Most of these homework problems are solved in two ways by first integrating with respect to x and then integrating with respect to y. However, problems 55, 56 and 58 cannot be solved in both ways so only one way is shown.
For some of these problems it is easier to set up the integral with respect to one particular variable. Later we
will see problems for which it is impossible to set up the integral with respect to a particular variable. Once the
integral (or integrals) are set up, you will see that some are easier than others to evaluate. The best thing for
you to do now is try to set it up both ways. After practicing this, you will become faster at recognizing the best
variable to use in your integration.
You can simplify my solutions by recognizing that the solid you are studying sometimes has twice the volume of
some other solid. Use these symmetries to simplify the necessary integration.
• (Section 6.2, #2) Let R be the finite region bounded by the graphs of y = 1 − x2 and y = 0. Find the
volume of the solid obtained when R is revolved around the x-axis.
Z 1
2
16π
π 1 − x2 dx =
1. V =
15
−1
Z 1
p
p
16π
1−y− − 1−y
dy =
2. V =
2πy
15
0
• (Section 6.2, #6) Let R be the finite region bounded by the graphs of y = ln x, y = 1, y = 2 and x = 0.
Find the volume of the solid obtained when R is revolved around the y-axis.
Z
e
Z
2πx (2 − 1) dx +
1. V =
0
Z
2. V =
1
e2
2πx (2 − ln x) dx = πe2 +
e
2
2
π (ey ) dy =
π 4
2e
π 4
2e
−
3π 2
2 e
=
π 4
2e
− π2 e2
− π2 e2
• (Section 6.2, #7) Let R be the finite region bounded by the graphs of y = x3 and y = x for x ≥ 0. Find
the volume of the solid obtained when R is revolved around the x-axis.
Z 1
2 4π
1. V =
πx2 − π x3
dx =
21
0
Z 1
4π
√
2. V =
2πy ( 3 y − y) dy =
21
0
• (Section 6.2, #9) Let R be the finite region bounded by the graphs of y 2 = x and x = 2y. Find the volume
of the solid obtained when R is revolved around the y-axis.
Z 4
√
64π
1. V =
2πx x − 12 x dx =
15
0
Z 2
2
64π
2
2. V =
π (2y) − π y 2
dy =
15
0
• (Section 6.2, #12) Let R be the finite region bounded by the graphs of y = e−x , y = 1 and x = 2. Find
the volume of the solid obtained when R is revolved around the line y = 2.
Z 2
π 5e4 + 8e2 − 1
2
−x 2
1. V =
π 2−e
− π (2 − 1) dx =
2e4
0
Z 1
π 5e4 + 8e2 − 1
2. V =
2π (2 − y) (2 − (− ln y)) dy =
2e4
e−2
• (Section 6.2, #14) Let R be the finite region bounded by the graphs of y = sin x and y = cos x for
0 ≤ x ≤ π/4. Find the volume of the solid obtained when R is revolved around the line y = −1.
√
Z π/4 π 4 2−3
2
2
1. V =
π (cos x − (−1)) − π (sin x − (−1)) dx =
2
0
√
Z
2/2
Z
2π (y − (−1)) arcsin y dy +
2. V =
1
√
0
2π (y − (−1)) arccos y dy =
2/2
√
√
√
√
√
π 4 2−3
−7π + 4 2π + 2π 2
π + 4 2π − 2π 2
+
=
4
4
2
• (Section 6.2, #16) Let R be the finite region bounded by the graphs of xy = 1, y = 0, x = 1 and x = 2.
Find the volume of the solid obtained when R is revolved around the line x = −1.
Z 2
2π (x − (−1)) x1 dx = 2π + 2π ln 2
1. V =
1
Z
1/2
2
2
π (2 − (−1)) − π (1 − (−1))
5π
π
= 2π + 2π ln 2
+ 2π ln 2 −
2
2
2. V =
0
Z
1
dy +
1/2
2
2
1
π y − (−1) − π (1 − (−1))
dy =
• (Section 6.2, #17) Let R be the finite region bounded by the graphs of x = y 2 and x = 1 − y 2 . Find the
volume of the solid obtained when R is revolved around the line x = 3.
√
√
√
Z 1/2
Z 1
√
√
√
√ 10 2π
9 2π 23 2π
+
=
1. V =
2π (3 − x) x − − x dx +
2π (3 − x) 1 − x − − 1 − x dx =
5
15
3
0
1/2
√
Z √2/2 2
2
10 2π
π 3 − y2 − π 3 − 1 − y2
2. V = √
dy =
3
− 2/2
• (Section 6.2, #33a) Let R be the finite region bounded by the graph of x2 + 4y 2 = 4. Find the volume of
the solid obtained when R is revolved around the line y = 2.
q
2
2 !
Z 2
q
1. V =
π 2 − − 1 − 14 x2
− π 2 − 1 − 14 x2
dx = 8π 2
−2
Z
1
2π (2 − y)
2. V =
p
p
4 − 4y 2 − − 4 − 4y 2
dy = 8π 2
−1
• (Section 6.2, #33b) Let R be the finite region bounded by the graph of x2 + 4y 2 = 4. Find the volume of
the solid obtained when R is revolved around the line x = 2.
q
q
Z 2
1 2
1 2
1. V =
2π (2 − x)
1 − 4x − − 1 − 4x
dx = 8π 2
−2
1
Z
2. V =
2
2 p
p
− π 2 − 4 − 4y 2
dy = 8π 2
π 2 − − 4 − 4y 2
−1
• (Section 6.2, #55) The base of the solid S is an elliptical region with boundary curve 9x2 + 4y 2 = 36.
Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base. Find
the volume of S.
q
2
Z 2 q
9 2
9 2
1
9 − 4x − − 9 − 4x
dx = 24
1. V =
4
−2
• (Section 6.2, #56) The base of the solid S is the triangular region with vertices (0, 0), (1, 0) and (0, 1).
Cross-sections perpendicular to the y-axis are equilateral triangles. Find the volume of S.
√
Z 1√
3
2
3
1. V =
4 (1 − y) dy = 12
0
• (Section 6.2, #58) The base of the solid S is the region enclosed by the parabola y = 1 − x2 and the x-axis.
Cross-sections perpendicular to the y-axis are squares. Find the volume of S.
Z 1 p
p
2
1. V =
1−y− − 1−y
dy = 2
0