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.


2 Answers 2


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.


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 $R = \sqrt {\dfrac {3}{2}}$

$V_2 - V_1 = \dfrac 43 \pi R^3 - \pi r^2 (2r) = ... > 0$

  • $\begingroup$ Why is $r=h$? There are many other solutions. For example $h=0$ and $r_1=\sqrt 2 r$. $\endgroup$
    – Andrei
    Jun 16, 2021 at 4:53
  • $\begingroup$ @Andrei First of all, if h = 0, the whole object has zero volume. I have skipped a result from calculus. We want $V_1$ to be as large as possible (for the comparison purpose). That will happen when $r_1 = h$. Then, we have $r_1^2 = r^2$. $\endgroup$
    – Mick
    Jun 16, 2021 at 9:12
  • $\begingroup$ The other answer has the maximum volume for the cylinder at $h=2r_1$ $\endgroup$
    – Andrei
    Jun 16, 2021 at 12:18
  • 1
    $\begingroup$ @Andrei You are right about that. Fixed. the logic will remains the same. $\endgroup$
    – Mick
    Jun 16, 2021 at 13:51

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