8.1 homework solutions - kevin quattrin, edd
TRANSCRIPT
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