# Volume of Spheres

• Jan 22nd 2007, 02:52 PM
symmetry
Volume 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, 03:39 PM
topsquark
Quote:

Originally Posted by symmetry
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) $r = \left ( \frac{S}{4 \pi} \right )^{1/2}$

Thus:
$V = \frac{4}{3} \pi \left [ \left ( \frac{S}{4 \pi} \right )^{1/2} \right ]^3$

$V = \frac{4 \pi}{3(4 \pi)^{3/2}} S^{3/2}$

$V = \frac{1}{3(4 \pi)^{1/2}} S^{3/2}$

b) If $S \to 2S$ then
$V \to \frac{1}{3(4 \pi)^{1/2}} (2S)^{3/2}$

$V \to 2^{3/2} \left ( \frac{1}{3(4 \pi)^{1/2}} S^{3/2} \right )$

$V \to 2^{3/2}V$

-Dan
• Jan 22nd 2007, 04:38 PM
symmetry
ok
Dan,

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

I did not follow that part.

Thanks!
• Jan 23rd 2007, 06:55 AM
topsquark
Quote:

Originally Posted by symmetry
Dan,

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

I did not follow that part.

Thanks!

$S = 4 \pi r^2$

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

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

Now, we need to take the square root of both sides. It was more convenient for me to express this as
$r = \left ( \frac{S}{4 \pi} \right )^{1/2}$
rather than
$r = \sqrt{\frac{S}{4 \pi}}$

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, 06:23 PM
symmetry
ok
I finally got it.

Thanks!