Washer Method About Other Lines — AP® Calculus AB Notes

Washers around any line

You now have both tools this topic needs. From the disc method about other lines you know that a radius is a distance from the axis of rotation, whatever line that is. From the washer method you know that a slice with a hole has area $\pi(R^2 - r^2)$ . Put them together: when a region is revolved about a line $y = k$ or $x = h$ that it does not touch, each slice is a washer centered on that line, and both radii are measured from it.

V = \pi\int_a^b \left(R^2 - r^2\right)\,dx, \qquad R = \text{axis to far edge}, \quad r = \text{axis to near edge}.

Left: the region between y = √x and y = 1 from x = 1 to 4, with the line y = −1 dashed below it as the axis of rotation, and arrows from the axis to the thin rectangle: the outer radius R up to the curve and the inner radius r up to the line y = 1. Right: the solid made by revolving the region about y = −1, a flared horn with a hole of radius 2 around the axis, with one washer highlighted in blue: R = √x + 1 and r = 2. Below: V = the integral from 1 to 4 of π[(√x + 1)² − 2²] dx = 47π/6. — **Figure 1.** The same region as in the last topic, now revolved about $y = -1$ . Measured from the new axis, the outer radius is $\sqrt{x} + 1$ and the inner radius is $2$ . LevlPrep original diagram.

Axis below the region

If the axis $y = k$ is below the region, the top curve is the far edge and the bottom curve is the near edge. Each radius is a curve's height minus $k$ . With $k$ negative, that means adding: about $y = -1$ , a height $f(x)$ becomes the radius $f(x) - (-1) = f(x) + 1$ .

Worked example. The region between $y = \sqrt{x}$ and $y = 1$ for $1 \le x \le 4$ is revolved about the line $y = -1$ . Find the volume.

Radii. Far edge, the curve: $R = \sqrt{x} - (-1) = \sqrt{x} + 1$ . Near edge, the line $y = 1$ : $r = 1 - (-1) = 2$ .

Integrand. $R^2 - r^2$ $= (x + 2\sqrt{x} + 1) - 4$ $= x$ $+ 2\sqrt{x} - 3$ .

Volume.

V = \pi\int_1^4 \left(x + 2\sqrt{x} - 3\right)\,dx = \pi\left[\frac{x^2}{2} + \frac{4}{3}x^{3/2} - 3x\right]_1^4 = \pi\left(\frac{20}{3} - \left(-\frac{7}{6}\right)\right) = \frac{47\pi}{6}.

About the $x$ -axis the same region gave $\frac{9\pi}{2} = \frac{27\pi}{6}$ . Moving the axis down puts every washer farther out, where it sweeps a wider ring.

Axis above the region: the edges trade roles

If the axis $y = k$ is above the region, the far edge is the bottom curve, because it is farther from a line overhead. So $R = k - g(x)$ (bottom) and $r = k - f(x)$ (top). This is the step students most often miss: "top curve is the outer radius" is only true when the axis is below.

Worked example. The region between $y = x$ and $y = x^2$ is revolved about the line $y = 1$ . Find the volume.

Picture. On $(0, 1)$ the line $y = x$ is on top and the parabola below. The axis $y = 1$ is above both.

Radii. Far edge, the parabola: $R = 1 - x^2$ . Near edge, the line: $r = 1 - x$ .

Volume.

V = \pi\int_0^1 \left[(1 - x^2)^2 - (1 - x)^2\right]\,dx = \pi\int_0^1 \left(x^4 - 3x^2 + 2x\right)\,dx = \pi\left(\frac{1}{5} - 1 + 1\right) = \frac{\pi}{5}.

Vertical lines

For an axis $x = h$ , the washers are horizontal: write the curves as $x$ in terms of $y$ and integrate in $dy$ . If the axis is to the left of the region, the right curve is the far edge, and each radius is $x(y) - h$ . If the axis is to the right, the left curve is the far edge, and each radius is $h - x(y)$ . About $x = -1$ , the region between $x = y$ and $x = \sqrt{y}$ gives $R = \sqrt{y} + 1$ and $r = y + 1$ , and $V = \frac{\pi}{2}$ .

Radii for every position of the axis

Axis	Outer radius $R$	Inner radius $r$
$y = k$ below the region (top $f$ , bottom $g$ )	$f(x) - k$	$g(x) - k$
$y = k$ above the region	$k - g(x)$	$k - f(x)$
$x = h$ left of the region (right $q$ , left $p$ )	$q(y) - h$	$p(y) - h$
$x = h$ right of the region	$h - p(y)$	$h - q(y)$

Three checks catch most errors: both radii are positive, $R > r$ between the limits, and $R - r$ equals the region's height (or width). If the last check fails, you measured one radius from the wrong place.

With a calculator, find the crossings, store them, and write the setup with each radius in parentheses, for example $\pi\int_a^b \left[(4 - e^x)^2 - (2 - x)^2\right]\,dx$ . Report three decimal places.