Compare volume of cylinder and sphere with same surface area. Given that, 
Volume of a  cylinder with surface area  a  = $V_1$ 
Volume of a  sphere  with surface area a  = $V_2$
Compare $V_1$ and $V_2$.
The correct answer for this question is $V_2 > V_1$. I'm not able to prove this. Can someone please help me here?
My approach : 
Assume  that, 
$r_1$ = radius  of  cylinder
$r_2$ = radius  of  sphere  and 
$h$ = height of  cylinder
$$a =  4\pi r_2^2 = 2\pi r_1(r_1 + h)$$
$$\implies 2r_2^2 = r_1(r_1+h) ...........(1)$$
Now based on equation (1), we need to compare
$$\frac{4}{3}\pi r_2^3 \ and \ \pi r_1^2h$$
I'm stuck at this point. I'm not sure how to deal with term $r_1(r_1 + h)$. Some help would be appreciated.
 A: I took a hint from this StackExchange answer. The volume of the sphere with a given surface area is fixed. The volume of the cylinder can vary based on variations in height and radius of the cylinder. We can calculate the maximum volume of a cylinder for a given surface area using ratio $\frac{h}{r} = 2$ as proved in the question I linked. If we use this relation, we have :
$$2r_2^2 = r_1(r_1+h)$$
$$\implies 2r_2^2 = 3r_1^2$$
$$\implies \frac{V_2}{V_1} = \frac{\frac{4}{3}\pi r_2^3}{\pi r_1^2h} = \frac{4}{3} \frac{r_2^3}{2r_1^3} = \sqrt{\frac{3}{2}} \gt 1$$
Thus, we can infer that the maximum volume of a cylinder with given surface area is also less than the volume of a sphere with that surface area. So, $V2 \gt V1$ can be inferred based on this argument.
A: For simplicity, I let the radii of the sphere and cylinder be R and r respectrvely.
From the question, we have $2\pi r(r + h) = 4 \pi R^2$.......(*)
For comparison purpose, we want to maximize the volume of the cylinder. As pointed out, its maximum volume occurs when h = 2r.
Then, (*) becomes $R = \sqrt {\dfrac {3}{2}}$
$V_2 - V_1 = \dfrac 43 \pi R^3 - \pi r^2 (2r) = ... > 0$
