198

8.1 Homework Solutions 1. y = x+1, y = 9− x2, x = −1, x = 2

A = 9− x2( )− x+1( )⎡

⎣⎢⎤⎦⎥−1

2

∫ dx = 9− x2 − x−1( )−1

2

∫ dx = 8− x− x2( )−1

2

∫ dx

= 8x− x2

2 − x3

3⎡

⎣⎢

⎤

⎦⎥−1

2

= 16− 42 −83

⎡⎣⎢

⎤⎦⎥− −8− 12 +

13

⎡⎣⎢

⎤⎦⎥=19.5

3. y = x2, y = 8− x3, x = −3, x = 3

A = 8− x3( )− x2⎡

⎣⎢⎤⎦⎥dx−3

1.716

∫ + x2 − 8− x3( )⎡⎣⎢

⎤⎦⎥dx1.716

3

∫ = 60.252

199

5. y = 5x− x2, y = x

A = 5x− x2( )− x⎡⎣⎢

⎤⎦⎥dx = 4x− x2⎡⎣ ⎤⎦dx0

4

∫0

4

∫ = 2x2 − 13 x3⎡

⎣⎢⎤⎦⎥0

4

= 32− 83 =32 3

7. 2, 2, 1, 1yx e x y y y= = − =− =

A = ey − y2 − 2( )⎡⎣⎢

⎤⎦⎥dy−1

1

∫ = ey − 13 y3 + 2y⎡

⎣⎢⎤⎦⎥−1

1

= e1 − 13+ 2⎡⎣⎢

⎤⎦⎥− 1e +13− 2

⎡⎣⎢

⎤⎦⎥= 5.684

200

9. 212, , 1, 2y x y x xx

= + = = =

A = 1x2

− x+ 2⎛⎝⎜

⎞⎠⎟dx

0

1.666

∫ + x+ 2 − 1x2

⎛⎝⎜

⎞⎠⎟dx

1.666

2

∫= lima→0+

x−2 − x+ 2( )12⎡⎣⎢

⎤⎦⎥dx

a

1.666

∫ + .554

= lima→0+

− 1x −23 x+ 2( )32

a

1.666

+ .554

= lima→0+

− 11.666 −

23 3.666( )32⎛

⎝⎜⎞⎠⎟− − 1a −

23 a+ 2( )32⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥+ .554

lima→0+

− 1a dne∴theintegraldiverges

11. y = x , y = e−2x , x =1

A = x − e−2x( )dx.301

1

∫ = 23 x

32 + 12 e−2x⎡

⎣⎢⎤⎦⎥.301

1

= .350

201

13. 2, 2xy x y= =

A = 2x − x2⎡⎣ ⎤⎦dx−.767

2

∫ = 2xln2 −

13 x

3⎡

⎣⎢

⎤

⎦⎥−.767

2

= 2.106

202

8.2 Homework Solutions 1. 2, 1, 0; about the axis.y x x y x= = = −

V =π r2 dx0

1

∫ =π x2( )2 dx0

1

∫ =π x4 dx0

1

∫ =π x55

⎡

⎣⎢⎢

⎤

⎦⎥⎥0

1

= π5

3. 1 , 1, 2, 0; about the -axis.y x x y xx

= = = =

V =π r2 dx1

2

∫ =π 1x

⎛⎝⎜

⎞⎠⎟

2

dx1

2

∫ =π x−2 dx =π x−1−1

⎡

⎣⎢

⎤

⎦⎥1

2

1

2

∫ =π 2

203

5. 2 , 0, 1; about the -axis.y x x y y= = =

V =π r2 dy0

1

∫ =π x2 dy0

1

∫ =π y2( )2 dy0

1

∫ =π y4 dy0

1

∫ =π y55

⎡

⎣⎢⎢

⎤

⎦⎥⎥0

1

= π5

7. 2 , 2 ; about the -axis.y x x y y= =

V =π R2 − r2( )dy0

2

∫ =π 2y( )2 − y( )2⎡⎣⎢

⎤⎦⎥dy

0

2

∫ π 4y2 − y4( )dy0

2

∫ =π 43 y

3 − 15 y5

0

2⎡

⎣⎢⎢

⎤

⎦⎥⎥= 64π15

204

9. 2 2, 4 , 2 2, 2; about the -axis.y x y x x x x= − = + = − =

V =π R2 − r2( )dx−2

2

∫ = 2π 2+ x2( )2 − 4 − x2( )2⎡

⎣⎢

⎤

⎦⎥dx0

2

∫= 2π x4 + 5x2⎡⎣ ⎤⎦dx0

2

∫

= 2π 15 x

5 + 53 x3

0

2⎡

⎣⎢⎢

⎤

⎦⎥⎥

= 592π15

205

11. 2, , 1; about the -axis.xy x y e x x−= = =

V =π R2 − r2( )dx.301

1

∫=π x( )2 − e−2x( )2⎡

⎣⎢⎤⎦⎥dx

.301

1

∫=π x− e−4x( )dx.301

1

∫=π 1

2 x2 + 14 e

−4x

.301

1⎡

⎣⎢⎢

⎤

⎦⎥⎥=1.207

206

8.3 Homework Solutions 1. Given the curves ( ) lnf x x= , ( ) xg x e−= and x = 4.

1 2 3 4 5

1

2

3

x

y

a. Find the area of the region bounded by three curves in the first quadrant. A = ln x− e−x( )dx1.3497996

4

∫ = 2.250

b. Find the volume of the solid generated by rotating the region around the line y = 2.

V =π ln x( )2 − e−x( )2⎛⎝⎜

⎞⎠⎟ dx1.3497996

4

∫ = 20.254

c. Find the volume of the solid generated by rotating the region around the line y = –1.

V =π ln x− −1( )⎡⎣ ⎤⎦2 − e−x − −1( )⎡⎣ ⎤⎦

2⎛⎝⎜

⎞⎠⎟dx

1.3497996

4∫ = 22.158

3. 1 , 0, 1, 3; about the line 1.y y x x yx

= = = = = −

V =π R2 − r2( )dx1

3

∫ =π 1+ 1x⎛⎝⎜

⎞⎠⎟

2

−12⎛

⎝⎜⎜

⎞

⎠⎟⎟dx

1

3

∫ = 8.997

5. 3, ; about the line 1.y x y x y= = =

V =π R2 − r2( )dx0

1

∫ =π x( )2 − x3( )2⎛⎝⎜

⎞⎠⎟dx

0

1

∫

=π x− x6( )dx0

1

∫ = 12 x

2 − 17 x7⎡

⎣⎢⎤⎦⎥0

1

= 5π14

207

7. y = ln x2 +1( ), y = cos x; about the line y =1.

V =π R2 − r2( )dx−.916

.916

∫ =π 1− y1( )2 − 1− y2( )2⎛⎝

⎞⎠ dx−.916

.916

∫=π 1− ln x2 +1( )( )2 − 1− cosx( )2⎛

⎝⎜⎞⎠⎟dx

−.916

.916

∫ = 3.447

9. y = 2+ sin x, y = 2, x = 0, x = 2π; about the line y = 2.

V =π r2 dx0

2π

∫ = 2π r2 dxπ

2π

∫ = 2π 2− 2+ sin x( )⎡⎣

⎤⎦2dx

π

2π

∫ = 9.870

208

8.4 Homework Solutions 1. sec , 1, 0, ; about the -axis.

6y x y x x yπ= = = =

V = 2πrl dx0

π6∫ = 2π ⋅x ⋅ secx−1( )dx

0

π6∫ = .064

3. 2

3

1 , 1, 8, 0; about the -axis.y x x y yx

= = = =

V = 2πrl dx1

8

∫ = 2π ⋅x ⋅ x−23⎛

⎝⎜⎞⎠⎟ dx1

8

∫ = 70.689

5. ( )21 and the -axis; about the -axis.y x x x y= −

V = 2πrl dx0

1

∫ = 2π ⋅x ⋅ydx0

1

∫ = 2πx x−1( )2 dx0

1

∫ = .209

7. 2, , 1; about the line 1.xy x y e x x−= = = =

V = 2πrl dx.301

1

∫ = 2π 1− x( ) x − e−2x( )dx.301

1

∫ = .554

9. 2, 2 ; about the line 1.xy x y x= = =−

V = 2πrl dx−.767

2

∫ + 2πrl dx2

4

∫= 2π 1+ x( ) 2x − x2( )dx−.767

2

∫ + 2π 1+ x( ) x2 − 2x( )dx2

4

∫= 55.428

209

8.5 Homework Solutions 1. y = x, y = e−2x, x =1; the cross-sections are semi-circles.

V = 12πr

2 dx.301

1⌠⌡⎮

= π2

x − e−2x2

⎛

⎝⎜⎞

⎠⎟

2

dx.301

1⌠

⌡⎮⎮

= .085

3. Given the curves

f x( ) = x3 −3x2 + 2x+ 4 ,

g x( ) = 2x 4− x and in the first

quadrant.

a. Find the area of the regions R, S, and T.

AreaR = x3 −3x2 + 2x+ 4( )− 2x 4− x( )⎡

⎣⎢⎤⎦⎥0

1.1418956⌠⌡⎮

dx = 2.463

AreaS = 2x 4− x( )− x3 −3x2 + 2x+ 4( )⎡

⎣⎢⎤⎦⎥1.1418956

2.5444495⌠⌡⎮

dx =1.551

AreaT = 2x 4− x⎡

⎣⎢⎤⎦⎥0

4⌠⌡⎮

dx− AreaS =15.515

b. Find the volume of the solid generated by rotating the curve ( ) 2 4g x x x= − around the line y = 8 on the interval 1,3x ⎡ ⎤⎣ ⎦∈ .

V =π 2x 4− x( )

1

3⌠⌡⎮

2

dx =184.308

210

c. Find the volume of the solid generated by rotating the region S around the line y = –1.

V = 2x 4− x − −1( )( )2 − x3 −3x2 + 2x+ 4− −1( )( )2⎡

⎣⎢

⎤

⎦⎥

1.1418956

2.5444495⌠

⌡⎮ dx = 5.752

d. Find the volume of the solid generated if R forms the base of a solid whose cross sections are squares whose bases are perpendicular to the x-axis

V = x3 −3x2 + 2x+ 4( )− 2x 4− x( )⎡

⎣⎢⎤⎦⎥0

1.1418956⌠⌡⎮

2

dx = 6.515

5. Let R and S be the regions bounded by the graphs ( )2 3x y y= − and

2y x= − +

a. Find the area of the regions R and S.

AR = 3y2 − y3( )− 2− y( )⎡

⎣⎢⎤⎦⎥.74589831

3.1149075⌠⌡⎮

dy = 6.185

AS = 2− y( )− 3y2 − y3( )⎡

⎣⎢⎤⎦⎥−.8608059

.74589831⌠⌡⎮

dy = 2.193

211

b. Find the volume of the solid when the region S is rotated around the line x = 3 .

VS =π 1+ y( )2 − 1−3y2 + y3( )2⎡

⎣⎢⎤⎦⎥−.8608059

.74589831⌠

⌡⎮ dy = 23.814

c. Find the volume of the solid when the region R is rotated around the line x = −2 .

V =π 3y2 − y3 + 2( )2 − 2− y+ 2( )2⎡

⎣⎢⎤⎦⎥.74589831

3.1149075⌠

⌡⎮ dy = 45.081

d. Find the volume of the solid if the region S forms the base of a solid whose cross sections are equilateral triangles with bases perpendicular to the y-axis.

(Hint: the area of an equilateral triangle is 2 34sA= ).

VS = 1+ y( )− 1−3y2 + y3( )⎡

⎣⎢⎤⎦⎥−.8608059

.74589831⌠⌡⎮

2

dy = 8.607

7. Let f and g be the functions given by f x( ) = 1

4 + sin πx( ) and g x( ) = e−x . Let

R be the region in the first quadrant enclosed by the y – axis and the graphs of f and g, and let S be the region in the first quadrant enclosed by the graphs of f and g, as shown in the figure below.

212

a. Find the area of R

AR = e−x − 1

4 + sin πx( )⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟0

.19436069⌠

⌡⎮ dx = .071

b. Find the area of S

AS =

14 + sin πx( )⎛

⎝⎜⎞⎠⎟− e−x⎛

⎝⎜⎞⎠⎟.19436069

.95723257⌠

⌡⎮ dx = .328

c. Find the volume of the solid generated when S is revolved around the horizontal line 2y = −

VS =π14 + sin πx( )+ 2⎛

⎝⎜⎞⎠⎟

2

− e−x + 2( )2⎛

⎝⎜⎜

⎞

⎠⎟⎟

.19436069

.95723257⌠

⌡

⎮⎮

dx = 5.837

d. The region R forms the base of a solid whose cross-sections are rectangles with bases perpendicular to the x-axis and heights equal to half the length of the base. Find the volume of the solid.

VS =π12

14 + sin πx( )⎛

⎝⎜⎞⎠⎟− e−x( )⎛

⎝⎜

⎞

⎠⎟

2

.19436069

.95723257⌠

⌡⎮⎮

dx = .007

213

8.6 Homework Set A Solutions Find the arc length of the curve. 1.

y =1+ 6x

32 on x ∈ 0, 1⎡⎣ ⎤⎦

′y = 9x12 → L = 1+ 81x dx

0

1

∫ = 6.103

3. x = 1

3y y − 3( ) on y ∈ 1, 9⎡⎣ ⎤⎦

x = 13 y32 − y

12 → ′x = 12 y12 − 12 y

−12

L = 14 y−

12 +

14 y

−1 dy1

9⌠

⌡⎮=10.177

5.

y = ln x

32 on x ∈ 1, 3⎡

⎣⎢⎤⎦⎥

′y = 1x32⋅ 32 x

12 = 32x→ L = 1+ 9

4x2 dx1

3⌠

⌡⎮=1.106

7. Find the perimeter of each of the two regions bounded by 2y x= and 2xy = .

L = 1+ 2x( )2 dx−.767

2

∫ + 1+ 2x ln2( )2 dx−.767

2

∫ + 1+ 2x( )2 dx2

4

∫ + 1+ 2x ln2( )2 dx2

4

∫=10.189+ 24.364 = 34.553

9. Find the length of the arc along ( ) 4

23 1 xf x t dt

−= −∫ on 2, 1x ⎡ ⎤⎣ ⎦∈ − .

′f x( ) = 3x4 −1→ L = 1+ 3x4 −1dx−2

1

∫ = 5.196