Topic 8.11 Volume with Washer Method: Revolving Around the x- or y-Axis: find the volume of a solid of revolution with a hole about a coordinate axis using the washer method.

What this topic is asking

The College Board (Topic 8.11) finds volumes of solids of revolution with a hole using the washer method. When the region revolved does not touch the axis, each cross section is a washer (an annulus), and you subtract the inner disc from the outer disc.

The washer method

A worked washer volume

Squaring each radius separately

The defining error of the washer method is computing $\pi\int (R - r)^2\,dx$ instead of $\pi\int (R^2 - r^2)\,dx$. These are not equal: $(R - r)^2 = R^2 - 2Rr + r^2$, which is the wrong expression. The geometry is outer disc area minus inner disc area, so you square each radius and then subtract: $R^2 - r^2$. Writing the two squared terms explicitly, $\pi\int (R^2 - r^2)$, before simplifying keeps this straight. This single distinction separates a correct washer setup from a common, heavily penalized mistake.

Identifying outer versus inner radius

The outer radius reaches the curve farther from the axis; the inner radius reaches the curve nearer the axis. About the $x$-axis with two positive curves, the upper curve gives the outer radius and the lower curve the inner radius. About the $y$-axis, the rightmost curve (largest $x$) is outer and the leftmost is inner, with everything written in terms of $y$. Sketching the region and a representative washer, then labelling which curve is outer and which is inner, prevents swapping them, a swap would make the integrand negative. With the radii correctly assigned and each squared before subtracting, the integral gives the volume.

The washer as a difference of two discs

The washer method is exactly the disc method applied twice and subtracted: the solid is the big solid generated by the outer curve minus the hole generated by the inner curve. So $\pi\int R^2$ is the outer solid's volume, $\pi\int r^2$ is the hole's volume, and the difference $\pi\int (R^2 - r^2)$ is the volume of what remains. Thinking of it as "outer disc volume minus inner disc volume" both explains why you square each radius separately, each is its own disc, and reinforces that you never square the difference of radii. It also tells you when the washer collapses to a disc: if the inner radius is zero (the region touches the axis), the hole vanishes and you are back to a single disc.

Deciding disc versus washer from the region

The choice between disc and washer is settled by one question: does the region being revolved touch the axis of revolution? If it touches, each cross section is a solid disc and you use $\pi\int R^2$; if there is a gap between the region and the axis, each cross section is a washer with a hole and you use $\pi\int (R^2 - r^2)$. Sketching the region against the axis answers this at a glance. The error of using a disc when a hole exists overstates the volume, while inventing an inner radius where the region actually touches the axis understates it. Confirming contact with the axis before choosing the method prevents both.