6.2. Volumes
6.2. Volumes: 6, 8, 13, 17, 50, 55, 61, 64, 66
6.
Write y = ln x as x = ey , the cross-sectional area is
A(y) = π(ey )2 = πe2y .
Then the volume is
V =
Z
2
A(y)dy = π
1
y=2
Z
2
e2y dy
1
π h 2y iy=2
πZ
e
e2y d(2y) =
=
y=1
2 y=1
2
4
2
i
h
π (e − e )
π 4
e − e2 =
.
=
2
2
1
8.
By solving y = 5 − x2 = x2 /4, we find the intersection points are (±2, 1).
So we want to integral from x = −2 to x = 2.
The radii are dependent on the value of y = 5 − x2 and y = x2 /4.
The cross-sectional area is
A(x) = π(outer radius)2 − π(inner radius)2

= π  5 − x2
2
x2
−
4
The volume is
V =
Z
!2 

2
A(x)dx = π
−2
Z
2
−2
15x4
.
= π 25 − 10x2 +
16
"
10x3 3x5
= π 25x −
+
3
16
#2
−2
80
80
50 −
+ 6 − −50 +
−6
3
3
176π
176
=
.
=π×
3
3
2
#
15x4
dx
25 − 10x +
16
2
"
=π
#
"
13.
Because sec x is period function, there are many regions.
But we just consider one and compute the volume.
First of all, we solve y = 1 + sec x = 3 to find that the intersection points are
(±π/3, 3).
So we want to integral from x = −π/3 to x = π/3.
The cross-sectional area is
A(x) = π(outer radius)2 − π(inner radius)2
h
= π (3 − 1)2 − (sec x)2
h
i
= π 4 − sec2 x
The volume is
V =
Z
π/3
A(x)dx = π
−π/3
Z
π/3
−π/3
i
h
i
4 − sec2 x dx.
(You may see the Reference page 6 in the textbook.)
π/3
= π [4x − tan x]−π/3
=π
4π √
4π √
− 3 − −
+ 3
3
3
3
4π √
− 3 .
= 2π
3
17.
√
√
We write y = x2 as x = y because we consider y ≥ 0. Note that x = y is
outer than x = y 2 while rotating. The cross-sectional area is
A(y) = π(outer radius)2 − π(inner radius)2
2 √
2
2
= π ( y + 1) − y + 1
h √
i
= π 2 y + y − 2y 2 − y 4
1
h
= π 2y 2 + y − 2y 2 − y 4
The volume is
V =
Z
1
A(y)dy = π
0
Z
0

1
i
i
√
2 y + y − 2y 2 − y 4 dy
h
1
3
2
y 2 2y 3 y 5 
4y
+
−
−
= π
3
2
3
5
0
4 1 2 1
=π
+ − −
3 2 3 5
29π
.
=
30
4
50.
The cross-sectional square has side a(1 − x/h) + bx/h = a + (b − a)x/h and
then area
A(x) = a + (b − a)
x
h
2
x2
x
= a2 + 2a(b − a) + (b − a)2 2 .
h
h
for 0 ≤ x ≤ h. Therefore, the volume of the solid is
V =
Z
h
A(x)dx =
0
"
Z
h
0
"
#
x
x2
a + 2a(b − a) + (b − a)2 2 dx
h
h
2
x2 (b − a)2 x3
= a x + a(b − a) +
h
3h2
2
2
#h
= a2 h + a(b − a)h +
5
2
0
(a − 2ab + b )h
(a2 + ab + b2 )h
(b − a) h
= abh +
=
.
3
3
3
If a = b, the solid is a parallelepiped with volume (b2 + bb + b2 )h/3 = b2 h.
If a = 0, the solid is a pyramid with volume (b2 )h/3.
It has be shown in example 8 on page 437 in the textbook.
2
55.
This problem, in some way, is like √
the example 7 on page 436 in textbook.
We write 9x2 + 4y 2 = 36 as y = ± 36 − 9x2 /2 for −2 ≤ x ≤ 2.
For every cross-sectional isosceles right triangle,the hypotenuse is
√
√
!
!
√
36 − 9x2
36 − 9x2
− −
= 36 − 9x2 .
2
2
√
√
Then the side lengths both are 36 − 9x2 / 2. And any cross-sectional
isosceles right triangle has area
√
!2
9
1
36 − 9x2
36 − 9x2
√
= 9 − x2 .
A(x) =
=
2
4
4
2
Hence, the volume is
V =
Z
2
A(x)dx =
2
−2
−2
3
= 9x − x3
4
Z
2
−2
9
9 − x2 dx
4
= (18 − 6) − (18 − 6) = 24.
6
61.
(a) For any cross-section, the outer radius is Router = R + r cos θ
and the inner radius is Rinner = R − r cos θ for −π/2 ≤ θ ≤ π/2.
So the cross-sectional area is
h
2
2
A(θ) = πRouter
− πRinner
= π (R + r cos θ)2 − (R − r cos θ)2
=π
h
R2 + r 2 cos2 θ + 2Rr cos θ − R2 + r 2 cos2 θ − 2Rr cos θ
= π [4Rr cos θ] = 4πRr cos θ.
The volume is
V =
Z
π/2
A(θ)d(r sin θ).
−π/2
(b) Now, we find the volume.
V =
Z
π/2
A(θ)d(r sin θ) = 4πRr 2
Z
π/2
cos2 θdθ
−π/2
−π/2
(see Reference Page 9)
= 4πRr
2
"
θ sin 2θ
+
2
4
#π/2
−π/2
7
= 4πRr 2 ×
π
= 2π 2 Rr 2 .
2
i
i
64.
√
Note that every cross-section is square of side 2 r 2 − z 2 for −r ≤ z ≤ r.
So the cross-sectional area is
2
√
A(z) = 2 r 2 − z 2 = 4(r 2 − z 2 ).
Therefore, the volume is
V =
Z
r
A(z)dz = 4
−r
"
z3
=4 r z−
3
2
Z
r
−r
#r
−r
"
(r 2 − z 2 )dz
#
2r 3 2r 3
16r 3
=4
=
+
.
3
3
3
8
66.
Let R = 15(cm) and r = 5(cm) be the radii.
Let Vbowl be summation of volumes of water and the ball in the water.
And let Vball be the volume of the ball in the water.
The volume of water is Vwater = Vbowl − Vball First, we find the Vbowl .
Every cross-sectional area is A(y) = π(R2 − y 2) where y denotes the depth.
So the volume is
Vbowl =
Z
−R+h
A(y)dy = π
−R
"
y3
=π R y−
3
=π
"
2
Z
−R+h
−R
#−R+h
−R
(−R + h)3
R (−R + h) −
3
2
"
(R2 − y 2)dy
!
R3
− −R +
3
3R2 h − 3Rh2 + h3
=π R h−
3
2
"
h3
= π Rh −
3
2
9
#
3
#
!#
We discuss in these two cases:
(1) the ball is completely in water (2r ≤ h ≤ R) and (2) not (h < 2r).
(1) Vball is the volume of a ball. We have
4
Vball = πr 3 .
3
So the volume of water is
Vwater = Vbowl − Vball
"
h3 4 3
− r
= π Rh −
3
3
2
#
(2) Similarly, we have
Vball = π
Z
−r+h
−r
"
(r 2 − x2 )dx
h3
= π rh −
3
2
#
So the volume of water is
Vwater = Vbowl − Vball
=π
"
h3
Rh2 −
3
!
h3
− rh2 −
3
!#
i
h
= π Rh2 − rh2 = π(R − r)h2 .
This turn out to be
Vwater = Vbowl − Vball
=
(
h
π 15h2 −
10πh2
h3
3
−
500
3
10
i
, if 10 ≤ h ≤ 15
.
, if h < 10