# Spheres and converting formulas

If the volume V of a sphere with radius r is V=(4/3)πr^3. If the surface area is s=4πr^2, how can I express the volume as a function of the surface area S? My first thought was to set them equal to each other but it doesn't seem like the right thing to do. Any hints/help would be appreciated

-edit- ok so from S=4πr^2, I got r=√(S/4π)...So now I am stuck as to what to do. Since I'm trying to do V(S)=?...I just replaced the r from V=(4/3)πr^3 with √(S/4π) but the answer doesn't correlate with the answer on the book. what am i doing wrong?

-edit-

V(S)=4/3π*(sqrt(S/4π)^3

V(S)=4/3π*((S^3/2)/(8π^3/2)

V(S)=(4Sπ^3/2)/(24π^3/2)

V(S)=(1Sπ^3/2)/(6π^3/2)

Answer should be:S/6*sqrt(S/π)...what did i do wrong?

Edit: V(S)=Sπ√π/6π√π

V(S)=S/6

Edit: V=4/3π*r^3

S=4π*r^2

r^2=S/4π

r=sqrt(S/4π)

V(S)=4/3π*(sqrt(S/4π)^3

V(S)=4/3π*S√S/8π√π

V(S)=S√S/3*2√π

V(S)=S√S/6√π

Special thanks to J.M for helping me figure this out

-
The straight forward method involves solving for $r$ in one of the equations, replacing the $r$ in the other equation with the expression for $r$ you've just derived, and then solve for the volume. Can you follow this? – J. M. Sep 21 '11 at 0:41
alright, I'm following you. so Since I want V(S): i should solve r for the S equation right? and then do the math? – Ronnie.j Sep 21 '11 at 1:00
That works. :) If you figure out the answer, you can post your solution to answer your question. – J. M. Sep 21 '11 at 1:02
can you look at my edits and see what i did wrong? – Ronnie.j Sep 21 '11 at 1:10
You did try simplifying after replacing the $r$? Note also that $\sqrt{\frac{S}{4\pi}}$ is the same as $\frac12\sqrt{\frac{S}{\pi}}$. – J. M. Sep 21 '11 at 1:14

$S=4r^2\pi \Rightarrow r^2=\frac{S}{4\pi} \Rightarrow r=\sqrt{\frac{S}{4\pi}}$
$V=\frac{4}{3}(\sqrt{\frac{S}{4\pi}})^3\pi \Rightarrow V=\frac{4}{3}\frac{S}{4\pi}\pi\sqrt{\frac{S}{4\pi}} \Rightarrow V=\frac{S}{3}\sqrt{\frac{S}{4\pi}} \Rightarrow V=\frac{S}{3}\sqrt{\frac{1}{4}}\sqrt{\frac{S}{\pi}}\Rightarrow V=\frac{S}{6}\sqrt{\frac{S}{\pi}}$