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.
Best Answer
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.