Volume of the hemisphere by rotation (two methods)

Text

Prove that the volume of the hemisphere of radius $R$ is $\tfrac23\pi R^3$: (a) as the solid of revolution about the $x$-axis of the curve $y=\sqrt{R^2-x^2}$ on $[0,R]$; (b) as the solid of revolution about the $y$-axis of the same curve.

Solution

(a) Rotation about the $x$-axis (disks in $x$). The disk radius at height $x$ is $y=\sqrt{R^2-x^2}$, so $y^2=R^2-x^2$: $V=\pi\int_0^R (R^2-x^2)\,dx=\pi\left[R^2 x-\frac{x^3}{3}\right]_0^R=\pi\left(R^3-\frac{R^3}{3}\right)=\frac23\pi R^3.$

(b) Rotation about the $y$-axis (disks in $y$). By symmetry the same curve, rotated about the $y$-axis, generates disks of radius $x$ with $x^2=R^2-y^2$: $V=\pi\int_0^R (R^2-y^2)\,dy=\pi\left[R^2 y-\frac{y^3}{3}\right]_0^R=\frac23\pi R^3.$ The two methods agree.

$\boxed{V=\frac23\pi R^3}$