Derive the formula for the volume of a sphere of radius R

Derive the formula for the volume of a sphere of radius $R$.

We derive the volume three independent ways: cross-sectional slicing, revolution of a semicircle, and a spherical-coordinates triple integral.

Method 1: Cross-sectional (disk) slicing

A sphere of radius $R$ centered at the origin satisfies $x^2+y^2+z^2=R^2$. For fixed $x\in[-R,R]$, the cross-section perpendicular to the $x$-axis is a circle $y^2+z^2=R^2-x^2$ of radius $\sqrt{R^2-x^2}$, with area:

\[A(x) = \pi(R^2-x^2)\]

Integrating these slices:

\[V = \int_{-R}^{R}\pi(R^2-x^2)\,dx = \pi\left[R^2x-\frac{x^3}{3}\right]_{-R}^{R} = \pi\left(\frac{2R^3}{3}-\left(-\frac{2R^3}{3}\right)\right) = \frac{4\pi R^3}{3}\]

Method 2: Revolving a semicircle

The upper semicircle $y=\sqrt{R^2-x^2}$, $-R\le x\le R$, revolved about the $x$-axis, sweeps out the same sphere. The disk method gives:

\[V = \pi\int_{-R}^{R}y^2\,dx = \pi\int_{-R}^{R}(R^2-x^2)\,dx = \frac{4\pi R^3}{3}\]

— identical to Method 1, as expected, since it's the same integral.

Method 3: Spherical coordinates

With $x=\rho\sin\phi\cos\theta$, $y=\rho\sin\phi\sin\theta$, $z=\rho\cos\phi$, the volume element is $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$, over $0\le\rho\le R$, $0\le\phi\le\pi$, $0\le\theta\le2\pi$:

\[V = \int_0^{2\pi}d\theta\int_0^{\pi}\sin\phi\,d\phi\int_0^{R}\rho^2\,d\rho = 2\pi\cdot[-\cos\phi]_0^{\pi}\cdot\left[\frac{\rho^3}{3}\right]_0^{R} = 2\pi\cdot2\cdot\frac{R^3}{3} = \frac{4\pi R^3}{3}\]

All three methods agree. ✓

Verification for a specific radius

For $R=3$: $V=\dfrac{4\pi\cdot27}{3}=36\pi\approx113.1$ cubic units.

Summary

\[\boxed{V = \frac{4\pi R^3}{3}}\]

Key insight: The three derivations — slicing, revolution, and spherical coordinates — are different ways of organizing the same underlying integral, and they necessarily agree. Differentiating $V$ with respect to $R$ gives $\frac{dV}{dR}=4\pi R^2$, exactly the sphere's surface area — the volume "grows" by adding a thin shell of that area at each radius.