MATH1120 Calculus II
Solutions to Selected Exercises in Notes 3
September 22, 2013
Exercise 1.1 We shall compare the oblique cone with the right cone of the same radius
and height. Using similar triangles, we find that the cross sections of both the oblique
cone and the right cone at the same horizontal level are disks of the same radius. Using
Cavalieri’s principle, we conclude that both cones have the same volume. Thus V = 13 r2 πh.
Exercise 1.2 Since we are to use shell method and the axis of rotation is the y-axis, we
should integrate with respect to x to get the volume of the jello. Note that −x2 + 6x − 8 =
0 =⇒ (x − 2)(x − 4) = 0 =⇒ x = 2 or 4. So the parabola cuts the x-axis at 2 and 4. The
Z 4
2
2πx(−x2 + 6x − 8)dx = 8π
shell radius is x while the shell height is −x + 6x − 8. V =
2
Exercise 1.3
(
(
(
x=2
x=3
y = −x2 + 6x − 8
=⇒
or
. (2, 0) and (3, 1) are two points of
y =x−2
y=0
y=1
intersection.
Z 3
Z 4
2
2
2
(1) V =
π((−x + 6x − 8 − (−1)) − (x − 2 − (−1)) )dx +
π((x − 2 − (−1))2 −
3
2
(−x2 + 6x − 8 − (−1))2 )dx √
(2) y = −x2 + 6x − 8 =⇒ x = 3 ± 1 − y using the quadratic formula. We shall divide
the region into three parts as shown in the diagram below, and set up three volume
integrals for the three parts respectively.
Z 1
Z 2
Z 1
p
p
V =
2π(y−(−1))(y+2−(3− 1 − y))dy+
2π(y−(−1))(4−(y+2))dy+
2π(y−(−1))(4−(3+ 1 − y))dy
0
1
1
0
2
Exercise 1.4
s
y
ey + e−y
e − e−y 2
1+
dy
Area = 2π
2
2
0
2
Z ln 2 y
e + e−y
= 2π
dy
2
0
Z ln 2 2y
e + 2 + e−2y
= 2π
dy
4
0
2y
ln 2
e
y e−2y
= 2π
+ −
8
2
8 0
15
=π
+ ln 2
16
Z
ln 2
Exercise 1.5
Z
2π(radius of cone frustum)ds
s
2
Z 2
dx
dy
=
2πy 1 +
dy
1
s
Z 2
1 2
3
=
2πy 1 + y − 3 dy
4y
1
Z 2
1 =
2πy y 3 + 3 dy
4y
1
Z 2
1
3
=
2πy y + 3 dy
4y
1
62π
=
5
Arc length =
C