#jsDisabledContent { display:none; } My Account | Register | Help

# Shell integration

Article Id: WHEBN0000330618
Reproduction Date:

 Title: Shell integration Author: World Heritage Encyclopedia Language: English Subject: Collection: Integral Calculus Publisher: World Heritage Encyclopedia Publication Date:

### Shell integration

A volume is approximated by a collection of hollow cylinders. As the cylinders get smaller the approximation gets better. The limit of this approximation is the shell integral.

Shell integration (the shell method in integral calculus) is a means of calculating the volume of a solid of revolution, when integrating along an axis perpendicular to the axis of revolution. This is in contrast to disk integration which integrates along the axis parallel to the axis of revolution.

• Definition 1
• Example 2
• References 4

## Definition

The shell method goes as follows: Consider a volume in three dimensions obtained by rotating a cross-section in the xy-plane around the y-axis. Suppose the cross-section is defined by the graph of the positive function f(x) on the interval [a, b]. Then the formula for the volume will be:

2 \pi \int_a^b x f(x) \mathrm{d}x

If the function is of the y coordinate and the axis of rotation is the x-axis then the formula becomes:

2 \pi \int_a^b y f(y) \mathrm{d}y

The formula is derived by computing the double integral in polar coordinates.

## Example

Consider the volume whose cross section on the interval [1, 2] is defined by:

y = (x-1)^2(x-2)^2
Our example in pictures.

In the case of disk integration we would need to solve for x given y. Because the volume is hollow in the middle we will find two functions, one that defines the inner solid and one that defines the outer solid. After integrating these two functions with the disk method we subtract them to yield the desired volume.

With the shell method all we need is the following formula:

2 \pi \int_1^2 x (x-1)^2(x-2)^2 \mathrm{d}x

By expanding the polynomial the integral becomes very simple. In the end we find the volume is \frac{\pi}{10}