Concept preview

What you will be able to do

1. 01Recognise when a solid of revolution has a hole through it, so that each cross-section is a washer (an annulus) rather than a full disc, and identify the outer radius $R$ and inner radius $r$ from the two bounding curves.
2. 02Set up and evaluate a washer-method volume integral about the $x$-axis or the $y$-axis, $V=\pi\int_a^b\left(R^2-r^2\right)\,dx$ (or $dy$), keeping every answer exact with $\pi$ carried symbolically.
3. 03Avoid the squared-difference trap by subtracting the squares of the radii, $R^2-r^2$, rather than squaring their difference, $(R-r)^2$, and explain geometrically why the two are not equal.

In the disc method the region being revolved is bordered by the axis on one side, so each cross-section is a filled circle. But most regions are trapped between two curves, and when you revolve such a region about an axis that lies outside it, the near curve sweeps out an empty tube while the far curve sweeps out the solid boundary. The slice that remains is a washer: an outer circle of radius $R$ with an inner circle of radius $r$ punched out. The letter $R$ always denotes the FARTHER boundary from the axis of revolution, and $r$ the NEARER boundary.