# 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.

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

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

• Why is $r=h$? There are many other solutions. For example $h=0$ and $r_1=\sqrt 2 r$. Jun 16, 2021 at 4:53
• @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 $r_1^2 = r^2$.
– Mick
Jun 16, 2021 at 9:12
• The other answer has the maximum volume for the cylinder at $h=2r_1$ Jun 16, 2021 at 12:18
• @Andrei You are right about that. Fixed. the logic will remains the same.
– Mick
Jun 16, 2021 at 13:51