Math 152 Class Notes
September 10, 2015
7.2 Volume by Slicing
Let S be a solid that ranges from x = a to x = b. Suppose that A(x) is the area of
the cross-section of S perpendicular to the x-axis and passing through the point x.
The Volume of S is dened by
ˆ
V =
b
A(x)dx
a
Example 1. Find the volume of the solid S whose base is the circle x2 + y 2 = 1. The
cross sections of S perpendicular to the x-axis are squares.
Example 2. Find the volume of the solid S whose base is the region bounded by y = x2 ,
y = 1. The cross sections of S perpendicular to the y -axis are equilateral triangles.
Solids of Revolution
Let R be the region bounded by y = f (x), x = a, x = b and y = 0. Suppose that S is
obtained by rotating the region R about the x-axis. Then S is a solid of revolution.
The cross-section at x is a disk with radius y = f (x) and so the cross-section area of
S at x is A(x) = π[f (x)]2 . Thus, the volume of S is given by
ˆ
b
π[f (x)]2 dx
V =
a
This method of nding volume is called the disk
method
.
Example
√ 3. Find the volume of the solid obtained by rotating the region bounded by
y = 1 − x2 and y = 0 about the x-axis.
When the region R is bounded by x = g(y), x = 0, y = c and y = d and is rotated
about the y -axis, the volume of the resulting solid is given by
ˆ
d
π[g(y)]2 dy
V =
c
Example 4. Find the volume of the solid obtained by rotating the region bounded by
y = x3 , y = 8 and x = 0 about the y -axis.
In general, if a solid of revolution is obtained by revolving a region about a line dierent
from the x-axis or the y -axis, then its volume can be calculated by
ˆ
V =
b
π · [radius]2 · [thickness]
a
where the [thickness] is dx or dy determined by the direction of the rotational axis.
Example
5. Find the volume of the solid obtained by rotating the region bounded by
√
y = x, y = 0, x = 4 about the x = 4.
Example 6. Find the volume of the solid obtained by rotating the region bounded by
y = x2 + 1, x = 0, y = 10 about the y = 10.
When the region is bounded between two curves y = f (x) and y = g(x) (f (x) ≥ g(x))
from x = a to x = b and rotated about x-axis, the cross-section area is A(x) =
π[f (x)2 − g(x)2 ]. Thus the volume of the resulting solid is given by
ˆ
b
π[f (x)2 − g(x)2 ]dx
V =
a
We call this method the
washer.
washer method
for the cross-section has the shape of a
Generally, we use the following general washer method to nd the volume when the
rotational axis is a line dierent from the x-axis or the y -axis.
ˆ
V =
b
π · [outer radius]2 − [inner radius]2 · [thickness]
a
where the [thickness] is dx or dy determined by the direction of the rotational axis.
Example 7. Find the volume of the solid obtained by rotating the region bounded by
y = x2 , y = 2x about the x-axis.
Example 8. Find the volume of the solid obtained by rotating the region bounded by
y = x2 , y = 2x about the y -axis.
Example 9. Find the volume of the solid obtained by rotating the region bounded by
y = x, y = x2 about the y = 2.
Example
10. Find the volume of the solid obtained by rotating the region bounded by
√
y = x, x = 4, y = 0 about the x = 5.