微積分作業解答 Chapter 7 Applications of Integration 7.1-7.4
7.1
1-4. Find the area of the shaded region.
1.
A
³
4
0
x3
2x 3
(5 x x ) x dy
2
4
2
0
32
3
3.
A
1
³
1
1
1
e y y3 2 y
3
1
e y ( y 2 2) dy
1 10
e e 3
5-16. Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x
or y. Draw a typical approximating rectangle and label its height and width. Then find the area of
the region.
9. y 12 x 2 , y
x2 6
先找出兩曲線交點作為積分上下限， 2 x 2 18 Ÿ x
則A
³
3
3
3, 3 ，
3
(12 x ) ( x 6) dx
2
2
2 º
ª
2 «18 x x 3 »
3 ¼0
¬
1
72
廖彥法 整理
微積分作業解答
11. x
2 y2, x y 1
先 找 出 兩 曲 線 交 點 作 為 積 分 上 下 限 ，
1
2 y2 y 1 0 Ÿ y
, 1 ，
2
則A
15. y
1/ 2
³
1
1
,y
x
則A
(1 y ) 2 y 2 dy
x, y
1
³
0
(x y
1
y
2
x
,x !0
4
2 1
1
1
x) dx ³ ( x)dx
1
4
x 4
21. Sketch the region that lies between the curves y
cos x and y sin 2 x and between x
0
29. Find the area of the crescent-shaped region (called a lune) bounded by arcs of circles with radii r
and R(see the figure).
由圖可看出，該斜線面積對y軸對稱，所以
A
則
令
微積分作業解答 Chapter 7 Applications of Integration 7.1-7.4
則³
r
0
R x dx
2
2
R
2
³
sin 1 ( r / R )
0
sin 1 ( r / R )
1
ª1
º
R « t sin 2t »
4
¬2
¼0
2
2
cos tdt
1 2 1 r r
R sin
R2 r 2
2
R 2
R2 r 2
且b
所以 A r R 2 r 2 Sr 2
r
R 2 sin 1
2
R
y
33. Find the number b such that the line y b divides the region bounded by the curves y
and y 4 into two regions with equal area.
x2
由圖可知對稱於y軸，故
2
A 2 ³ (4 x 2 ) dx
0
且 Ab
b
2 Ab Ÿ Ab
2³ (b x 2 )dx
0
[4 x 4 3/ 2
b Ÿb
3
37. For what values of m do the line y
x3 2
]0
3
16
3
4 2 / 3 | 2.52
mx and the curve y
x
enclose a region? Find the
x 1
2
area of the region.
由圖可知兩曲線相交時，交點 x
A
r
1
1
1 存在， ! 1 Ÿ 0 m 1
m
m
微積分作業解答 Chapter 7 Applications of Integration 7.1-7.4
7.2
1-12. Find the volume of the solid obtained by rotating the region bounded by the given curves
about the specified line. Sketch the region, the solid, and a typical disk or washer.
1. y 1 / x, x 1, x
2, y
0; about the x axis
3.
5.
,
,
1
V
7. y 2
1
1
S ³ ( y1 y2 ) dx S ³ ( x x ) dx S ( x 3 x 7 )
0
3
7
0
2
x, x
2
1
2
4S
21
6
2 y; about the y axis
2
V
4
1
S ³ ( x1 x2 ) dy S ³ ( 4 y y ) dy S ( y 3 y 5 )
0
3
5
0
2
2
2
2
4
4
64S
15
廖彥法 整理
微積分作業解答 Chapter 7 Applications of Integration 7.1-7.4
9. y
x, y
x ; about y 1
1
V
1
S ³ ( y1 y2 ) dx S ³ [(1 x) 2 (1 x ) 2 ]dx S (
2
2
0
3 2 1 3 4 3/ 2
x x x )
2
3
3
0
S
6
1
3
)
0
13. The region enclosed by the curves y
(a) S ³
S /2
0
(b)為 y
1
0
cos x,0 d x d
x與y
x is rotated about the line x 1 . Find the
(b) S ³ ( y 4 y 8 )dy
cos 2 xdx
(a) 為 y
x 3 and y
29S
30
4
S
對x軸旋轉所得體積。
2
x 所交會的區域對y軸旋轉所得體積。
5
廖彥法 整理
微積分作業解答 Chapter 7 Applications of Integration 7.1-7.4
25-37. Find the volume of the described solid S.
27. A cap of a sphere with radius r and height h.
9x 4 y
2
2
36 Ÿ y
36 9 x 2
r
,2 d x d 2 取正，為橢圓上半部的曲線函數
2
41. (a) Set up an integral for the volume of a solid torus (the donut-shaped solid shown in the figure)
with radii r and R.
(b) By interpreting the integral as an area, find the volume of the torus.
(a) 該圓形函數為 ( x R ) 2 y 2
r2
分成左右兩邊的函數：左函數 xL
r
R r 2 y 2 ；右函數 xR
R r2 y2 ， 0 d y d r
微積分作業解答 Chapter 7 Applications of Integration 7.1-7.4
45. Find the volume common to two spheres, each with radius r, if the center of each sphere lies on
the surface of the other sphere.
r
左邊圓形函數為 ( x ) 2 y 2
2
而斜線部分的函數為 y
V
S ³ y dx S ³
2
r/2
0
r2
r
r
r 2 ( x )2 ， 0 d x d
2
2
r/2
3
3
1
1
( r 2 x 2 xr ) dx S ( r 2 x x 3 x 2 r )
4
4
3
2
0
其全部體積為斜線部分對x軸旋轉體積的兩倍，即
7
5 3
Sr
24
5 3
Sr
12
廖彥法 整理
微積分作業解答 Chapter 7 Applications of Integration 7.1-7.4
7.3
1. Let S be the solid obtained by rotating the region shown in the figure about the y-axis. Explain
why it is awkward to use slicing to find the volume V of S. Sketch a typical approximating shell.
What are its circumference and height? Use shells to find V.
若使用圓環法(Washer)，則必須找出 y x( x 1) 2 的頂點，將其分成左右兩邊的函數，以
進行體積積分，對於該函數而言，難以化整為好積分的函數。
V
2
3.
5.
,
,
,
1
˜
2 x 2
x2
1
8
1
廖彥法 整理
微積分作業解答 Chapter 7 Applications of Integration 7.1-7.4
9-14. Use the method of cylindrical shells to find the volume of the solid obtained by rotating the
region bounded by the given curves about the x-axis. Sketch the region and a typical shell.
9. x 1 y 2 , x
0, y 1, y
2
V
11. y
2S xydy
x3 , y
2S
8, x
2
(1 y 2 ) ˜ ydy
1
1
2S ( y 2 y 4 )
2
21S
0
8
V
2S ³ xydy
2S ³
13. y
4 x 2 ,2 x y
6
8
0
3
y ˜ ydy
6 7/3
Sy
7
0
768S
7
先找出兩函數交點
­ y 4x2
Ÿx
®
2
x
y
6
¯
V
3
,1 Ÿ y
2
2S ³ y ( x1 x2 ) dy
4
4
2S ³ y[
0
9,4
9
y
y
y
y
(
)]dy 2S ³ y[( 3) (
)]dy
4
2
2
2
2
9
ª y3 3 2 1 5/ 2 º
4 5/ 2
Sy
2S « y y »
5
5
0
¬ 6 2
¼4
250
S
3
9
廖彥法 整理
15-20. Use the method of cylindrical shells to find the volume generated by rotating the region
bounded by the given curves about the specified axis. Sketch the region and a typical shell.
15. y
x2
0 x 1x
2 about x 1
x2, y
0, x 1, x
2; about x
V
17. y
4
2
V
19. y
2S ³ (4 x ) ydx
x 1, y
2
2S ³ ( 4 x) x dx
2
1
ª x3 x 4 º
2S « »
4 ¼1
¬3
67
S
6
0, x 5; about y 3
2
V
2S ³ (3 y )(5 x) dy
2
2S ³ (3 y )(5 y 1) dy
2
0
ª
y4 º
2S «12 y 2 y 2 y 3 »
4 ¼0
¬
24S
21-26. Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the
region bounded by the given curves about the specified axis.
10
廖彥法 整理
微積分作業解答 Chapter 7 Applications of Integration 7.1-7.4
x4, y
23. y
1
sin(Sx / 2); about x
1
29-32. Each integral represents the volume of a solid. Describe the solid.
3
29.
³ 2Sx dx
5
0
3
3
³ 2Sx dx
2S ³ x ˜ x 4 dx ，即 0 d y d x 4 ， 0 d x d 3 之區域對 y 軸旋轉所得之體積。
5
0
0
1
31.
³ 2S (3 y)(1 y
0
1
³
0
2
) dy
1
2S ³ ( y 3)( y 2 1) dy ，即(i) x
2S (3 y )(1 y 2 ) dy
0
對 y 3 軸旋轉所得之體積。(ii) x
體積。
y 2 、 x 1、 y
y2 1 、 x
0、 y
0 所圍成之區域
0 所圍成之區域對 y
3 軸旋轉所得之
33-38. The region bounded by the given curves is rotated about the specified axis. Find the volume
of the resulting solid by any method.
33. y
x 2 x 2, y
0; about x axis
y
x 2 x 2, y
0Ÿ x
V
ª x5 x 4
º
S ³ y dx S ³ ( x x 2) dx S « x 3 2 x 2 4 x »
2
2
¬5
¼ 2
2,1
1
35. y
2
1
2
x ( 4 / x); about x
5, y
2
81
S
10
1
4
4
2S ³ [ x ( 1)][5 ( x )]dx
1
x
4
V
ª x3
º
2S « 2 x 2 x 4 ln x »
¬ 3
¼1
11
8S (3 ln 4)
廖彥法 整理
微積分作業解答 Chapter 7 Applications of Integration 7.1-7.4
7.4
1. Use the arc length formula (3) to find the length of the curve y 2 3x,2 d x d 1 . Check your
answer by noting that the curve is a line segment and calculating its length by the distance formula.
³
L
1
³
1 ( y ' ) 2 dx
對於線段 y
1 ( 3) 2 dx 3 10
2
2 3x,2 d x d 1 ，其端點為 (2,8), (1,1) ，
所以長度為 [1 ( 2)]2 [(1) 8]2
3 10
3-14. Find the length of the curve.
3. y 1 6 x 3 / 2 ,0 d x d 1
³
L
1
³
1 ( y ' ) 2 dx
0
令 t 1 81x Ÿ dt
則 ³ 1 81x dx
0
³
1 81x dx
0
81dx
82
1 82 1/ 2
t dt
81 ³1
1
1
1 (9 x1/ 2 ) 2 dx
2 3/ 2
t
243
1
2
(82 82 1)
243
x5
1
,1 d x d 2
6 10 x 3
5. y
³
L
1 ( y ' ) 2 dx
³
5
3
( x 4 x 4 ) 2 dx
6
10
2
1
³
2
1
5 4 3 4
x x dx
6
10
2
ª 1 5 3 3 º
«¬ 6 x 10 x »¼
1
1261
240
ln(sec x ),0 d x d S / 4
9. y
³
L
13. y
1 ( y ' ) 2 dx
³
S /4
0
³
(sec x) 2 dx
S /4
sec xdx
0
S /4
ln(sec x tan x) 0
ln( 2 1)
e x ,0 d x d 1
L
³
1 ( y ' ) 2 dx
令 tan t
ex Ÿ x
則 ³ 1 e 2 x dx
令u
cos t Ÿ du
則³
1
dt
sin t cos 2 t
1
³
1 e 2 x dx
0
ln | tan t |Ÿ dx
1
³ sin t cos
2
t
1
dt
sin t cos t
dt
sin tdt
³ (u
2
1
du
1)u 2
§
1
³ ¨© u
1 1 u 1
ln
C
u 2 u 1
2
1/ 2 1/ 2 ·
¸ du
u 1 u 1 ¹
1 1
1
ln | u 1 | ln | u 1 | C
u 2
2
1
1 cos t 1
ln
C
cos t 2 cos t 1
12
廖彥法 整理
微積分作業解答 Chapter 7 Applications of Integration 7.1-7.4
1
1 e
1
所以 ³ 1 e 2 x dx
0
2x
1
2x
1
1
e
ln
C
1
2
1
1 e2x
ª
1 1 1 e2x
« 1 e 2 x ln
2 1 1 e2x
«¬
1 1 1 e2x
1 e 2 x ln
C
2 1 1 e2x
1
º
»
»¼
0
1 1 1 e2
1 1 2
1 e ln
2 ln
2
2 1 1 e
2 1 2
2
17. x
y y 3 ,1 d y d 4 ,Set up, but do not evaluate, an integral for the length of the curve.
x
y y3 Ÿ
L
³
dx
dy
1 3y 2
1 ( x' ) 2 dy
³
4
2 6 y 2 9 y 4 dy
1
25. Sketch the curve with equation x 2 / 3 y 2 / 3 1 and use symmetry to find its length.
L
y
L (t )
³
1 ( y ' ) 2 dx
令 u 1 9 x Ÿ du
t
則 ³ 1 9 x dx
1
³
t
1
1 9 x dx
9dx
1 1 9 t 1 / 2
u du
9 ³10
1 9 t
2 3/ 2
u
27
10
2
(1 9t ) .3 / 2 10 10
27
13
廖彥法 整理