Volumes by Washers

A region that misses the axis

The line y = x and the parabola $y=x^2$y = x² cross at x = 0 and x = 1, and between them the line is on top. Spin the region between them about the x-axis. Below the parabola there is a gap between the region and the axis, and that gap sweeps out a hole through the middle of the solid.

y = x and $y=x^2$y = x², crossing at (0, 0) and (1, 1). The region between them stops short of the x-axis everywhere except at the origin, so the solid it sweeps has a hole.

Each slice is a ring

Cut the solid across the axis at x. The slice is a ring, called a washer: its outer edge is swept by the upper curve, at distance R from the axis, and its hole by the lower curve, at distance r.

Its face is the disc of radius R with the disc of radius r removed, so its area is $\pi R^2-\pi r^2=\pi \left(R^2-r^2\right)$πR² − πr² = π(R² − r²).

One slice: a disc of radius R with a disc of radius r taken out of the middle. Its area is $\pi R^2-\pi r^2$πR² − πr².

Square each radius first

$R^2-r^2$R² − r² and ${\left(R-r\right)}^2$(R − r)² are different numbers. With R = 3 and r = 2, $R^2-r^2=9-4=5$R² − r² = 9 − 4 = 5, but ${\left(R-r\right)}^2=1$(R − r)² = 1. The second is the area of a circle whose radius is the ring’s width, and it is far too small: the ring is long as well as narrow.

The washer formula

A slice of thickness dx has volume $\pi \left(R^2-r^2\right)dx$π(R² − r²) dx, so $V=\pi \int \left(R^2-r^2\right)dx$V = π ∫ (R² − r²) dx. R and r are both distances from the axis of the spin, both measured at the same x.

It is the disc method twice: the solid swept by the upper curve, $\pi \int R^{2}dx$π ∫ R² dx, less the hole swept by the lower one, $\pi \int r^{2}dx$π ∫ r² dx.

y = x and $y=x^2$y = x²

For the region above, R = x and $r=x^2$r = x² on [0, 1]. Then $R^2-r^2=x^2-x^4$R² − r² = x² − x⁴, and $V=\pi$V = π times the integral of $x^2-x^4$x² − x⁴ from 0 to 1, which is $\pi \left(\frac{1}{3}-\frac{1}{5}\right)=\frac{2\pi }{15}$π(1/3 − 1/5) = 2π/15, about 0.419 cubic units.

Squaring the gap instead gives $\pi$π times the integral of ${\left(x-x^2\right)}^2=x^2-2x^3+x^4$(x − x²)² = x² − 2x³ + x⁴, which is $\pi \left(\frac{1}{3}-\frac{1}{2}+\frac{1}{5}\right)=\frac{\pi }{30}$π(1/3 − 1/2 + 1/5) = π/30, a quarter of the true volume. Ignoring the hole gives $\pi$π times the integral of $x^2$x², which is $\frac{\pi }{3}$π/3, the solid swept by the line alone.

About another line

Spin the same region about the line y = 2. Each radius is now the distance from y = 2 down to a curve. The line y = x is the nearer, at distance 2 − x, and the parabola $y=x^2$y = x² the farther, at $2-x^2$2 − x². So the outer radius is $R=2-x^2$R = 2 − x² and the inner is r = 2 − x: the curves have changed roles.

$R^2-r^2=\left(4-4x^2+x^4\right)-\left(4-4x+x^2\right)=x^4-5x^2+4x$R² − r² = (4 − 4x² + x⁴) − (4 − 4x + x²) = x⁴ − 5x² + 4x, and $V=\pi$V = π times its integral from 0 to 1, which is $\pi \left(\frac{1}{5}-\frac{5}{3}+2\right)=\frac{8\pi }{15}$π(1/5 − 5/3 + 2) = 8π/15, about 1.676 cubic units.

The same region with the line y = 2 above it. Measured from y = 2, the parabola is farther away than the line y = x, so it sweeps the outer edge of each ring and the line sweeps the hole.

About the y-axis

Spin the region about the y-axis and the slices are stacked in dy. At height y the line is at x = y and the parabola at $x=\sqrt{y}$x = √y, which is farther from the y-axis. So $R=\sqrt{y}$R = √y, r = y, and $V=\pi$V = π times the integral of $y-y^2$y − y² from 0 to 1, which is $\pi \left(\frac{1}{2}-\frac{1}{3}\right)=\frac{\pi }{6}$π(1/2 − 1/3) = π/6.

The usual mistakes

Squaring the difference. $\pi {\left(R-r\right)}^2$π(R − r)² gives $\frac{\pi }{30}$π/30 for y = x and $y=x^2$y = x², not $\frac{2\pi }{15}$2π/15.

Adding the hole. $\pi \left(R^2+r^2\right)$π(R² + r²) counts the hole as solid, and more besides.

Leaving the hole in. $\pi \int R^{2}dx$π ∫ R² dx is the disc method, $\frac{\pi }{3}$π/3 here.

Measuring from the wrong axis. About y = 2 a radius is 2 minus the height of the curve, not the height itself and not 2 plus it.

The wall of a vase

In the application below, the wall of a turned vase lies between an outside radius $\sqrt{y+9}$√(y + 9) and an inside radius $\sqrt{y+4}$√(y + 4). Each slice of the wall is a washer, and squaring the radii before subtracting leaves the same area, $5\pi$5π, at every height.