Solutions for Homework 6
Math 1206: Sections 7.1 and 6.1
March 16, 2014
Problem 7.1.8.
Answer . Let u = ln(secx + tanx). Then du =
integral becomes:
Z
du
√ =
u
Z
secx tanx + sec2 xdx
= secxdx. Then the
secx + tanx
p
1
1
u− 2 du = 2u 2 + C = 2 ln(secx + tanx) + C.
Problem 7.1.34.
Answer . Let u = lnx. Then du =
Z
1
dx. Then the integral becomes:
x
2u du =
2u
2lnx 2 2ln2 − 1
=[
] =
.
ln2
ln2 1
ln2
Problem 6.1.7 a.
Answer . For this problem we use the method of ’slices’. Note that the cross-sectional
rectangles length is the distance between the lines y = 6 and y = 3x. Thus, the length is
6 − 3x. Clearly, the limits of integration are x = 0 and x = 2 since the intersection of the
lines y = 3x and y = 6 occurs at x = 2. As a result, the area of the cross-sectional rectangles
can be expressed as
A(x) = 10(6 − 3x)
. Thus, using the method of slices we obtain:
Z 2
Z 2
3x2 2
V =
A(x)dx =
10(6 − 3x)dx = 10[6x −
] = 10(12 − 6) = 60.
2 0
0
0
Problem 6.1.16.
Answer . For this problem we use the method of ’disks’. Clearly, the limits of integration
are 0 and 2. We integrate with respect to y. Thus, using the method of disks we obtain:
Z 2
Z 2
3y 2
9π 2
9π y 3 2
V =
π( ) dy =
y dy =
[ ] = 6π.
2
4
4 3 0
0
0
1
Problem 6.1.42.
Answer . For this problem we use the method of ’washers’. First, we find the limits of
integration, which, in this case, will be the x-coordinates of the intersection points of the
lines y = 4 − x2 and y = 2 − x. Thus, the limits of integration will be the solutions of the
equation:
4 − x2 = 2 − x ⇔ x2 − x − 2 = 0 ⇔ (x − 2)(x + 1) = 0 ⇔ x = 2; x = −1.
Therefore, the limits of integration are -1 and 2. Using the washer method we obtain:
Z 2
Z 2
2 2
2
V =
π[(4 − x ) − (2 − x) ]dx = π
16 − 8x2 + x4 − 4 + 4x − x2 dx
−1
−1
Z 2
=π
x4 − 9x2 + 4x + 12dx
−1
x5
= π[ − 3x3 + 2x2 + 12x]2−1
5
108
=
π.
5
Problem 6.1.49.
Answer . For this problem we use the method of ’washers’. We integrate with respect to
√
y. The outer radius of a washer is 2 while the inner radius is given by y + 1. Thus, the
method of washers yields:
Z 1
Z 1
√
√
2
2
V =
π[2 − ( y + 1) ]dy = π
4 − y + 2 y − 1dy
0
Z0 1
√
=π
−y + 2 y + 3dy
0
3
−y 2 4y 2
= π[
+
+ 3y]10
2
3
23
= π.
6
2