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?

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- Jan 22nd 2007, 01:52 PMsymmetryVolume of Spheres
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? - Jan 22nd 2007, 02:39 PMtopsquark
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 - Jan 22nd 2007, 03:38 PMsymmetryok
Dan,

Hello. How did you get r to equal (s/4pi)^(1/2)?

I did not follow that part.

Thanks! - Jan 23rd 2007, 05:55 AMtopsquark
$\displaystyle S = 4 \pi r^2$

$\displaystyle \frac{S}{4 \pi} = \frac{4 \pi r^2}{4 \pi}$

$\displaystyle r^2 = \frac{S}{4 \pi}$

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 - Jan 23rd 2007, 05:23 PMsymmetryok
I finally got it.

Thanks!