6
APPLICATIONS OF INTEGRATION
6.1 Areas Between Curves
1. D =] {=4 {=0 (|W |E) g{ = ]4 0 (5{ {2) {g{ =] 4 0 (4{ { 2) g{ =2{21 3{3 4 0= 32 64 3 (0) = 32 3 3. D =] |=1 |=1({U {O) g| = ] 1 1 h| (|2 2)g| =] 1 1 h| |2+ 2g| =h|1 3|3+ 2| 1 1= h11 3+ 2 h1+1 3 2 = h 1h + 103 5. D = ] 2 1 (9 {2) ({ + 1)g{ =] 2 1(8 { { 2) g{ = 8{ {2 2 { 3 3 2 1 =16 2 8 3 8 1 2 +13 = 22 3 +1 2 =392
7. The curves intersect when{ = {2 {2 { = 0 {({ 1) = 0 { = 0 or 1. D =] 1 0 ({ { 2) g{ =1 2{213{3 1 0=12 13 = 16 9. D =] 2 1 1 { 1{2 g{ = ln { + 1{ 2 1 =ln 2 +1 2 (ln 1 + 1) = ln 2 1 2 0=19 11. D = ] 1 0 { {2g{ =k2 3{3@213{3 l1 0 =2 313 = 13 263
264 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION 13. 12 {2= {2 6 2{2= 18 {2= 9 { = ±3, so D = ] 3 3 (12 {2) ({2 6)g{ = 2] 3 0 18 2{2g{ [by symmetry] = 218{ 2 3{3 3 0= 2 [(54 18) 0] = 2(36) = 72
15. The curves intersect whentan { = 2 sin { (on [ @3> @3]) sin { = 2 sin { cos { 2 sin { cos { sin { = 0 sin { (2 cos { 1) = 0 sin { = 0 or cos { = 1
2 { = 0 or { = ± 3. D =] @3 @3(2 sin { tan {) g{ = 2 ] @3 0 (2 sin { tan {) g{ [by symmetry] = 2k2 cos { ln |sec {|l @3 0 = 2 [(1 ln 2) (2 0)] = 2(1 ln 2) = 2 2 ln 2 17. 1 2{ = { 1 4{2= { {2 4{ = 0 {({ 4) = 0 { = 0 or 4, so D = ] 4 0 { 1 2{ g{ + ] 9 4 1 2{ {g{ =k2 3{3@214{2 l4 0+ k 1 4{223{3@2 l9 4 =16 3 4 0+81 4 18 4 16 3 =81 4 +323 26 =5912
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19.2|2= 4 + |2 |2= 4 | = ±2, so D =] 2 2 (4 + |2) 2|2g| = 2] 2 0 (4 | 2) g| [by symmetry] = 24| 1 3|3 2 0= 2 8 8 3 =32 3
21.The curves intersect when1 |2 = |2 1 2 = 2|2 |2= 1 | = ±1. D = ] 1 1 (1 |2) (|2 1)g| =] 1 12(1 | 2) g| = 2 · 2] 1 0 (1 | 2) g| = 4| 1 3|3 1 0= 4 1 1 3 =8 3
23.Notice thatcos { = sin 2{ = 2 sin { cos {
2 sin { cos { cos { = 0 cos { (2 sin { 1) = 0 2 sin { = 1 or cos { = 0 { = 6 or 2. D =] @6 0 (cos { sin 2{) g{ + ] @2 @6 (sin 2{ cos {) g{ =sin { +1 2cos 2{ @6 0 + 1 2cos 2{ sin { @2 @6 = 1 2+12·12 0 +1 2· 1 +1 2 1 1 2·12 12 =1 2
25.The curves intersect when{2= 2 {2+ 1 {4+ {2= 2 {4+ {2 2 = 0 ({2+ 2)({2 1) = 0 {2= 1 { = ±1. D =] 1 1 2 {2+ 1 {2 g{ = 2] 1 0 2 {2+ 1 {2 g{ = 2k2 tan1{ 1 3{3 l1 0= 2 2 · 4 13 = 2 3 2=47
266 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION 27. 1@{ = { 1 = {2 { = ±1 and 1@{ =1 4{ 4 = {2 { = ±2, so for { A 0, D =] 1 0 { 14{ g{ +] 2 1 1 { 14{ g{ =] 1 0 3 4{ g{ +] 2 1 1 { 14{ g{ =3 8{2 1 0+ ln |{| 1 8{2 2 1 =3 8+ ln 2 1 2 0 1 8 = ln 2
29. An equation of the line through(0> 0) and (2> 1) is | = 12{; through (0> 0) and(1> 6) is | = 6{; through (2> 1) and (1> 6) is | = 53{ +133. D =] 0 1 5 3{ +133 (6{)g{ +]2 0 5 3{ +133 1 2{ g{ = ] 0 1 13 3{ +133 g{ + ] 2 0 13 6{ +133 g{ =13 3 ] 0 1({ + 1) g{ + 13 3 ] 2 0 1 2{ + 1 g{ =13 3 1 2{2+ { 0 1+133 1 4{2+ { 2 0 =13 3 0 1 2 1 +13 3[(1 + 2) 0] = 133 ·12 +133 · 1 =132
31. The curves intersect whensin { = cos 2{ (on [0> @2]) sin { = 1 2 sin2{ 2 sin2{ + sin { 1 = 0 (2 sin { 1)(sin { + 1) = 0 sin { = 1
2 { = 6. D =] @2 0 |sin { cos 2{| g{ = ] @6 0 (cos 2{ sin {) g{ + ] @2 @6 (sin { cos 2{) g{ =1 2sin 2{ + cos { @6 0 + cos { 1 2sin 2{ @2 @6 =1 4 3 +1 2 3 (0 + 1) + (0 0) 1 2 3 1 4 3 =3 2 3 1 33. Leti({) = cos2 {
4 sin2 { 4 and{ = 1 0 4 . The shaded area is given by
D =U01i({) g{ P4 = 1 4 i1 8 + i3 8 + i5 8 + i7 8 0=6407
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35. From the graph, we see that the curves intersect at{ = 0 and { = d 0=896, with { sin({2) A {4on(0> d). So the area D of the region bounded by the curves is
D =] d 0 { sin({2) {4g{ =1 2cos({2) 15{5 d 0 = 1 2cos(d2) 15d5+12 0=037
37. From the graph, we see that the curves intersect at
{ = d r 1=11> { = e r 1=25> and { = f r 2=86> with
{3 3{ + 4 A 3{2 2{ on (d> e) and 3{2 2{ A {3 3{ + 4 on(e> f). So the area of the region bounded by the curves is
D =] e d ({3 3{ + 4) (3{2 2{)g{ +] f e (3{2 2{) ({3 3{ + 4)g{ = ] e d({ 3 3{2 { + 4) g{ +] f e ({ 3+ 3{2+ { 4) g{ =1 4{4 {312{2+ 4{ e d+ 1 4{4+ {3+12{2 4{ f er 8=38
39.As the gure illustrates, the curves| = { and | = {5 6{3+ 4{ enclose a four-part region symmetric about the origin (since {5 6{3+ 4{ and { are odd functions of {). The curves intersect at values of{ where {5 6{3+ 4{ = {; that is, where
{({4 6{2+ 3) = 0. That happens at { = 0 and where {2= 6 ±36 12
2 = 3 ±
6; that is, at { = s3 +6, s3 6, 0,s3 6, ands3 +6. The exact area is
2 ] 3+6 0 ({ 5 6{3+ 4{) {g{ = 2] 3+ 6 0 { 5 6{3+ 3{g{ = 2] 3 6 0 ({ 5 6{3+ 3{) g{ + 2] 3+ 6 36 ({ 5+ 6{3 3{) g{ CAS = 126 9
268 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION 41. 1 second = 1
3600 hour, so10 s = 3601 h. With the given data, we can takeq = 5 to use the Midpoint Rule. w =1@3600
5 =18001 , so
distanceKelly distanceChris= U1@360 0 yNgw U1@360 0 yFgw = U1@360 0 (yN yF) gw P5= 18001 [(yN yF)(1) + (yN yF)(3) + (yN yF)(5) + (yN yF)(7) + (yN yF)(9)] = 1 1800[(22 20) + (52 46) + (71 62) + (86 75) + (98 86)] = 1 1800(2 + 6 + 9 + 11 + 12) =18001 (40) = 451 mile, or 11713feet
43. Letk({) denote the height of the wing at { cm from the left end.
D r P5= 200 05 [k(20) + k(60) + k(100) + k(140) + k(180)] = 40(20=3 + 29=0 + 27=3 + 20=5 + 8=7) = 40(105=8) = 4232 cm2
45. We know that the area under curveD between w = 0 and w = { isU0{yD(w) gw = vD({), where yD(w) is the velocity of car A andvAis its displacement. Similarly, the area under curveE between w = 0 and w = { is
U{
0 yB(w) gw = vB({). (a) After one minute, the area under curveD is greater than the area under curve E. So car A is ahead after one minute. (b) The area of the shaded region has numerical valuevA(1) vB(1), which is the distance by which A is ahead of B after
1 minute.
(c) After two minutes, car B is traveling faster than car A and has gained some ground, but the area under curveD from w = 0 tow = 2 is still greater than the corresponding area for curve E, so car A is still ahead.
(d) From the graph, it appears that the area between curvesD and E for 0 w 1 (when car A is going faster), which corresponds to the distance by which car A is ahead, seems to be about3 squares. Therefore, the cars will be side by side at the time{ where the area between the curves for 1 w { (when car B is going faster) is the same as the area for 0 w 1. From the graph, it appears that this time is { 2=2. So the cars are side by side when w 2=2 minutes. 47. To graph this function, we must rst express it as a combination of explicit
functions of|; namely, | = ±{{ + 3. We can see from the graph that the loop extends from{ = 3 to { = 0, and that by symmetry, the area we seek is just twice the area under the top half of the curve on this interval, the equation of the top half being| = {{ + 3. So the area is D = 2U30 {{ + 3g{. We substitutex = { + 3, so gx = g{ and the limits change to 0 and 3, and we get
D = 2U03[(x 3)x ] gx = 2U03(x3@2 3x1@2) gx = 2k2 5x5@2 2x3@2 l3 0= 2 2 5 323 233= 24 5 3
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49. By the symmetry of the problem, we consider only the rst quadrant, where | = {2 { =s|. We are looking for a number e such that
] e 0 s | g| =]4 e s | g| 2 3 k |3@2le 0= 2 3 k |3@2l4 e e3@2= 43@2 e3@2 2e3@2= 8 e3@2= 4 e = 42@3 2=52.
51.We rst assume thatf A 0, since f can be replaced by f in both equations without changing the graphs, and if f = 0 the curves do not enclose a region. We see from the graph that the enclosed areaD lies between { = f and { = f, and by symmetry, it is equal to four times the area in the rst quadrant. The enclosed area is
D = 4U0f(f2 {2) g{ = 4f2{ 1 3{3 f 0= 4 f31 3f3 = 42 3f3 =8 3f3 SoD = 576 83f3= 576 f3= 216 f =3216 = 6. Note thatf = 6 is another solution, since the graphs are the same.
53.The curve and the line will determine a region when they intersect at two or more points. So we solve the equation{@({2+ 1) = p{
{ = {(p{2+ p) {(p{2+ p) { = 0 {(p{2+ p 1) = 0 { = 0 or p{2+ p 1 = 0 { = 0 or {2 = 1 p p { = 0 or { = ± u 1
p 1. Note that if p = 1, this has only the solution { = 0, and no region is determined. But if1@p 1 A 0 1@p A 1 0 ? p ? 1, then there are two solutions. [Another way of seeing this is to observe that the slope of the tangent to| = {@({2+ 1) at the origin is |0(0) = 1 and therefore we must have 0 ? p ? 1.] Note that we cannot just integrate between the positive and negative roots, since the curve and the line cross at the origin. Sincep{ and {@({2+ 1) are both odd functions, the total area is twice the area between the curves on the interval k
0>s1@p 1l. So the total area enclosed is 2] 1@p1 0 { {2+ 1 p{ g{ = 21 2ln({2+ 1) 12p{2 1@p1 0 = [ln(1@p 1 + 1) p(1@p 1)] (ln 1 0) = ln(1@p) 1 + p = p ln p 1
270 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
6.2 Volumes
1. A cross-section is a disk with radius2 12{> so its area is D({) = 2 12{2. Y = ] 2 1 D({) g{ = ] 2 1 2 1 2{ 2 g{ = ]2 1 4 2{ +1 4{2 g{ = 4{ {2+ 1 12{3 2 1 = 8 4 + 8 12 4 1 + 1 12 = 1 + 7 12 = 19 12
3. A cross-section is a disk with radius1@{, so its area is D({) = (1@{)2. Y =] 2 1 D({) g{ = ] 2 1 1 { 2 g{ = ]2 1 1 {2g{ = 1{ 2 1 = 1 2 (1) = 2
5. A cross-section is a disk with radius2s|, so its area is D(|) = 2s|2. Y = ] 9 0 D(|) g| = ] 9 0 2s|2g| = 4 ] 9 0 | g| = 4 1 2|2 9 0= 2 (81) = 162
7. A cross-section is a washer (annulus) with inner radius{3and outer radius{, so its area is D({) = ({)2 ({3)2= ({2 {6). Y =] 1 0 D({) g{ = ] 1 0 ({ 2 {6) g{ = 1 3{317{7 1 0= 1 3 17 = 4 21
9.A cross-section is a washer with inner radius|2 and outer radius2|, so its area is
D(|) = (2|)2 (|2)2= (4|2 |4). Y =]2 0 D(|) g| = ] 2 0 (4| 2 |4) g| = 4 3|315|5 2 0= 32 3 325 = 64 15
11.A cross-section is a washer with inner radius1 { and outer radius 1 {, so its area is D({) = (1 {)2 1 {2 = k(1 2{ + {2) 1 2{ + {l = 3{ + {2+ 2{. Y =]1 0 D({) g{ = ] 1 0 3{ + {2+ 2{g{ = k3 2{2+13{3+43{3@2 l1 0= 3 2+53 = 6
13.A cross-section is a washer with inner radius(1 + sec {) 1 = sec { and outer radius 3 1 = 2, so its area is D({) = 22 (sec {)2= (4 sec2{). Y = ] @3 @3D({) g{ = ] @3 @3 (4 sec 2{) g{ = 2 ] @3 0 (4 sec 2{) g{ [by symmetry] = 2 k4{ tan {l @3 0 = 2 4 3 3 0 = 2 4 3 3 15. Y =]1 1 (1 | 2)2g| = 2]1 0 (1 | 2)2g| = 2 ] 1 0 (1 2| 2+ |4) g| = 2 | 2 3|3+15|5 1 0 = 2 · 8 15= 1615
272 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
17. | = {2 { = | for { 0. The outer radius is the distance from { = 1 to { =s| and the inner radius is the distance from{ = 1 to { = |2. Y = ] 1 0 ks | (1)l2|2 (1)2 g| = ] 1 0 s | + 12 (|2+ 1)2g| = ] 1 0 | + 2s| + 1 |4 2|2 1g| = ]1 0 | + 2s| |4 2|2g| = k1 2|2+43|3@215|523|3 l1 0= 1 2+431523 = 29 30
19. R1aboutRD (the line | = 0): Y = ] 1 0 D({) g{ = ] 1 0 ({ 3)2g{ = ] 1 0 { 6g{ = 1 7{7 1 0= 7
21. R1aboutDE (the line { = 1): Y = ] 1 0 D(|) g| = ] 1 0 1 s3 |2g| = ] 1 0 (1 2| 1@3+ |2@3) g| = k| 3 2|4@3+35|5@3 l1 0 = 1 3 2 +35 = 10
23. R2aboutRD (the line | = 0): Y = ] 1 0 D({) g{ = ] 1 0 (1)2 {2g{ = ] 1 0 (1 {) g{ = { 1 2{2 1 0= 1 1 2 = 2
25. R2aboutDE (the line { = 1): Y =] 1 0 D(|) g| = ] 1 0 (1)2 (1 |2)2g| = ]1 0 1 (1 2|2+ |4)g| = ] 1 0 (2| 2 |4) g| = k2 3|315|5 l1 0= 2 315 = 7 15
27. R3aboutRD (the line | = 0): Y =] 1 0 D({) g{ = ] 1 0 {2 ({3)2g{ = ] 1 0 ({ { 6) g{ = 1 2{217{7 1 0= 1 2 17 = 5 14 .
Note: Let R = R1+ R2+ R3. If we rotateR about any of the segments RD, RF, DE, or EF, we obtain a right circular cylinder of height1 and radius 1. Its volume is u2k = (1)2· 1 = . As a check for Exercises 19, 23, and 27, we can add the answers, and that sum must equal . Thus, 7 + 2 +5 14 =2 + 7 + 514 = .
29.R3aboutDE (the line { = 1): Y =]1 0 D(|) g| = ]1 0 (1 |2)2 1 s3 |2g| = ]1 0 k (1 2|2+ |4) (1 2|1@3+ |2@3)lg| = ] 1 0 (2| 2+ |4+ 2|1@3 |2@3) g| = k2 3|3+15|5+32|4@335|5@3 l1 0= 2 3+15+3235 =13 30
Note: See the note in Exercise 27. For Exercises 21, 25, and 29, we have
10+7 15+13 30 = 3 + 14 + 13 30 = . 31.Y = ] @4 0 (1 tan 3{)2g{ 33.Y = ] 0 (1 0)2 (1 sin {)2g{ = ] 0 12 (1 sin {)2g{ 35.Y = ] 8 8 [3 (2)]2ks|2+ 1 (2)l2 g| = ] 2 2 22 52s1 + |2+ 22 g|
37. | = 2 + {2cos { and | = {4+ { + 1 intersect at
{ = d r 1=288 and { = e r 0=884= Y = ] e
d [(2 + {
274 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION 39. Y = ] 0 q sin2{ (1)2 [0 (1)]2rg{ CAS = 11 8 2 41. U @2
0 cos2{ g{ describes the volume of the solid obtained by rotating the region R =
({> |) | 0 {
2> 0 | cos { of the{|-plane about the {-axis.
43. ] 1 0 (|
4 |8) g| = ] 1 0
(|2)2 (|4)2g| describes the volume of the solid obtained by rotating the region R = ({> |) | 0 | 1> |4 { |2 of the{|-plane about the |-axis.
45. There are 10 subintervals over the 15-cm length, so we’ll useq = 10@2 = 5 for the Midpoint Rule. Y =U15
0 D({) g{ P5= 1505 [D(1=5) + D(4=5) + D(7=5) + D(10=5) + D(13=5)] = 3(18 + 79 + 106 + 128 + 39) = 3 · 370 = 1110 cm3
47. (a)Y =U210 [i({)]2g{ r 10 24 [i(3)]2+ [i(5)]2+ [i(7)]2+ [i(9)]2 r 2 (1=5)2+ (2=2)2+ (3=8)2+ (3=1)2r 196 units3
(b)Y =U04 (outer radius)2 (inner radius)2g| r 4 0
4
(9=9)2 (2=2)2+(9=7)2 (3=0)2+(9=3)2 (5=6)2+(8=7)2 (6=5)2 r 838 units3
49. We’ll form a right circular cone with heightk and base radius u by revolving the line| = uk{ about the {-axis.
Y = ]k 0 u k{ 2 g{ = ] k 0 u2 k2 {2g{ = u 2 k2 1 3{3 k 0 = uk22 1 3k3 = 13 u2k
Another solution: Revolve { = u
k| + u about the |-axis. Y = ]k 0 u k| + u 2 g|= ] k 0 u2 k2|2 2u 2 k | + u2 g| = u2 3k2|3 u 2 k|2+ u2| k 0 = 1 3u2k u2k + u2k =1 3 u2k Or use substitution withx = u u
k| and gx = ukg| to get ] 0 u x 2k u gx = ku 1 3x3 0 u= ku 13u3= 1 3 u2k.
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51.{2+ |2= u2 {2= u2 |2 Y = ] u uk u2 |2g| = u2| |3 3 u uk = u3 u3 3 u2(u k) (u k)3 3 = 2 3u313(u k) 3u2 (u k)2 =1 3 2u3 (u k)3u2u2 2uk + k2 =1 3 2u3 (u k)2u2+ 2uk k2 =1 3 2u3 2u3 2u2k + uk2+ 2u2k + 2uk2 k3 =1 3 3uk2 k3= 1
3 k2(3u k), or, equivalently, k2
u k3
53.For a cross-section at height|, we see from similar triangles that @2
e@2 = k |k , so = e
1 |k. Similarly, for cross-sections having2e as their base and replacing , = 2e1 |
k . So Y = ]k 0 D(|) g| = ]k 0 k e1 |klk2e1 |klg| =]k 0 2e 21 | k 2 g| = 2e2] k 0 1 2|k + |k22 g| = 2e2| |2 k + | 3 3k2 k 0 = 2e 2k k +1 3k =2
3e2k [ =13Ek where E is the area of the base, as with any pyramid.]
55.A cross-section at height} is a triangle similar to the base, so we’ll multiply the legs of the base triangle, 3 and 4, by a proportionality factor of(5 })@5. Thus, the triangle at height } has area
D(}) = 12· 3 5 } 5 · 4 5 } 5 = 61 }52, so Y = ]5 0 D(}) g} = 6 ] 5 0 1 }52g} = 6 ]0 1 x 2(5 gx) x = 1 }@5, gx = 1 5g} = 301 3x3 0 1= 30 1 3 = 10 cm3
57.Ifo is a leg of the isosceles right triangle and 2| is the hypotenuse, theno2+ o2= (2|)2 2o2= 4|2 o2= 2|2. Y =U2 2D({) g{ = 2 U2 0 D({) g{ = 2 U2 0 12(o)(o) g{ = 2 U2 0 |2g{ = 2U2 0 14(36 9{2) g{ = 92 U2 0(4 {2) g{ =9 2 4{ 1 3{3 2 0=92 8 8 3 = 24
276 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
59. The cross-section of the base corresponding to the coordinate{ has length | = 1 {. The corresponding square with side v has area
D({) = v2= (1 {)2= 1 2{ + {2. Therefore, Y =] 1 0 D({) g{ = ] 1 0 (1 2{ + { 2) g{ ={ {2+1 3{3 1 0= 1 1 +1 3 0 = 1 3 Or: ] 1 0 (1 {) 2g{ =] 0 1 x 2(gx) [x = 1 {] =1 3x3 1 0=13
61. The cross-section of the basee corresponding to the coordinate { has length 1 {2. The heightk also has length 1 {2, so the corresponding isosceles triangle has areaD({) =12ek =12(1 {2)2. Therefore,
Y =] 1 1 1 2(1 {2)2g{ = 2 · 1 2 ] 1 0 (1 2{ 2+ {4) g{ [by symmetry] ={ 2 3{3+15{5 1 0= 1 2 3+15 0 = 8 15
63. (a) The torus is obtained by rotating the circle({ U)2+ |2= u2about the|-axis. Solving for {, we see that the right half of the circle is given by { = U +su2 |2 = i(|) and the left half by { = U su2 |2= j(|). So Y = Uuu [i(|)]2 [j(|)]2 g| = 2 U0ukU2+ 2Usu2 |2+ u2 |2U2 2Usu2 |2+ u2 |2lg| = 2 U0u4Usu2 |2g| = 8 UUu 0 s u2 |2g|
(b) Observe that the integral represents a quarter of the area of a circle with radiusu, so 8 UU0usu2 |2g| = 8 U ·1
4 u2= 2 2u2U.
65. (a) Volume(V1) =U0kD(}) g} = Volume(V2) since the cross-sectional area D(}) at height } is the same for both solids. (b) By Cavalieri’s Principle, the volume of the cylinder in the gure is the same as that of a right circular cylinder with radiusu
and heightk, that is, u2k.
67.The volume is obtained by rotating the area common to two circles of radiusu, as shown. The volume of the right half is
Yright= Uu@2 0 |2g{ = Uu@2 0 k u21 2u + { 2lg{ = ku2{ 1 3 1 2u + { 3lu@2 0 = 1 2u313u3 0 1 24u3 = 5 24 u3 So by symmetry, the total volume is twice this, or125 u3.
Another solution: We observe that the volume is the twice the volume of a cap of a sphere, so we can use the formula from Exercise 51 withk =12u: Y = 2 · 13 k2(3u k) =23 21u23u 12u=125 u3.
69.Take the{-axis to be the axis of the cylindrical hole of radius u. A quarter of the cross-section through|> perpendicular to the |-axis, is the rectangle shown. Using the Pythagorean Theorem twice, we see that the dimensions of this rectangle are
{ =sU2 |2and} =su2 |2, so 1 4D(|) = {} = s u2 |2sU2 |2, and Y =Uuu D(|) g| =Uuu 4su2 |2sU2 |2g| = 8Uu 0 s u2 |2sU2 |2g|
71. (a) The radius of the barrel is the same at each end by symmetry, since the function| = U f{2is even. Since the barrel is obtained by rotating the graph of the function| about the {-axis, this radius is equal to the value of| at { = 12k, which is U f12k2= U g = u.
(b) The barrel is symmetric about the|-axis, so its volume is twice the volume of that part of the barrel for { A 0. Also, the barrel is a volume of rotation, so
Y = 2] k@2 0 | 2g{ = 2 ]k@2 0 U f{22 g{ = 2 U2{ 2 3Uf{3+15f2{5 k@2 0 = 2 1 2U2k 121Ufk3+1601 f2k5
Trying to make this look more like the expression we want, we rewrite it asY = 13 k2U2+U212Ufk2+803f2k4. ButU212Ufk2+803f2k4 =U 14fk22401f2k4= (U g)22514fk22 = u225g2.
