241
Worksheet 6.3.
Volumes of Revolution
1. Find the volume of the solid obtained by rotating the region under the graph of the
function about the x-axis over the given interval.
1
a. f (x) = 2 , [1, 4]
x
y
x
b. f (x) =
√
π
cos x sin x, [0, ]
2
y
x
242
2. Sketch the region enclosed by the two curves and find the volume of the solid obtained
by rotating the region about the x-axis.
a. y = x2 and y = 2x + 3
10
8
6
4
2
-2
-1
1
2
3
b. y = sec x, y = csc x, y = 0, x = 0, and x =
y
2
1
Π
€€€€
4
Π
€€€€
2
x
π
2
243
3. Find the volume of the solid obtained by rotating the region enclosed by the graph about
the given line.
5
1
a. y = , y = − x, about y = −1
x
2
y
2
1.5
1
0.5
0.5
1
1.5
2
x
-0.5
-1
b. y = 16 − x, y = 3x + 12, x = 0 about x = 2
y
16
14
12
10
8
6
4
2
0.5
1
1.5
2
2.5
3
3.5
4
x
244
Solutions to Worksheet 6.3
1. Find the volume of the solid obtained by rotating the region under the graph of the
function about the x-axis over the given interval.
1
a. f (x) = 2 , [1, 4]
x
y
x
The volume of the solid of revolution is
4
Z 4
Z 4
21π
1 −3 −2 2
−4
=
π
(x ) dx = π
x dx = π − x
3
64
1
1
1
√
π
b. f (x) = cos x sin x, [0, ]
2
y
x
245
The volume of the solid of revolution is
Z
Z π/2 √
2
π
( cos x sin x) dx = π
π/2
(cos x sin x) dx
1
Z 1
π
1 2 = π
u du = π
u =
2
2
0
0
0
0
2. Sketch the region enclosed by the two curves and find the volume of the solid obtained
by rotating the region about the x-axis.
a. y = x2 and y = 2x + 3
10
8
6
4
2
-2
-1
1
2
3
2
(1) Setting x = 2x + 3 yields
0 = x2 − 2x − 3 = (x − 3)(x + 1).
The two curves therefore intersect at x = −1 and x = 3. The region enclosed by
the two curves is shown in the figure.
(2) When the region is rotated about the x-axis, each cross section is a washer with
outer radius R = 2x + 3 and inner radius r = x2 .
(3) The volume of the solid of revolution is
Z 3
Z 3
2
2 2
π
(2x + 3) − (x ) dx = π
(4x2 + 12x + 9 − x4 ) dx
−1
−1
= π
3
1088π
4 3
1 5 2
x + 6x + 9x − x =
3
5
15
−1
246
b. y = sec x, y = csc x, y = 0, x = 0, and x =
π
2
!!!!
2
1
Π
€€€€
4
Π
€€€€
2
(1) The region in question is shown in the figure.
(2) When the region is rotated about the x-axis, cross sections for x ∈ [0, π/4] are
circular disks with radius R = sec x, whereas cross sections for x ∈ [π/4, π/2] are
circular disks with radius R = csc x.
(3) The volume of the solid of revolution is
Z
0
π/4
2
π sec x dx +
Z
π/2
π/4
π/2
+ π (− cot x) |π/4
π csc2 x dx = π (tan x) |π/4
0
= π (1) + π (1)
= 2π
3. Find the volume of the solid obtained by rotating the region enclosed by the graphs about
the given line.
a. y =
5
1
, y = − x, about y = −1
x
2
247
y
2
1.5
1
0.5
0.5
1
1.5
2
x
-0.5
-1
The region enclosed by the two curves is shown in the figure. Rotating the region
about the line y = −1 produces a solid whose cross sections have outer radius R =
7
1
1
1
5
5
− x − (−1) = − x and inner radius r = − (−1) = + 1. Setting = − x and
2
2
x
x
x
2
5
1
1
solving for x, the curves y = and y = − x intersect at x = and x = 2. The volume
x
2
2
of the solid of revolution is
π
Z
2
1
2
2 2 !
Z 2
7
45
1
2
1
2
dx
−x −
+1
dx = π
− 7x + x − 2 −
1
2
x
4
x
x
2
2
7 2 1 3 1
45
x − x + x + − 2 ln |x| =π
4
2
3
x
1
2
39
=π
− 4 ln 2
8
b. y = 16 − x, y = 3x + 12, x = 0 about x = 2.
248
y
16
14
12
10
8
6
4
2
0.5
1
1.5
2
2.5
3
3.5
4
x
Rotating the region enclosed by y = 16 − x, y = 3x + 12 and the y-axis about x = 2
produces a solid with two different cross sections. For each y ∈ [12, 15], the cross section
1
1
is a washer with outer radius R = 2 and inner radius r = 2 − (y − 12) = 6 − y; for
3
3
each y ∈ [15, 16], the cross section is a washer with outer radius R = 2 and inner radius
r = 2 − (16 − y) = y − 14. The volume of the solid of revolution is
2 !
Z 16
Z 15
1
2
dy + π
(2)2 − (y − 14)2 dy
π
(2) − 6 − y
3
15
12
Z 16
Z 15 1 2
(−y 2 + 28y − 192) dy
= π
− y + 4y − 32 dy + π
9
15
12
15
16
1 3
1 3
2
2
= π − y + 2y − 32y + π − y + 14y − 192y 27
3
12
20
=
π
3
15