5.3

5.3 Volumes by Cylindrical

Shells

Volumes by Cylindrical Shells

Let’s consider the problem of finding the volume of the solid obtained by rotating about the y-axis the region bounded by y = 2x² – x³ and y = 0. (See Figure 1.)

If we slice perpendicular to the y-axis, we get a washer.

Figure 1

Volumes by Cylindrical Shells

But to compute the inner radius and the outer radius of the washer, we’d have to solve the cubic equation y = 2x² – x³ for x in terms of y; that’s not easy.

Fortunately, there is a method, called the method of cylindrical shells, that is easier to use in

such a case.

Figure 2 shows a cylindrical shell with inner radius r₁, outer radius r₂,

and height h. ^{Figure 2}

Volumes by Cylindrical Shells

Its volume V is calculated by subtracting the volume V₁ of the inner cylinder from the volume V₂ of the outer cylinder:

V = V₂ – V₁

= πr₂²h – πr₁²h = π(r₂² – r₁²)h

= π(r₂ + r₁)(r₂ – r₁)h

= 2π h(r₂ – r₁)

Volumes by Cylindrical Shells

If we let ∆r = r₂ – r₁ (the thickness of the shell) and r = (r₂ + r₁) (the average radius of the shell), then this

formula for the volume of a cylindrical shell becomes

and it can be remembered as

V = [circumference] [height] [thickness]

Volumes by Cylindrical Shells

Now let S be the solid obtained by rotating about the y-axis the region bounded by y = f(x) [where f(x) ≥ 0], y = 0, x = a, and x = b, where b > a ≥ 0. (See Figure 3.)

We divide the interval [a, b] into n subintervals [x_{i – 1}, x_i] of equal width ∆x and let be the midpoint of the ith

subinterval.

Figure 3

Volumes by Cylindrical Shells

If the rectangle with base [x_{i – 1}, x_i] and height is

rotated about the y-axis, then the result is a cylindrical shell with average radius , height , and thickness ∆x.

(See Figure 4.) So by Formula 1 its volume is

Figure 4

Volumes by Cylindrical Shells

Therefore an approximation to the volume V of S is given by the sum of the volumes of these shells:

This approximation appears to become better as n → .

But, from the definition of an integral, we know that

Volumes by Cylindrical Shells

Thus the following appears plausible:

The best way to remember Formula 2 is to think of a typical shell, cut and flattened as in Figure 5, with radius x,

circumference 2πx, height f(x), and thickness ∆x or dx:

This type of reasoning will be helpful in other situations, such as when we rotate about lines other than the y-axis.

Volumes by Cylindrical Shells

Figure 5

Example 1

Find the volume of the solid obtained by rotating about the y-axis the region bounded by y = 2x² – x³ and y = 0.

Solution:

From the sketch in Figure 6 we see that a typical shell has radius x, circumference 2πx, and height f(x) = 2x² – x³.

Example 1 – Solution

So, by the shell method, the volume is

It can be verified that the shell method gives the same answer as slicing.

cont’d

Disks and Washers versus

Cylindrical Shells

Disks and Washers versus Cylindrical Shells

When computing the volume of a solid of revolution, how do we know whether to use disks (or washers) or cylindrical shells? There are several considerations to take into

account: Is the region more easily described by top and bottom boundary curves of the form y = f(x) or by left and right boundaries x = g (y)?

Which choice is easier to work with? Are the limits of

integration easier to find for one variable versus the other?

Does the region require two separate integrals when using x as the variable but only one integral in y? Are we able to evaluate the integral we set up with our choice of variable?

Disks and Washers versus Cylindrical Shells

If we decide that one variable is easier to work with than the other, then this dictates which method to use. Draw a sample rectangle in the region, corresponding to a

cross-section of the solid.

The thickness of the rectangle, either ∆x or ∆y,

corresponds to the integration variable. If you imagine the rectangle revolving, it becomes either a disk (washer) or a shell.