 0 g{ = U1 0 12{ ₁ 2{2 2{ +32  g{ =1 4 U1 0  {3_{ 4{}2_{+ 3{}_{g{ =}1 4 k {4 4  4{ 3 3 + 3{ 2 2 l1 0 = 5 48  0=1042

29. (a)i({> |)  0, so i is a joint density function ifUU_R2i({> |) gD = 1. Here, i({> |) = 0 outside the rst quadrant, so UU R2i({> |) gD = U 0 U 0 0=1h(0=5{ + 0=2|)g| g{ = 0=1 U 0 U 0 h0=5{h0=2|g| g{ = 0=1 U 0 h0=5{g{ U 0 h0=2|g| = 0=1 lim_wU₀wh0=5{_{g{ lim} w Uw 0h0=2|g| = 0=1 lim_w  2h0=5{w 0wlim  5h0=2|w 0 = 0=1 lim w  2(h0=5w_{ 1)}_lim w  5(h0=2w_{ 1)}_{= (0=1) · (2)(0  1) · (5)(0  1) = 1}

Thusi({> |) is a joint density function. (b) (i) No restriction is placed on[, so

S (\  1) =U_ U₁i({> |) g| g{ =U₀U₁0=1h(0=5{+0=2|)_{g| g{} = 0=1U 0 h0=5{g{ U_ 1 h0=2|g| = 0=1 lim_w U_w 0h0=5{g{ lim_w U_w 1h0=2|g| = 0=1 lim_w2h0=5{w 0 wlim  5h0=2|w 1 = 0=1 limw  2(h0=5w_{ 1)} _lim w  5(h0=2w_{ h}0=2₎ (0=1) · (2)(0  1) · (5)(0  h0=2_{) = h}0=2_{ 0=8187} (ii)S ([  2> \  4) =U_2 U_4 i({> |) g| g{ =U₀2U₀40=1h(0=5{+0=2|)g| g{ = 0=1U₀2h0=5{_g{U4 0 h0=2|g| = 0=1  2h0=5{2 0  5h0=2|4 0 = (0=1) · (2)(h1_{ 1) · (5)(h}0=8_{ 1)} = (h1_{ 1)(h}0=8_{ 1) = 1 + h}1=8_{ h}0=8_{ h}1_{ 0=3481}

(c) The expected value of[ is given by 1=UUR2{ i({> |) gD = U 0 U 0 { k 0=1h(0=5{+0=2|)l_{g| g{} = 0=1U₀{h0=5{_g{U 0 h0=2|g| = 0=1 lim_w Uw 0{h0=5{g{ lim_w Uw 0h0=2|g|

SECTION 16.5 APPLICATIONS OF DOUBLE INTEGRALS ET SECTION 15.5 ¤ 239 in the Table of Integrals):U{h0=5{g{ = 2{h0=5{U2h0=5{g{ = 2{h0=5{ 4h0=5{= 2({ + 2)h0=5{. Thus 1= 0=1 lim_w  2({ + 2)h0=5{w 0wlim  5h0=2|w 0 = 0=1 lim w(2)  (w + 2)h0=5w_{ 2} _lim w(5)  h0=2w_{ 1} = 0=1(2)  lim w w + 2 h0=5w  2 

(5)(1) = 2 [by l’Hospital’s Rule] The expected value of\ is given by

2= UU R2| i({> |) gD = U 0 U 0 | k 0=1h(0=5 +0=2|)l_{g| g{} = 0=1U₀h0=5{_g{U 0 |h0=2|g| = 0=1 lim_w Uw 0h0=5{g{ lim_w Uw 0|h0=2|g|

To evaluate the second integral, we integrate by parts withx = | and gy = h0=2|g| (or again we can use Formula 96 in the Table of Integrals) which givesU|h0=2|g| = 5|h0=2|+U5h0=2|g| = 5(| + 5)h0=2|. Then

2= 0=1 lim_w2h0=5{w0 _wlim  5(| + 5)h0=2|w 0 = 0=1 lim w  2(h0=5w_{ 1)} _lim w  5(w + 5)h0=2w_{ 5} = 0=1(2)(1) · (5)  lim w w + 5 h0=2w  5 

= 5 [by l’Hospital’s Rule] 31. (a) The random variables[ and \ are normally distributed with ₁= 45, ₂= 20, ₁= 0=5, and ₂= 0=1.

The individual density functions for[ and \ , then, are i1({) = 1

0=52h({45) 2_@0=5 and i2(|) = 1 0=12h (|20)2_@0=02

. Since[ and \ are independent, the joint density function is the product i({> |) = i1({)i2(|) = 1 0=52h({45) 2_@0=5 1 0=12h(|20) 2_@0=02 = 10 h2({45) 2_50(|20)2 = ThenS (40  [  50, 20  \  25) =U₄₀50U₂₀25i({> |) g| g{ =10 U₄₀50U₂₀25h2({45)250(|20)2g| g{. Using a CAS or calculator to evaluate the integral, we getS (40  [  50, 20  \  25)  0=500.

(b)S (4([  45)2+ 100(\  20)2 2) =UU_G 10h2({45)250(|20)2gD, where G is the region enclosed by the ellipse 4({  45)2_{+ 100(|  20)}2_{= 2. Solving for | gives | = 20 ±} 1

10

s

2  4({  45)2_{, the upper and lower halves of the}

ellipse, and these two halves meet where| = 20 [since the ellipse is centered at (45> 20)]  4({  45)2= 2  { = 45 ±_1 2. Thus ]] G 10 h2({45) 2_50(|20)2 gD = 10  ]45+1@2 451@2 ] 20+1 102  4({45)2 201 10  2  4({45)2 h 2({45)2_50(|20)2 g| g{. Using a CAS or calculator to evaluate the integral, we getS (4([  45)2+ 100(\  20)2 2)  0=632.

33. (a) Ifi(S> D) is the probability that an individual at D will be infected by an individual at S , and n gD is the number of infected individuals in an element of areagD, then i(S> D)n gD is the number of infections that should result from exposure of the individual atD to infected people in the element of area gD. Integration over G gives the number of infections of the person atD due to all the infected people in G. In rectangular coordinates (with the origin at the city’s

TX 10

center), the exposure of a person atD is H = ]] Gni(S> D) gD = n ]] G 20  g(S> D) 20 gD = n ]] G % 1  s ({  {0)2+ (|  |0)2 20 & g{ g| (b) IfD = (0> 0), then H = n ]] G  1  1₂₀s{2_{+ |}2  g{ g| = n ] 2 0 ] 10 0  1  u₂₀u gu g = 2n  u2 2  u 3 60 10 0 = 2n50 50 3  =200 3 n  209n

ForD at the edge of the city, it is convenient to use a polar coordinate system centered at D. Then the polar equation for the circular boundary of the city becomesu = 20 cos  instead of u = 10, and the distance from D to a point S in the city is againu (see the gure). So

H = n] @2 @2 ] 20 cos  0  1  u 20  u gu g = n] @2 @2  u2 2  u 3 60 u=20 cos  u=0 g = nU_@2@2 200 cos2_{ }400 3 cos3  g = 200nU_@2@2 1 2 +12cos 2 23  1  sin2__{cos }_g = 200n1

2 +14sin 2 23sin  +23·13sin3

@2 @2= 200n  4 + 0 23 +29 +4 + 0 23 +29  = 200n 2 89   136n

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.

16.6

Triple Integrals

ET 15.6

1. UUU_E{|}2gY =U₀1U₀3 U_12 {|}2g| g} g{ =U₀1U₀31₂{|2}2|=2_|=1g} g{ =U₀1U₀33₂{}2g} g{ =U₀11 2{}3 }=3 }=0g{ = U1 0 272 { g{ = 274{2 1 0=274 3. U₀1U₀}U₀{+}6{} g| g{ g} =U₀1U₀}6{|}|={+}_|=0 g{ g} =U₀1U₀}6{}({ + }) g{ g} =U₀12{3_{} + 3{}2_}2{=} {=0g} = U1 0(2}4+ 3}4) g} = U1 0 5}4g} = }5 1 0= 1 5. U₀3U₀1U ₀ 1}2}h|g{ g} g| =U₀3U₀1{}h|{=  1}2 {=0 g} g| = U3 0 U1 0 }h|  1  }2_{g} g|} =U₀3k1 3(1  }2)3@2h| l}=1 }=0g| = U3 0 13h|g| = 13h| 3 0=13(h3 1) 7. U₀@2U₀|U₀{cos({ + | + })g} g{ g| =U₀@2U₀|sin({ + | + })}={_}=0g{ g| =U@2 0 U| 0 [sin(2{ + |)  sin({ + |)] g{ g| =U₀@21 2cos(2{ + |) + cos({ + |) {=| {=0g| =U@2 0  1

2cos 3| + cos 2| +12cos |  cos |

 g| =1

6sin 3| +12sin 2|  12sin |

@2 0 =16 12 = 13 9. UUU_H 2{ gY =U₀2U ₀ 4|2U₀|2{ g} g{ g| =U₀2U ₀ 4|22{}}=|_}=0g{ g| =U₀2U ₀ 4|22{| g{ g| =U₀2{2_|{=4|2 {=0 g| = U2 0(4  |2)| g| =  2|2_1 4|4 2 0 = 4

SECTION 16.6 TRIPLE INTEGRALS ET SECTION 15.6 ¤ 241 11.HereH = {({> |> }) | 0  {  1> 0  | {> 0  }  1 + { + |}, so UUU H 6{| gY = U1 0 U_{ 0 U1+{+| 0 6{| g} g| g{ = U1 0 U_{ 0  6{|}}=1+{+|_}=0 g| g{ =U₀1U₀{ 6{|(1 + { + |) g| g{ =U₀13{|2_{+ 3{}2_|2_{+ 2{|}3|={ |=0 g{ = U1 0 (3{2+ 3{3+ 2{5@2) g{ = k {3₊3 4{4+47{7@2 l1 0= 65 28

13. H is the region below the parabolic cylinder } = 1  |2and above the

square[1> 1] × [1> 1] in the {|-plane. UUU H{2h|gY = U1 1 U1 1 U1|2 0 {2h|g} g| g{ =U_11 U_11 {2_h|_{(1  |}2_{) g| g{} =U_11 {2_g{U1 1(h| |2h|) g| =1 3{3 1 1  h|_{ (|}2_{ 2| + 2)h}|1 1  integrate by parts twice  =1 3(2)[h  h  h1+ 5h1] =3h8 15. HereW = {({> |> }) | 0  {  1> 0  |  1  {> 0  }  1  {  |}, so UUU W {2gY = U1 0 U1{ 0 U1{| 0 {2g} g| g{ = U1 0 U1{ 0 {2(1  {  |)g| g{ =U₀1U₀1{({2_{ {}3_{ {}2_{|)g| g{ =}U1 0  {2_{|  {}3_{| }1 2{2|2 |=1{ |=0 g{ =U₀1{2_{(1  {)  {}3_{(1  {) }1 2{2(1  {)2  g{ =U₀11 2{4 {3+12{2  g{ =1 10{514{4+16{3 1 0 = 1 1014 +16 = 601

17. The projectionH on the |}-plane is the disk |2+ }2  1. Using polar

coordinates| = u cos  and } = u sin , we get UUU H{ gY = UU G kU₄ 4|2_{+ 4}}2{ g{ l gD =1 2 UU G  42_{ (4|}2_{+ 4}}2₎2_gD = 8U₀2U₀1(1  u4_{) u gu g = 8}U2 0 g U1 0(u  u5) gu = 8(2)1 2u216u6 1 0 =163

19.The plane2{ + | + } = 4 intersects the {|-plane when 2{ + | + 0 = 4  | = 4  2{, so H = {({> |> }) | 0  {  2, 0  |  4  2{, 0  }  4  2{  |} and Y =U2 0 U42{ 0 U42{| 0 g} g| g{ = U2 0 U42{ 0 (4  2{  |) g| g{ =U₀24|  2{| 1 2|2 |=42{ |=0 g{ =U₀24(4  2{)  2{(4  2{) 1 2(4  2{)2  g{ =U2 0 (2{2 8{ + 8) g{ = ₂ 3{3 4{2+ 8{ 2 0=163

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21. Y = ] 3 3 ] 9{2 9{2 ] 5| 1 g} g| g{ = ] 3 3 ] 9{2 9{2(5  |  1) g| g{ = ] 3 3 k 4| 1 2|2 l|=9{2 |=9{2 g{ =U_33 89  {2_{g{ = 8}{ 2  9  {2₊9 2sin1 _{ 3 3 3 

using trigonometric substitution or Formula 30 in the Table of Integrals

 = 89 2sin1(1) 92sin1(1)  = 36 2    2  = 36 Alternatively, use polar coordinates to evaluate the double integral: ] 3 3 ] 9{2 9{2(4  |) g| g{ = ] 2 0 ]3 0 (4  u sin ) u gu g =U₀22u2_1 3u3sin  u=3 u=0 g =U₀2(18  9 sin ) g = 18 + 9 cos l2 0 = 36

23. (a) The wedge can be described as the region

G =({> |> }) | |2_{+ }}2_{ 1, 0  {  1, 0  |  {}

=q({> |> }) | 0  {  1, 0  |  {, 0  } s1  |2r

So the integral expressing the volume of the wedge is UUU G gY = U1 0 U{ 0 U 1  |2 0 g} g| g{. (b) A CAS givesU₀1U₀{U ₀ 1  |2 g} g| g{ = ₄ 1₃. (Or use Formulas 30 and 87 from the Table of Integrals.)

25. Herei({> |> }) = 1

ln(1 + { + | + })andY = 2 · 4 · 2 = 16, so the Midpoint Rule gives UUU E i({> |> }) gY  o S l=1 p S m=1 q S n=1i  {l> |m> }nY

= 16[i(1> 2> 1) + i(1> 2> 3) + i(1> 6> 1) + i(1> 6> 3) + i(3> 2> 1) + i(3> 2> 3) + i(3> 6> 1) + i(3> 6> 3)] = 16 1

ln 5+ln 71 +ln 91 +ln 111 +ln 71 +ln 91 +ln 111 +ln 131



 60=533 27. H = {({> |> }) | 0  {  1, 0  }  1  {, 0  |  2  2}},

the solid bounded by the three coordinate planes and the planes } = 1  {, | = 2  2}.

SECTION 16.6 TRIPLE INTEGRALS ET SECTION 15.6 ¤ 243 29.

IfG₁,G₂,G₃are the projections ofH on the {|-, |}-, and {}-planes, then

G1=({> |) | 2  {  2, 0  |  4  {2=({> |) | 0  |  4, 4  |  { 4  | G2=(|> }) | 0  |  4, 1₂4  |  } 1₂4  |=(|> }) | 1  }  1, 0  |  4  4}2 G3=({> }) | {2+ 4}2 4 Therefore H =q({> |> }) | 2  {  2, 0  |  4  {2_{, }1 2 s 4  {2_{ |  } } 1 2 s 4  {2_{ |}r =q({> |> }) | 0  |  4, 4  |  { 4  |, 1 2 s 4  {2_{ |  } } 1 2 s 4  {2_{ |}r =q({> |> }) | 1  }  1, 0  |  4  4}2_{, }s_{4  |  4}}2_{ { }s_{4  |  4}}2r =q({> |> }) | 0  |  4, 1 2  4  |  } 1 2  4  |, s4  |  4}2_{ { }s_{4  |  4}}2r =q({> |> }) | 2  {  2, 1 2  4  {2_{ } } 1 2  4  {2_{,0  |  4  {}2_{ 4}}2r =q({> |> }) | 1  }  1, 4  4}2_{ { }_{4  4}}2_{,0  |  4  {}2_{ 4}}2r Then UUU Hi({> |> }) gY = U2 2 U4{2 0 U 4{2_|@2 4{2_|@2i({> |> }) g} g| g{ = U4 0 U_4| 4| U 4{2_|@2 4{2_|@2i({> |> }) g} g{ g| =U_11 U₀44}2U _4|4}2 4|4}2i({> |> }) g{ g| g} = U4 0 U 4|@2 4|@2 U 4|4}2 4|4}2i({> |> }) g{ g} g| =U_22 U 4{2@2 4{2_@2 U4{2_4}2 0 i({> |> }) g| g} g{ = U1 1 U 44}2 44}2 U4{2_4}2 0 i({> |> }) g| g{ g}

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31.

IfG1,G2, andG3are the projections ofH on the {|-, |}-, and {}-planes, then G1= q ({> |) | 2  {  2> {2_{ |  4}r₌q_{({> |) | 0  |  4> }s_{|  { }s_|r_, G2= q (|> }) | 0  |  4> 0  }  2 1 2| r =q(|> }) | 0  }  2> 0  |  4  2}r, and G3= q ({> }) | 2  {  2> 0  }  2 1 2{2 r =q({> }) | 0  }  2> 4  2}  { 4  2}r Therefore _{H =}q_{({> |> }) | 2  {  2, {}2_{ |  4, 0  }  2 }1 2| r =q({> |> }) | 0  |  4, s|  { s|, 0  }  2 1 2| r =q({> |> }) | 0  |  4, 0  }  2 1 2|,  s |  { s|r =q({> |> }) | 0  }  2, 0  |  4  2}, s|  { s|r =q({> |> }) | 2  {  2, 0  }  2 1 2{2,{2 |  4  2} r =q({> |> }) | 0  }  2, 4  2}  { 4  2}, {2_{ |  4  2}}r Then UUU Hi({> |> }) gY = U2 2 U4 {2U₀2|@2i({> |> }) g} g| g{ =U₀4U_|| U₀2|@2i({> |> }) g} g{ g| =U4 0 U2|@2 0 U| |i({> |> }) g{ g} g| = U2 0 U42} 0 U| |i({> |> }) g{ g| g} =U_22 U₀2  {2@2U_{42}2 i({> |> }) g| g} g{ = U2 0 U 42} 42} U42} {2 i({> |> }) g| g{ g}

SECTION 16.6 TRIPLE INTEGRALS ET SECTION 15.6 ¤ 245 33.

The diagrams show the projections ofH on the {|-, |}-, and {}-planes. Therefore U1 0 U_1 { U1  | 0 i({> |> }) g} g| g{ = U1 0 U|2 0 U1| 0 i({> |> }) g} g{ g| = U1 0 U1} 0 U|2 0 i({> |> }) g{ g| g} =U1 0 U_1| 0 U_|2 0 i({> |> }) g{ g} g| = U₁ 0 U_1 { 0 U_1}  { i({> |> }) g| g} g{ =U1 0 U(1})2 0 U_1} { i({> |> }) g| g{ g} 35. U1 0 U1 | U| 0 i({> |> }) g} g{ g| = UUU Hi({> |> }) gY where H = {({> |> }) | 0  }  |, |  {  1, 0  |  1}.

IfG1,G2, andG3are the projections ofH on the {|-, |}- and {}-planes then G1= {({> |) | 0  |  1, |  {  1} = {({> |) | 0  {  1, 0  |  {},

G2= {(|> }) | 0  |  1, 0  }  |} = {(|> }) | 0  }  1, }  |  1}, and

G3= {({> }) | 0  {  1, 0  }  {} = {({> }) | 0  }  1, }  {  1}.

Thus we also have

H = {({> |> }) | 0  {  1, 0  |  {, 0  }  |} = {({> |> }) | 0  |  1, 0  }  |, |  {  1} = {({> |> }) | 0  }  1, }  |  1, |  {  1} = {({> |> }) | 0  {  1, 0  }  {, }  |  {} = {({> |> }) | 0  }  1, }  {  1, }  |  {} . Then U1 0 U1 | U| 0 i({> |> }) g} g{ g| = U1 0 U{ 0 U| 0 i({> |> }) g} g| g{ = U1 0 U| 0 U1 | i({> |> }) g{ g} g| =U1 0 U1 } U1 |i({> |> }) g{ g| g} = U1 0 U{ 0 U{ } i({> |> }) g| g} g{ =U₀1U_}1U_}{i({> |> }) g| g{ g}

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37. p =UUU_H ({> |> }) gY =U₀1U  { 0 U1+{+| 0 2 g} g| g{ = U1 0 U { 0 2(1 + { + |) g| g{ =U₀12| + 2{| + |2|={ |=0 g{ = U1 0  2{ + 2{3@2_{+ {}_{g{ =}k4 3{3@2+45{5@2+12{2 l1 0= 79 30 P|} =UUU_H {({> |> }) gY =U₀1U  { 0 U1+{+| 0 2{ g} g| g{ = U1 0 U { 0 2{(1 + { + |) g| g{ =U₀12{| + 2{2_{| + {|}2|={ |=0 g{ = U1 0(2{3@2+ 2{5@2+ {2) g{ = k 4 5{5@2+47{7@2+13{3 l1 0= 179 105 P{}=UUU_H |({> |> }) gY =U₀1U  { 0 U_1+{+| 0 2| g} g| g{ = U₁ 0 U { 0 2|(1 + { + |) g| g{ =U₀1|2_{+ {|}2₊2 3|3 |={ |=0 g{ = U1 0  { + {2₊2 3{3@2  g{ =k1 2{2+13{3+154{5@2 l1 0= 11 10 P{| =UUU_H}({> |> }) gY =U₀1U _{ 0 U1+{+| 0 2} g} g| g{ = U1 0 U_{ 0  }2}=1+{+| }=0 g| g{ = U1 0 U_{ 0 (1 + { + |)2g| g{ =U₀1U₀{(1 + 2{ + 2| + 2{| + {2_{+ |}2_{) g| g{ =}U1 0  | + 2{| + |2_{+ {|}2_{+ {}2_{| +}1 3|3 |={ |=0 g{ =U1 0 _{{ +}7 3{3@2+ { + {2+ {5@2  g{ =k2 3{3@2+1415{5@2+12{2+13{3+27{7@2 l1 0= 571 210

Thus the mass is79₃₀and the center of mass is({> |> }) =  P|} p > Pp{}> Pp{|  =  358 553> 3379> 571553  . 39. p =U₀dU₀dU₀d({2+ |2+ }2) g{ g| g} =U₀dU₀d1₃{3+ {|2+ {}2{=d_{=0 g| g} =U₀dU₀d1₃d3+ d|2+ d}2g| g} =U₀d1 3d3| +13d|3+ d|}2 |=d |=0g} = Ud 0 ₂ 3d4+ d2}2  g} =2 3d4} +13d2}3 d 0 =23d5+13d5= d5 P|} =U₀dU₀dU₀d{3+ {(|2+ }2)g{ g| g} =U₀dU₀d1₄d4+1₂d2(|2+ }2)g| g} =U₀d1 4d5+16d5+12d3}2  g} = 1 4d6+13d6=127d6= P{}= P{|by symmetry ofH and ({> |> }) Hence({> |> }) =₁₂7d>₁₂7d>₁₂7d. 41. L{=U₀OU₀OU₀On(|2+ }2) g} g| g{ = nU₀OU₀OO|2+1₃O3g| g{ = nU₀O2₃O4g{ =2₃nO5. By symmetry,L{= L|= L}= 2₃nO5. 43. L} =UUU_H({2+ |2) ({> |> }) gY = UU {2_+|2_d2 kUk 0 n({2+ |2) g} l gD = UU {2_+|2_d2n({ 2_{+ |}2_{)k gD} = nkU₀2U₀d(u2_{) u gu g = nk}U2 0 g Ud 0 u3gu = nk(2) ₁ 4u4 d 0 = 2nk ·14d4=12nkd4 45. (a)p =U_33 U 9{2 9{2 U5| 1 s {2_{+ |}2_{g} g| g{} (b)({> |> }) =  P|} p > Pp{}> Pp{|  where P|} =U_33 U 9{2 9{2 U5| 1 { s {2_{+ |}2_{g} g| g{, P{}=}U3 3 U 9{2 9{2 U5| 1 | s {2_{+ |}2_{g} g| g{, and} P{|=U_33 U 9{ 2 9{2 U5| 1 } s {2_{+ |}2_{g} g| g{.} (c)L}=U_33 U 9{2 9{2 U5| 1 ({2+ |2) s {2_{+ |}2_{g} g| g{ =}U3 3 U 9{2 9{2 U5| 1 ({2+ |2)3@2g} g| g{

SECTION 16.6 TRIPLE INTEGRALS ET SECTION 15.6 ¤ 247 47. (a)p =U₀1U ₀ 1{2U₀|(1 + { + | + }) g} g| g{ = 3₃₂ +11₂₄ (b) ({> |> }) =  p1U1 0 U 1{2 0 U| 0 {(1 + { + | + }) g} g| g{> p1U1 0 U 1{2 0 U| 0 |(1 + { + | + }) g} g| g{> p1U1 0 U 1{2 0 U| 0 }(1 + { + | + }) g} g| g{  =  28 9 + 44> 30 + 12845 + 220> 45 + 208135 + 660  (c)L_}= ]1 0 ] 1{2 0 ] | 0 ({ 2_{+ |}2_{)(1 + { + | + }) g} g| g{ = 68 + 15} 240

49. (a)i({> |> }) is a joint density function, so we knowUUU_R3i({> |> }) gY = 1. Here we have UUU R3i({> |> }) gY = U  U  U i({> |> }) g} g| g{ = U2 0 U2 0 U2 0 F{|} g} g| g{ = FU2 0 { g{ U₂ 0 | g| U₂ 0 } g} = F ₁ 2{2 2 0 ₁ 2|2 2 0 ₁ 2}2 2 0= 8F

Then we must have8F = 1  F = 1₈.

(b)S ([  1> \  1> ]  1) =U_1 U_1 U_1 i({> |> }) g} g| g{ =U₀1U₀1U₀11₈{|} g} g| g{ =1 8 U1 0 { g{ U1 0 | g| U1 0 } g} = 18 ₁ 2{2 1 0 ₁ 2|2 1 0 ₁ 2}2 1 0= 18 ₁ 2 3₌ 1 64

(c)S ([ + \ + ]  1) = S (([> \> ]) H) where H is the solid region in the rst octant bounded by the coordinate planes and the plane{ + | + } = 1. The plane { + | + } = 1 meets the {|-plane in the line { + | = 1, so we have

S ([ + \ + ]  1) =UUU_Hi({> |> }) gY =U₀1U₀1{U₀1{|1 8{|} g} g| g{ = 1 8 U1 0 U1{ 0 {| ₁ 2}2 }=1{| }=0 g| g{ =161 U1 0 U1{ 0 {|(1  {  |)2g| g{ = 1 16 U1 0 U1{ 0 [({3 2{2+ {)| + (2{2 2{)|2+ {|3] g| g{ = 1 16 U1 0  ({3_{ 2{}2_{+ {)}1 2|2+ (2{2 2{)13|3+ { ₁ 4|4 |=1{ |=0 g{ = 1 192 U1 0({  4{2+ 6{3 4{4+ {5) g{ = 1921 ₁ 30  = 1 5760 51.Y (H) = O3  iave= 1 O3 ] O 0 ] O 0 ]O 0 {|} g{ g| g} = 1O3 ]O 0 { g{ ] O 0 | g| ] O 0 } g} = 1_O3  {2 2 O 0  |2 2 O 0  }2 2 O 0 = 1O3 O2 2 O 2 2 O 2 2 = O 3 8

53.The triple integral will attain its maximum when the integrand1  {2 2|2 3}2is positive in the regionH and negative everywhere else. For ifH contains some region I where the integrand is negative, the integral could be increased by excluding I from H, and if H fails to contain some part J of the region where the integrand is positive, the integral could be increased by includingJ in H. So we require that {2+ 2|2+ 3}2  1. This describes the region bounded by the ellipsoid

16.7

Triple Integrals in Cylindrical Coordinates

ET 15.7

1. (a)

{ = 2 cos ₄ =2, | = 2 sin ₄ =2, } = 1, so the point is2>2> 1in rectangular coordinates.

(b) { = 4 cos 3  = 2, | = 4 sin 3  = 23, and} = 5, so the point is2> 23> 5in rectangular coordinates.

3. (a)u2= {2+ |2= 12+ (1)2= 2 so u =2; tan  = |

{= 11 = 1 and the point (1> 1) is in the fourth quadrant of the{|-plane, so  = 7₄ + 2q; } = 4. Thus, one set of cylindrical coordinates is2>7₄ > 4.

(b)u2= (1)2+32= 4 so u = 2; tan  = _13 =3 and the point1> 3is in the third quadrant of the {|-plane, so  = 4

3 + 2q; } = 2. Thus, one set of cylindrical coordinates is

 2>4

3> 2

 .

5. Since =₄ butu and } may vary, the surface is a vertical half-plane including the }-axis and intersecting the {|-plane in the half-line| = {, {  0.

7. } = 4  u2= 4  ({2+ |2) or 4  {2 |2, so the surface is a circular paraboloid with vertex(0> 0> 4), axis the }-axis, and opening downward.

9. (a){2+ |2= u2, so the equation becomes} = u2=

(b) Substituting{2+ |2= u2and| = u sin , the equation {2+ |2= 2| becomes u2= 2u sin  or u = 2 sin .

11. 0  u  2 and 0  }  1 describe a solid circular cylinder with

radius2, axis the }-axis, and height 1, but @2    @2 restricts the solid to the rst and fourth quadrants of the{|-plane, so we have a half-cylinder.

13. We can position the cylindrical shell vertically so that its axis coincides with the}-axis and its base lies in the {|-plane. If we use centimeters as the unit of measurement, then cylindrical coordinates conveniently describe the shell as6  u  7, 0    2, 0  }  20.

SECTION 16.7 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES ET SECTION 15.7 ¤ 249

15. The region of integration is given in cylindrical coordinates by

H = {(u> > }) | 0    2, 0  u  4, u  }  4}. This represents the solid region bounded below by the cone} = u and above by the horizontal plane } = 4.

U4 0 U2 0 U4 u u g} g gu = U4 0 U2 0  u}}=4_}=u g gu =U₀4U₀2u(4  u) g gu =U₀4(4u  u2_{) gu}U2 0 g =  2u2_1 3u3 4 0  2₀ =32 64 3  (2) =64 3

17.In cylindrical coordinates,H is given by {(u> > }) | 0    2> 0  u  4> 5  }  4}. So UUU H s {2_{+ |}2_{gY =}U2 0 U4 0 U4 5  u2_{u g} gu g =}U2 0 g U4 0 u2gu U4 5g} =2₀ 1 3u3 4 0  }4_5= (2)64 3  (9) = 384

19.In cylindrical coordinatesH is bounded by the paraboloid } = 1 + u2, the cylinderu2= 5 or u =5, and the {|-plane, soH is given by(u> > }) | 0    2> 0  u 5> 0  }  1 + u2. Thus

UUU H h}gY = U2 0 U 5 0 U1+u2 0 h}u g} gu g = U2 0 U 5 0 u  h}}=1+u2 }=0 gu g = U2 0 U 5 0 u(h1+u 2  1) gu g =U₀2 gU₀5uh1+u2  ugu = 2k1 2h1+u 2 1 2u2 l 5 0 = (h 6_{ h  5)}

21.In cylindrical coordinates,H is bounded by the cylinder u = 1, the plane } = 0, and the cone } = 2u. So H = {(u> > }) | 0    2> 0  u  1> 0  }  2u} and

UUU H{2gY = U₂ 0 U₁ 0 U_2u 0 u2cos2 u g} gu g = U₂ 0 U₁ 0  u3_cos2_{ }}}=2u }=0 gu g = U₂ 0 U₁ 0 2u4cos2 gu g =U2 0 ₂ 5u5cos2 u=1 u=0g = 25 U₂ 0 cos2 g = 2₅ ] 2 0 1 + cos 2 2 g = 15   + 1₂sin 2 2 0 = 25

23. (a) The paraboloids intersect when{2+ |2= 36  3{2 3|2  {2+ |2= 9, so the region of integration isG = ({> |) | {2+ |2 9. Then, in cylindrical coordinates,

H =(u> > }) | u2_{ }  36  3u}2_{,0  u  3, 0    2and} Y =U₀2U₀3U_u36  3u2 2u g} gu g = U2 0 U3 0  36u  4u3_{gu g =}U2 0  18u2_{ u}4u=3 u=0g = U2 0 81 g = 162.

(b) For constant densityN, p = NY = 162N from part (a). Since the region is homogeneous and symmetric, P|}= P{}= 0 and P{|=U₀2U₀3U363u 2 u2 (}N) u g} gu g = N U2 0 U3 0 u ₁ 2}2 }=363u2 }=u2 gu g = N 2 U2 0 U3 0 u((36  3u2)2 u4) gu g =N2 U2 0 g U3

0(8u5 216u3+ 1296u) gu

= N 2(2) ₈ 6u62164 u4+12962 u2 3 0= N(2430) = 2430N Thus({> |> }) =  P|} p > Pp{}> Pp{|  =0> 0>2430N 162N  = (0> 0> 15).

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25. The paraboloid} = 4{2+ 4|2intersects the plane} = d when d = 4{2+ 4|2or{2+ |2 =1₄d. So, in cylindrical coordinates,H =(u> > }) | 0  u 1₂d> 0    2> 4u2 }  d. Thus

p =]2 0 ] _d@2 0 ] d 4u2Nu g} gu g = N ] 2 0 ] _d@2 0 (du  4u 3_{) gu g} = N]2 0 ₁ 2du2 u4 u=d@2 u=0 g = N ] 2 0 1 16d2g = 18d2N

Since the region is homogeneous and symmetric,P|} = P{}= 0 and

P{|= ] 2 0 ] _d@2 0 ] d 4u2Nu} g} gu g = N ] 2 0 ] _d@2 0 ₁ 2d2u  8u5  gu g = N] 2 0 ₁ 4d2u243u6 u=d@2 u=0 g = N ] 2 0 1 24d3g =121d3N Hence({> |> }) =0> 0>2₃d.

27. The region of integration is the region above the cone} =s{2+ |2, or} = u, and below the plane } = 2. Also, we have 2  |  2 with s4  |2_{ { }s_{4  |}2_{which describes a circle of radius 2 in the{|-plane centered at (0> 0). Thus,}

] ₂ 2 ] _4|2 4|2 ]₂  {2_+|2{} g} g{ g| = ] ₂ 0 ] ₂ 0 ] ₂ u (u cos ) } u g} gu g = ] ₂ 0 ] ₂ 0 ] ₂ u u 2_{(cos ) } g} gu g} =U₀2U₀2u2_{(cos )}1 2}2 }=2 }=u gu g =12 U2 0 U2 0 u2(cos )  4  u2_{gu g} =1 2 U2 0 cos  g U2 0  4u2_{ u}4_{gu =} 1 2[sin ]20 ₄ 3u315u5 2 0= 0

29. (a) The mountain comprises a solid conical regionF. The work done in lifting a small volume of material Y with density j(S ) to a height k(S ) above sea level is k(S )j(S ) Y . Summing over the whole mountain we get

Z =UUU_Fk(S )j(S ) gY .

(b) HereF is a solid right circular cone with radius U = 62,000 ft, height K = 12,400 ft, and densityj(S ) = 200 lb@ft3at all pointsS in F. We use cylindrical coordinates: Z =U₀2U₀KU₀U(1}@K)} · 200u gu g} g = 2U₀K200}1 2u2 u=U(1}@K) u=0 g} = 400 ] K 0 } U 2 2  1  }_K2 g} = 200U2] K 0  }  2}_K2 + }_K3₂  g} = 200U2}2 2  2} 3 3K + } 4 4K2 K 0 = 200U 2K2 2  2K 2 3 + K 2 4  = 50 3U2K2= 503(62,000)2(12,400)2 3=1 × 1019ft-lb u U= K  }K = 1  }K

SECTION 16.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES ET SECTION 15.8 ¤ 251

16.8

Triple Integrals in Spherical Coordinates

ET 15.8

1. (a)

{ =  sin ! cos  = (1) sin 0 cos 0 = 0, | =  sin ! sin  = (1) sin 0 sin 0 = 0, and } =  cos ! = (1) cos 0 = 1 so the point is (0> 0> 1) in rectangular coordinates. (b) { = 2 sin 4cos3 =  2 2 ,| = 2 sin4sin3 =  6 2 , } = 2 cos 4 =  2 so the point is2 2 >  6 2 >  2in rectangular coordinates. 3. (a) =s{2+ |2+ }2 =1 + 3 + 12 = 4, cos ! = }  = 2  3 4 =  3 2  ! = 6, and cos  =_{sin !}{ =_{4 sin(@6)}1 = 1₂   = ₃ [since| A 0]. Thus spherical coordinates are



4> ₃> ₆. (b) =0 + 1 + 1 =2, cos ! = 1

2  ! = 34 , andcos  = 2 sin(3@4)0 = 0   = 32 [since| ? 0]. Thus spherical coordinates are2> 3

2 > 34 

.

5.Since! =₃, the surface is the top half of the right circular cone with vertex at the origin and axis the positive}-axis. 7. = sin  sin !  2 =  sin  sin ! {2+ |2+ }2= | {2+ |2 | +1₄ + }2 =1₄

{2_{+ (| }1

2)2+ }2=14. Therefore, the surface is a sphere of radius12 centered at

 0>1

2> 0

 . 9. (a){ =  sin ! cos , | =  sin ! sin , and } =  cos !, so the equation }2 = {2+ |2becomes

( cos !)2_{= ( sin ! cos )}2_{+ ( sin ! sin )}2_or2_cos2_{! =}2_sin2_{!. If  6= 0, this becomes cos}2_{! = sin}2_{!. ( = 0}

corresponds to the origin which is included in the surface.) There are many equivalent equations in spherical coordinates, such astan2! = 1, 2 cos2! = 1, cos 2! = 0, or even ! = ₄,! = 3₄ .

(b){2+ }2= 9 ( sin ! cos )2+ ( cos !)2 = 9 2sin2! cos2 + 2cos2! = 9 or 2_sin2_{! cos}2_{ + cos}2_!_{= 9.}

11. = 2 represents a sphere of radius 2, centered at the origin, so   2 is this sphere and its interior. 0  !  ₂ restricts the solid to that portion of the region that lies on or above the{|-plane, and 0   ₂ further restricts the solid to the rst octant. Thus the solid is the portion in the rst octant of the solid ball centered at the origin with radius 2.

13.   1 represents the solid sphere of radius 1 centered at the origin.

3

4  !   restricts the solid to that portion on or below the cone ! = 34 .

15. } s{2+ |2because the solid lies above the cone. Squaring both sides of this inequality gives}2 {2+ |2  2}2_{ {}2_{+ |}2_{+ }}2₌2 _{ }}2 ₌2_cos2_{! }1

22  cos2! 12. The cone opens upward so that the inequality is

cos ! _1

2, or equivalently0  !  4. In spherical coordinates the sphere} = {2+ |2+ }2is cos ! = 2 

 = cos !. 0    cos ! because the solid lies below the sphere. The solid can therefore be described as the region in spherical coordinates satisfying0    cos !, 0  ! ₄.

17. The region of integration is given in spherical coordinates by

H = {(> > !) | 0    3> 0    @2> 0  !  @6}. This represents the solid region in the rst octant bounded above by the sphere = 3 and below by the cone ! = @6. U@6 0 U@2 0 U3 0 2sin ! g g g! = U@6 0 sin ! g! U@2 0 g U3 0 2g = cos !@6₀ @2₀ 1 33 3 0 =  1   3 2 2  (9) = 9₄ 2 3 19. The solidH is most conveniently described if we use cylindrical coordinates:

H =(u> > }) | 0     2> 0  u  3> 0  }  2  . Then UUU Hi({> |> }) gY = U@2 0 U3 0 U2

0 i(u cos > u sin > }) u g} gu g.

21. In spherical coordinates,E is represented by {(> > !) | 0    5> 0    2> 0  !   }. Thus UUU E({2+ |2+ }2)2gY = U 0 U2 0 U5 0(2)22sin ! g g g! = U 0 sin ! g! U2 0 g U5 0 6g = cos !₀ 2₀ 1 77 5 0= (2)(2) _78,125 7  =312,500 7   140,249=7

23. In spherical coordinates,H is represented by(> > !)1    2>0    ₂> 0  !  ₂. Thus UUU H} gY = U@2 0 U@2 0 U2 1( cos !) 2sin ! g g g! = U@2 0 cos ! sin ! g! U@2 0 g U2 1 3g =1 2sin2! @2 0  @2₀ 1 44 2 1= ₁ 2  2 ₁₅ 4  =15 16

SECTION 16.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES ET SECTION 15.8 ¤ 253 25.UUU_H {2gY =U₀U₀U₃4( sin ! cos )22sin ! g g! g =U₀cos2 gU₀sin3! g!U₃44g

=1 2 +14sin 2  0  1 3(2 + sin2!) cos !  0 ₁ 55 4 3 =  2 ₂ 3 +23 ₁ 5(45 35) =156215 

27.The solid region is given byH =(> > !) | 0    d> 0    2>₆  !  ₃and its volume is Y =UUU_HgY =U_@6@3U₀2U₀d2_{sin ! g g g! =}U@3 @6 sin ! g! U2 0 g Ud 0 2g = [ cos !]@3 @6[]20 ₁ 33 d 0=  1 2+  3 2  (2)1 3d3  =31 3 d3

29. (a) Since = 4 cos ! implies 2= 4 cos !, the equation is that of a sphere of radius 2 with center at (0> 0> 2). Thus Y =U₀2U₀@3U₀4 cos !2_{sin ! g g! g =}U2 0 U@3 0 ₁ 33 =4 cos ! =0 sin ! g! g = U2 0 U@3 0 ₆₄ 3 cos3!  sin ! g! g =U₀216 3 cos4! !=@3 !=0 g = U2 0 163 ₁ 16 1  g = 5l2 0 = 10

(b) By the symmetry of the problemP_|}= P_{}= 0. Then

P{|=U₀2U₀@3U₀4 cos !3cos ! sin ! g g! g =U₀2U₀@3cos ! sin !64 cos4!g! g

=U₀2641 6cos6! !=@3 !=0 g = U2 0 212 g = 21 Hence({> |> }) = (0> 0> 2=1).

31.By the symmetry of the region,P{|= 0 and P|}= 0. Assuming constant density N,

p =UUU_H NY = NU₀U₀U₃4 2_{sin ! g g! g = N}U 0 g U 0 sin ! g! U4 3 2g = N cos !₀1 33 4 3= 2N ·373 = 743N

and P{}=UUU_H | N gY = NU₀U₀U₃4( sin ! sin ) 2sin ! g g! g = NU₀sin  gU₀sin2! g!U₃4 3g

= N  cos ₀ 1 2! 14sin 2!  0 ₁ 44 4 3= N(2)  2 ₁ 4(256  81) = 1754 N

Thus the centroid is({> |> }) =  P|} p > Pp{}> Pp{|  =  0> 175N@4_74N@3 > 0  =0>525 296> 0  .

33. (a) The density function is({> |> }) = N, a constant, and by the symmetry of the problem P{}= P|}= 0. Then

P{|=U₀2U₀@2U₀dN3sin ! cos ! g g! g =1₂Nd4U₀@2sin ! cos ! g! =1₈Nd4. But the mass isN(volume of the hemisphere) = 2₃Nd3, so the centroid is0> 0>3₈d.

(b) Place the center of the base at(0> 0> 0); the density function is ({> |> }) = N. By symmetry, the moments of inertia about any two such diameters will be equal, so we just need to ndL_{:

L{=U₀2U₀@2U₀d(N2sin !) 2(sin2! sin2 + cos2!) g g! g

= NU₀2U₀@2(sin3_{! sin}2_{ + sin ! cos}2_!)1 5d5  g! g =1 5Nd5 U2 0  sin2__{ cos ! +}1 3cos3!  +1 3cos3! !=@2 !=0 g =15Nd5 U2 0 ₂ 3sin2 +13  g =1 5Nd5 ₂ 3 ₁ 2 14sin 2  +1 3 2 0 =15Nd5 ₂ 3(  0) +13(2  0)  = 4 15Nd5

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