MATH 101 HOMEWORK 8 - SOLUTIONS
1. A bowl is shaped like a hemisphere with diameter 30 cm. A ball with diameter 10 cm
is placed in the bowl and water is poured into the bowl to a depth of h centimeters. Find
the volume of water in the bowl. (Note: you may have to consider the cases h ≤ 10 and
h > 10 separately.)
In a coordinate system as above, the equations for the large ball and the small ball,
respectively, are
(y − 15)2 + x2 = 152 , (y − 5)2 + x2 = 52 .
The volume inside the large ball up to height h, using slices, is
Z h
Z h
y 3 h
h3 2
2
π(15 − (y − 15) )dy =
π(30y − y 2 )dy = π 15y 2 −
.
= π 15h2 −
3 0
3
0
0
If h < 10, the volume inside the small ball up to height h is
Z h
Z h
y 3 h
h3 2
2
2
2
2
.
π(5 − (y − 5) )dy =
π(10y − y )dy = π 5y −
= π 5h −
3 0
3
0
0
Thus, if h < 10, the volume of the water inside the bowl is
h3 h3 2
2
π 15h −
− π 5h −
= 10πh2 .
3
3
If h ≥ 10, the volume of the water is
500 h3 4
h3
π 15h2 −
− π · 53 = π 15h2 −
−
.
3
3
3
3
2. A notch is cut to the centre of a cylindrical log having radius 20 cm. One plane face
of the notch is perpendicular to the axis of the log. The other one intersects it along a
diameter of the log at an angle of 30 deg. (See Figure 7.20 in the textbook.) What volume
of wood was removed from the log by cutting the notch?
The cross-sections of the notch perpendicular to the diameter are right triangles, with
the angle at the diameter equal to 30 degrees. Consider a cross-section at a distance x
1
from the√center of the log. Then one leg of the triangle is
tan(30◦ ) 202 − x2 . Thus the volume of the notch is
Z
2
1
0 tan(30◦ ) · (202 − x2 )dx =
−20 2
Z
2
0
0
√
202 − x2 , and the other is
400 − x2
√
dx
3
1 x3 20 16000
= √ 400x −
= √ ≈ 9237.6cm3 .
3 0
3
3 3
3. Set up and evaluate the integrals for:
(a) the length of the curve y = ex + e−x between x − 0 and x = a
Z
L=
ap
Z
1+
(ex
−
e−x )2
a
dx =
0
p
e2x + e−2x + 1 dx
0
(there is not much more that one can do with this.)
(b) the area of the surface obtained by rotating the curve y = e2x , 0 ≤ x ≤ a, around the
x-axis.
Z
Z
p
2πe
1 + 4e4x dx =
a
2x
S=
Z
a
0
a
π
p
1 + 4u2 du
x=0
πp
1 + tan2 θ sec2 θ dθ =
2
Z
a
π
sec3 θ dθ
2
x=0
x=0
a
π
= (sec θ tan θ + ln | sec θ + tan θ|)
4
x=0
a
p
p
π
2x
2x 4x
4x
= ( 4e + 1 · 2e + ln( 4e + 1 + 2e )
4
x=0
p
√
√
π p 4a
= ( 4e + 1 · 2e2a + ln( 4e4a + 1 + 2e2a ) − 2 5 − ln( 5 + 2)).
4
=
We substituted u = e2x , du = 2e2x , and then 2u = tan θ, 2du = sec2 θdθ.
2