§6-2 Volumes + §6-3 Volumes by Cylindrical Shells
Homework(§6-2)：7,11,18,19,21,22,28,35,41,47,49,50,56,64,65,67
\Homework(§6-3)：3,13,17,20,22,23,29,30,39,41
公式：
z
⎧
V2 ( 2維截面)＝ ⎨
⎩
V3 = ∫ V2 dx
§6.2 (截面為環狀)
§6.3 (截面為圓柱狀)
4
Example 1：Prove that the volume of the sphere with radius r is π r 3 .
3
Solution：
r
r
0
0
(
)
Ｖ　= 2∫ π r ′2 dx = 2∫ π r 2 − x 2 dx
4
= π r 3 .....................................
3
r
r
0
0
3 r
2 2
(§6.2)
V = 2 ∫ 2π xh dx = ∫ 2π x r 2 − x 2 dx
=−
4π 2
r −x
3
(
)
0
4
= π r 3 ......... (§6.3)
3
Example 2：Find the volume of the cone with base radius r and the height h.
Solution：
x r′
=
h r
r
⇒ r ′ = x.
h
Ｖ　= ∫0 π r ′2 dx = ∫0
h
1
= π r 2 h.
3
h
π r2
h
2
x 2 dx
Example 3：A frustum of a pyramid with square base of side b, square top of side
a, and
height h.
Solution：
Ｖ = ∫0 r 2 dx
h
k
a
ah
=
⇒k=
h b−a
b−a
x+k r
⎛b−a⎞
= ⇒ r = a+⎜
⎟ x.
k
a
⎝ h ⎠
b−a ⎞
h⎛
⇒ Ｖ = ∫0 ⎜ a +
x ⎟ dx
h
⎝
⎠
2
h
b−a ⎞
⎛
=
x⎟
⎜a +
h
3(b − a ) ⎝
⎠
Example 4：
{
3
h
0
=
(
)
h 2
a + ab + b 2 .
3
y= x
y = x2
（由此 2 曲線所圍區域繞著 x 軸所旋轉出來的體積）
Solution：
方法 1（截面為環狀）
1
∫ πr
=∫ πx
V=
0
1
2
−π r2 dx
2
− π x 4 dx
1
0
=
2
r1 = x, r2 = x 2
2π
.
15
方法 2（截面為圓柱狀）
1
V = ∫ 2π r h dy
0
1
= ∫ 2π y
0
2π
=
.
15
(
)
y − y dy
⎧ y = x3
⎪
Example 5： ⎨ y = 8
⎪x = 0
⎩
Solution：
2
(
)
V = ∫ 61 − x 6 π dx（截面為環狀）
0
8
1
= ∫ 2π y 3 y dy
0
768
=
π.
7
Example 6：
{
y = x − x2
y =0
（截面為圓柱狀）
x=2
Solution：
Ｖ　= ∫ 0 2π ( 2 − x ) ( x − x 2 ) dx
1
π
= .
2
Example 7： A wedge is cut out of a circular cylinder of radius 4 by two planes.
One plane is perpendicular to the axis of the cylinder. The other
intersects the first at an angle of 30 Ο along a diameter of the
cylinder. Find the volume of the wedge.
Solution：
(Method 1)
V = ∫ (長方形截面)dx
4
0
4
⎛ x ⎞
2
x2 ⎜
= ∫ 2
4
−
⎟dx
0
3
⎝
長
⎠
寬
128
=
.
3 3
(Method 2)
V = 2 ∫ (ΔABC截面 ) dx
4
0
= ∫ (AB × BC)dx
4
0
=∫
4
16 − x 2
0
=
dx
3
128
3 3
.
Example 8：Find the volume common to two spheres, each with radius r , if
the center of each sphere lies on the surface of other .
Solution：
r
2
0
r
2
0
V = 2 ∫ π y 2 dx
2
⎛ 2 ⎛
r ⎞ ⎞⎟
⎜
= 2 ∫ π r − ⎜ x + ⎟ dx
⎜
2 ⎠ ⎟⎠
⎝
⎝
5
=
π r 3.
24
or
0
V = 2 ∫ r π y 2 dx
−
2
2
⎛ 2 ⎛
r ⎞ ⎞⎟
⎜
= 2 ∫ r π r − ⎜ x − ⎟ dx
⎜
−
2 ⎠ ⎟⎠
⎝
2 ⎝
5
=
π r 3.
24
0