Shell integration



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

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:


 * $$2 \pi \int_a^b x f(x)\, dx$$

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


 * $$2 \pi \int_a^b y f(y)\, dy$$

If the function is rotating around the line $x = h$ then the formula becomes:


 * $$\begin{cases}

\displaystyle 2 \pi \int_a^b (x-h) f(x)\,dx, & \text{if}\ h \le a < b\\ \displaystyle 2 \pi \int_a^b (h-x) f(x)\,dx, & \text{if}\ a < b \le h, \end{cases}$$ and for rotations around $y = k$ it becomes
 * $$\begin{cases}

\displaystyle 2 \pi \int_a^b (y-k) f(y)\,dy, & \text{if}\ k \le a < b\\ \displaystyle 2 \pi \int_a^b (k-y) f(y)\,dy, & \text{if}\ a < b \le k. \end{cases}$$

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

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


 * $$y = (x-1)^2(x-2)^2$$

With the shell method we simply use the following formula:


 * $$V = 2 \pi \int_1^2 x ((x-1)^2(x-2)^2) \,dx $$

By expanding the polynomial, the integration is easily done giving $\pi⁄10$ cubic units.

Comparison With Disc Integration
Much more work is needed to find the volume if we use disc integration. First, we would need to solve $$y = (x-1)^2(x-2)^2$$ for $x$. Next, because the volume is hollow in the middle, we would need two functions: one that defined an outer solid and one that defined the inner hollow. After integrating each of these two functions, we would subtract them to yield the desired volume.