Prove that the volume of the solid of revolution obtained

Prove that the volume of the solid of revolution obtained by rotating a plane figure $F$ about an axis in its plane that does not intersect $F$ equals the area $S$ of the figure multiplied by the circumference of the circle described by the centroid of $F$.

Step 1 · Setup

Let $F$ lie in the $xy$-plane with $x\ge0$ throughout (so the $y$-axis, the axis of rotation, does not intersect $F$). Let $S=\iint_FdA$ and $\bar x=\dfrac1S\iint_Fx\,dA$.

Step 2 · Volume via the shell method

A vertical strip of $F$ at position $x$, width $dx$, height $\ell(x)$, generates a cylindrical shell of volume $dV=2\pi x\,\ell(x)\,dx$, so:

$$ V = 2\pi\int x\,\ell(x)\,dx = 2\pi\iint_Fx\,dA. $$

Step 3 · Identify the centroid integral

By definition of the centroid, $\iint_Fx\,dA=S\bar x$, so:

$$ V = 2\pi S\bar x. $$

Step 4 · Interpretation and general axis

The quantity $2\pi\bar x$ is the circumference of the circle traced by the centroid, so $V=S\cdot C$. For a general axis $\ell$ not intersecting $F$, with $\bar d$ the distance from the centroid to $\ell$, the same argument (replacing $x$ with the perpendicular distance to $\ell$) gives $V=2\pi\bar d\cdot S$. As checks: rotating a disk of radius $r$ centered at distance $R>r$ from the axis gives the torus volume $V=2\pi R\cdot\pi r^2=2\pi^2r^2R$; rotating a semicircle of radius $a$ about its diameter ($\bar x=4a/3\pi$, $S=\pi a^2/2$) gives $V=2\pi\cdot\frac{4a}{3\pi}\cdot\frac{\pi a^2}2=\frac{4\pi a^3}3$, the sphere's volume, as expected.

Additional Notes

• This derivation matches the source exactly. This closes out the series of moment/inertia/centroid problems in this collection with the classical theorem that connects centroids back to volumes of revolution.