Area and Volume Revisited — Question 10

Question 10 – Solution

Strategy. Rebuild both standard sphere formulas from integration, then use their ratio to recover the radius.

Step 1: Volume In spherical coordinates, $$\begin{align*} V&=\int_0^{2\pi}\int_0^\pi\int_0^R \rho^2\sin\phi\,d\rho\,d\phi\,d\theta =2\pi(2)\frac{R^3}{3}=\frac{4\pi R^3}{3}. \end{align*}$$

Step 2: Surface area For the upper graph $z=\sqrt{R^2-r^2}$, the area factor is $R/\sqrt{R^2-r^2}$. Its area is $$\begin{align*} S_{\mathrm{upper}}&=\int_0^{2\pi}\int_0^R \frac{Rr}{\sqrt{R^2-r^2}}\,dr\,d\theta=2\pi R^2. \end{align*}$$ Doubling for both hemispheres gives $S=4\pi R^2$.

Step 3: Use the ratio $$\frac{S}{V}=\frac{4\pi R^2}{4\pi R^3/3}=\frac 3R=\frac 32, \qquad \boxed{R=2}.$$ Therefore $$\boxed{S=16\pi},\qquad \boxed{V=\frac{32\pi}{3}}.$$

Verification Surface area divided by volume has units of inverse length, and $3/R$ has those units. Substituting $R=2$ reproduces the stated ratio.