Volume of the cone by rotation (two methods)

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Prove the formula for the volume of the cone of base radius $R$ and height $H$: (a) as the solid of revolution about the $x$-axis of the line $y=-\tfrac{R}{H}(x-H)$ (which equals $R$ at $x=0$ and $0$ at $x=H$); (b) as the solid of revolution about the $y$-axis of the line $y=-\tfrac{H}{R}(x-R)$.

Solution

(a) Rotation about the $x$-axis (disks in $x$). The disk radius at height $x$ is $y=\tfrac{R}{H}(H-x)$, so $V=\pi\int_0^H y^2\,dx=\pi\frac{R^2}{H^2}\int_0^H(H-x)^2\,dx=\pi\frac{R^2}{H^2}\cdot\frac{H^3}{3}=\frac13\pi R^2 H.$

(b) Rotation about the $y$-axis (disks in $y$). The line $y=-\tfrac{H}{R}(x-R)$ gives $x=R\left(1-\tfrac{y}{H}\right)$, the disk radius at height $y$: $V=\pi\int_0^H x^2\,dy=\pi R^2\int_0^H\left(1-\frac{y}{H}\right)^2 dy=\pi R^2\cdot\frac{H}{3}=\frac13\pi R^2 H.$ Both methods give the same result.

$\boxed{V=\frac13\pi R^2 H}$