Problem 54
Medium Volumes and Geometry

Derive the formula for the volume of a spherical cap

Derive the formula for the volume of a spherical cap of height $h$.

Explanation of Method

Place a sphere of radius $R$ with the cap occupying $x\in[R-h,R]$ along a diameter, integrate the circular cross-sectional areas over that range, and check the result against the known hemisphere and full-sphere special cases.

We slice the cap into circular cross-sections and integrate, then check special cases.

Step 1: Set up the geometry

For a sphere $x^2+y^2+z^2=R^2$, a cap of height $h$ cut off by a plane occupies $x\in[R-h,R]$.

Step 2: Cross-sectional area

At $x\in[R-h,R]$, the circular cross-section has $y^2+z^2=R^2-x^2$, so $A(x)=\pi(R^2-x^2)$.

Step 3: Integrate

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

Step 4: Evaluate at the limits

At $x=R$: $\pi\left(R^3-\frac{R^3}{3}\right)=\frac{2\pi R^3}{3}$.

At $x=R-h$: expanding $(R-h)^3=R^3-3R^2h+3Rh^2-h^3$,

\[\pi\left[R^2(R-h)-\frac{(R-h)^3}{3}\right] = \pi\left[\frac{2R^3}{3}-Rh^2+\frac{h^3}{3}\right]\]

Step 5: Subtract

\[V = \frac{2\pi R^3}{3}-\pi\left[\frac{2R^3}{3}-Rh^2+\frac{h^3}{3}\right] = \pi Rh^2-\frac{\pi h^3}{3} = \pi h^2\left(R-\frac h3\right)\]

\[\boxed{V = \frac{\pi h^2(3R-h)}{3}}\]

Step 6: Equivalent form in terms of the cap's base radius $a$

Since $a^2=R^2-(R-h)^2=2Rh-h^2$, we get $3a^2=6Rh-3h^2$, and:

\[V = \frac{\pi h}{6}(3a^2+h^2)\]

Step 7: Special cases

Hemisphere ($h=R$): $V=\frac{\pi R^2(3R-R)}{3}=\frac{2\pi R^3}{3}$, exactly half a full sphere. ✓

Full sphere ($h=2R$): $V=\frac{\pi(2R)^2(3R-2R)}{3}=\frac{4\pi R^3}{3}$. ✓

Thin cap ($h\ll R$): $V\approx\pi Rh^2$, matching a thin cylinder of radius $\sqrt{Rh}$ and height $h$.

Summary

\[\boxed{V = \frac{\pi h^2(3R-h)}{3} = \frac{\pi h}{6}(3a^2+h^2)}\]

Key insight: The cap formula interpolates smoothly between a thin sliver ($V\approx\pi Rh^2$ for small $h$) and the full sphere ($h=2R$), passing through the hemisphere exactly at the midpoint $h=R$.


Result

Final answer:

\[\boxed{V = \frac{\pi h^2(3R-h)}{3}}\]