The volume of a sphere of radius r is V = (4/3)(pi)(r^3); the surface area S of this sphere is S = 4(pi)(r^2).
(a) Express the volume V as a function of the surface area S.
(b) If the surface area doubles, how does the volume change?
a) $\displaystyle r = \left ( \frac{S}{4 \pi} \right )^{1/2}$
Thus:
$\displaystyle V = \frac{4}{3} \pi \left [ \left ( \frac{S}{4 \pi} \right )^{1/2} \right ]^3 $
$\displaystyle V = \frac{4 \pi}{3(4 \pi)^{3/2}} S^{3/2}$
$\displaystyle V = \frac{1}{3(4 \pi)^{1/2}} S^{3/2}$
b) If $\displaystyle S \to 2S$ then
$\displaystyle V \to \frac{1}{3(4 \pi)^{1/2}} (2S)^{3/2}$
$\displaystyle V \to 2^{3/2} \left ( \frac{1}{3(4 \pi)^{1/2}} S^{3/2} \right ) $
$\displaystyle V \to 2^{3/2}V$
-Dan
$\displaystyle S = 4 \pi r^2$
$\displaystyle \frac{S}{4 \pi} = \frac{4 \pi r^2}{4 \pi}$
$\displaystyle r^2 = \frac{S}{4 \pi}$
Now, we need to take the square root of both sides. It was more convenient for me to express this as
$\displaystyle r = \left ( \frac{S}{4 \pi} \right )^{1/2}$
rather than
$\displaystyle r = \sqrt{\frac{S}{4 \pi}}$
Both of these forms are completely equivalent, though you would probably recognize the second as being more familiar. Perhaps I should have mentioned this when I solved the problem.
-Dan