A sphere and a cube have the same surface. Show that the ratio of the volume of the sphere to that of the cube is $\sqrt{6} : \sqrt{π}$.

Let the side of the cube be a cm and let radius of sphere be r cm.

Surface area of sphere = 4πr².

Surface area of cube = 6a².

Given,
surface area of sphere = surface area of cube.

∴ 4πr² = 6a²

⇒ $\dfrac{r^2}{a^2} = \dfrac{6}{4π}$

⇒ $\dfrac{r}{a} = \sqrt{\dfrac{6}{4π}}$.

Volume of sphere = $\dfrac{4}{3}πr^3$.

Volume of cube = a³.

Ratio of volume of sphere to volume of cube is

$\Rightarrow \dfrac{\text{Volume of sphere}}{\text{Volume of cube}} = \dfrac{\dfrac{4}{3}πr^3}{a^3} \\[1em] = \dfrac{4πr^3}{3a^3} \\[1em] = \dfrac{4π}{3} \times \dfrac{r^3}{a^3} \\[1em] = \dfrac{4π}{3} \times \Big(\dfrac{r}{a}\Big)^3 \\[1em] = \dfrac{4π}{3} \times \Big(\sqrt{\dfrac{6}{4π}}\Big)^3 \\[1em] = \dfrac{4π}{3} \times \dfrac{6}{4π} \times \sqrt{\dfrac{6}{4π}} \\[1em] = \dfrac{4π}{3} \times \dfrac{6}{4π} \times \dfrac{1}{2}\sqrt{\dfrac{6}{π}} \\[1em] = \dfrac{24π}{24π}\sqrt{\dfrac{6}{π}} \\[1em] = \sqrt{\dfrac{6}{π}}.$

Hence proved that the ratio is $\sqrt{6} : \sqrt{π}$.