**Introduction to Volumes of Solids of Revolution**

A useful way—and one that we encounter frequently in everyday life—for generating solids is by revolving a planar region about an axis. For example, we can think of a ball (the interior of a sphere) as the solid obtained by rotating a disk about an axis (Fig. 8.12). We can think of a cylinder as the solid obtained by rotating a rectangle about an adjacent axis (Fig. 8.13). We can think of a tubular solid as obtained by rotating a rectangle around a non-adjacent axis (Fig. 8.14).

There are two main methods for calculating volumes of solids of revolution: the method of washers and the method of cylinders. The first of these is really an instance of volume by slicing, just as we saw in the last section. The second uses a different geometry; instead of slices one uses cylindrical shells. We shall develop both technologies by way of some examples.

**The Method of Washers**

#### Examples

**Example 1**

A solid is formed by rotating the triangle with vertices (0,0), (2,0), and (1,1) about the *x* -axis. See Fig. 8.15. What is the resulting volume?

**Solution 1**

For 0 ≤ *x* ≤ 1, the upper edge of the triangle has equation *y = x*. Thus the segment being rotated extends from (*x* , 0) to (*x, x*). Under rotation, it will generate a disk of radius *x* , and hence area *A*(*x*) = π*x*^{2}. Thus the volume generated over the segment 0 ≤ *x* ≤ 1 is

Similarly, for 1 ≤ *x* ≤ 2, the upper edge of the triangle has equation *y* = 2 − *x*. Thus the segment being rotated extends from (*x* , 0) to ( *x* , 2 − *x* ). Under rotation, it will generate a disk of radius 2 − *x*, and hence area *A* ( *x* ) = π (2 − *x* )^{2}. Thus the volume generated over the segment 1 ≤ *x* ≤ 2 is

In summary, the total volume of our solid of revolution is

**Example 2**

The portion of the curve *y = x*^{2} between *x* = 1 and *x* = 4 is rotated about the *x* -axis (Fig. 8.16). What volume does the resulting surface enclose?

**Solution 2**

At position *x* , the curve is *x*^{2} units above the *x* -axis. The point ( *x, x*^{2} ), under rotation, therefore generates a circle of radius *x*^{2}. The disk that the circle bounds has area *A* ( *x* ) = π · ( *x*^{2} )^{2}. Thus the described volume is

**Math Note** : The reasoning we have used in the last two examples shows this: If the curve *y* = *f* ( *x* ), *a* ≤ *x* ≤ *b* , is rotated about the *x* -axis then the volume enclosed by the resulting surface is

**You Try It**: Calculate the volume enclosed by the surface obtained by rotating the curve , about the *x* -axis.

**Example 3**

The curve *y* = *x*^{3} , 0 ≤ *x* ≤ 3, is rotated about the *y* -axis. What volume does the resulting surface enclose?

**Solution 3**

It is convenient in this problem to treat *y* as the independent variable and *x* as the dependent variable. So we write the curve as *x* = *y* ^{1/3}. Then, at position *y* , the curve is distance *y*^{1/3} from the axis so the disk generated under rotation will have radius *y* ^{1/3} ( Fig. 8.17). Thus the disk will have area *A* ( *y* ) = π · [ *y*^{1/3}]^{2}. Also, since *x* ranges from 0 to 3 we see that *y* ranges from 0 to 27. As a result, the volume enclosed is

**Math Note**: The reasoning we have used in the last example shows this: If the curve *x *= *g *(*y*),* c* ≤* y* ≤* d*, is rotated about the *y*-axis then the volume enclosed by the resulting surface is

**You Try It**: Calculate the volume enclosed when the curve *y* = *x* ^{1/3}, 32 ≤ *x* ≤ 243, is rotated about the *y* -axis.

**Example 4**

Set up, but do not evaluate, the integral that represents the volume generated when the planar region between *y* = *x*^{2} + 1 and *y* = 2*x* + 4 is rotated about the *x* -axis.

**Solution 4**

When the planar region is rotated about the *x* -axis, it will generate a donut-shaped solid. Notice that the curves intersect at *x* = −1 and *x* = 3; hence the intersection lies over the interval [−1,3]. For each *x* in that interval, the segment connecting (*x, x*^{2} + 1) to ( *x* , 2*x* + 4) will be rotated about the *x* -axis. *It will generate a washer*. See Fig. 8.18. The area of that washer is

*A* ( *x* ) = π · [2 *x* + 4]^{2} − π · [ *x*^{2} + 1].

[ *Notice that we calculate the area of a washer by subtracting the areas of two circles—not by subtracting the radii and then squaring* .]

It follows that the volume of the solid generated is

Find practice problems and solutions for these concepts at: Applications of the Integral Practice Test.

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