Solids of Revolution 1: Disks

The hard part about getting volumes from integration is figuring out the cross-sectional area $A(x)$. In general, this requires some knowledge of geometry. But if we have a solid of revolution, the geometry is easy:


Each slide is a disk of radius $f(x)$, area $A(x) = \pi f(x)^2$, and thickness $\Delta x$. This makes the total volume

$$\int_a^b A(x)\, dx = \pi \int_a^b f(x)^2\, dx.$$