A cube and a sphere have equal volume. What is the ratio of their surface areas?

Question

The answer is supposed to be $$ \sqrt[3]{6} : \sqrt[3]{\pi} $$

Since $$ \ a^3 = \frac{4}{3} \pi r^3 $$ I have expressed it as: $$ \ a = \sqrt[3]{ \frac{4}{3} \pi r^3} $$

and,

$$ \ 6 \left( \sqrt[3]{ \frac{4}{3} \pi r^3 } \right) ^2 : 4 \pi r^2 $$

But I am not really sure how to arrive at the desired result. I have tried to simplify it, but apparently I am missing some step in the process and come to a result that is far from the correct one.

Could you please help? Thank you.

Answer 1

A sphere with equal volume to a cube of side $a$ must have radius $r$:

$$\frac{4\pi r^3}{3} = a^3$$

So

$$r = \sqrt[3]{\frac{3}{4 \pi}}a$$

Now just take

$$\frac{4\pi r^2}{6a^2} = \frac{4\pi\sqrt[3]{\frac{9}{16 \pi^2}}a^2}{6a^2} = \frac{2}{3}\pi\sqrt[3]{\frac{9}{16 \pi^2}} = \sqrt[3]{\frac{9\cdot 8 \pi^3}{27 \cdot 16 \pi^2}} = \sqrt[3]{\pi/6}.$$

Answer 2

Another way to approach this is to write the surface areas of each solid in terms of its volume:

cube -- $$ S_c \ = \ 6 \ a^2 \ \ , \ \ V_c \ = \ a^3 \ \ \Rightarrow \ \ a \ = \ V_c^{1/3} \ \ \Rightarrow \ \ S_c \ = \ 6 \ (V^{1/3})^2 \ = \ 6 \ V_c^{2/3} \ \ ; $$

sphere -- $$ S_s \ = \ 4 \ \pi \ r^2 \ \ , \ \ V_s \ = \ \frac{4}{3} \ \pi \ r^3 \ \ \Rightarrow \ \ r \ = \ \left(\frac{3 \ V_s}{4 \ \pi} \right)^{1/3} $$ $$ \Rightarrow \ \ S_s \ = \ 4 \ \pi \ \left[\left(\frac{3 \ V_s}{4 \ \pi} \right)^{1/3} \right]^2 \ = \ (4 \ \pi)^{1/3} \cdot 3^{2/3} \cdot V_s^{2/3} \ = \ \ (36 \ \pi)^{1/3} \ V_s^{2/3} \ \ . $$

The ratio of the surface areas is then

$$ \frac{S_c}{S_s} \ \ = \ \ \frac{6 \ V_c^{2/3}}{(36 \ \pi)^{1/3} \ V_s^{2/3}} \ \ ; $$

upon equating the volumes, we have

$$ \frac{S_c}{S_s} \ \ = \ \ \frac{6 \ V ^{2/3}}{(36 \ \pi)^{1/3} \ V ^{2/3}} \ \ = \ \ \frac{6 }{6^{2/3} \ \pi^{1/3} } \ \ = \ \ \left(\frac{6 }{ \pi } \right)^{1/3} \ \ = \ \ \frac{6^{1/3} }{ \pi^{1/3} } \ \ \text{or} \ \ \frac{\sqrt[3]{6} }{ \sqrt[3]{ \pi} } \ \ . $$

Answer 3

Unit volume assumed: $$ 1 = \frac{4}{3}\pi r^3 = a^3 $$ So we have $$ r = \sqrt[3]{\frac{3}{4\pi}} \\ \quad a = 1 $$ So the surface ratio is $$ A_s = 4 \pi r^2 = 4 \pi \left( \frac{3}{4\pi} \right)^{2/3} = \sqrt[3]{4\pi \cdot 9} =\sqrt[3]{\pi} \, 6^{2/3} \\ A_c = 6 $$ So we get $$ A_s : A_c = \sqrt[3]{\pi} : \sqrt[3]{6} $$

Answer 4

For convenience, we will compare two shapes of volume $\dfrac16$.

For a cube, $V=\dfrac16=c^3,S=6c^2$, so that $S=6^{1/3}$.

For a sphere, $V=\dfrac16=\dfrac43\pi r^3,S=4\pi r^2$ so that $S=4\pi\left(\dfrac1{8\pi}\right)^{2/3}=\pi^{1/3}$.