Washer Method About the Axes

Slices with a hole

The disc method worked because every rectangle in the region stood on the axis. Spin it, and it sweeps out a solid disc. Now take a region that does not reach the axis, such as the area between $y = \sqrt{x}$ and $y = 1$ for $1 \le x \le 4$. Every vertical rectangle starts at height 1 and goes up to $\sqrt{x}$. Spin that rectangle about the $x$-axis and it sweeps out a ring: a disc of radius $\sqrt{x}$ with a hole of radius 1 cut out of the middle. That ring is a washer.

Two radii, both from the axis

A washer has two radii, and both start at the axis of rotation:

• the outer radius $R$ reaches the edge of the region farther from the axis;
• the inner radius $r$ reaches the edge nearer to the axis, where the hole ends.

The washer's area is the big disc minus the hole: $\pi R^2 - \pi r^2$. So the washer method says

About the $x$-axis, with $f$ on top and $g$ on the bottom (both nonnegative), $R = f(x)$ and $r = g(x)$. A disc is just a washer with $r = 0$.

Difference of squares, not square of the difference

The most common washer error is writing $\pi\int (R - r)^2\,dx$. It treats the region's height as the radius of a solid disc, as if the region had been slid down to the axis. In the example it gives $\pi\int_1^4 (\sqrt{x} - 1)^2\,dx$$= \frac{7\pi}{6}$, about a quarter of the true volume. The ring is far from the axis, so it sweeps out much more than a disc of the same width would.

About the $y$-axis

Revolving about the $y$-axis gives horizontal washers stacked in $y$. Write each curve as $x$ in terms of $y$. The curve farther from the $y$-axis (larger $x$) gives $R$, the nearer one gives $r$, and the limits are $y$-values.

Checklist

1. Sketch the region and the axis. Does the region touch the axis all along? If not, use washers.
2. Draw one rectangle perpendicular to the axis. Label its far end and near end.
3. $R$ = distance from the axis to the far end; $r$ = distance to the near end.
4. Write $\pi\int (R^2 - r^2)$ with the right variable and limits, then integrate.

On a calculator, type $R$ and $r$ in parentheses, square each, subtract, and multiply by $\pi$, for example $\pi\int_{-b}^{b} \left[(2 + \cos x)^2 - (x^2)^2\right]\,dx$. Store the crossings and give three decimal places.