Surface area of sphere and total surface area of cube are same. Show that the ratio of the volume of the sphere to that of the cube is \(\sqrt{6} : \sqrt{\pi}\).

Proof:

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

Surface area of the sphere = 4πr²; total surface area of the cube = 6a².

Given 4πr² = 6a², so r² = \(\dfrac{6a^2}{4\pi} = \dfrac{3a^2}{2\pi}\) and r = \(a\sqrt{\dfrac{3}{2\pi}}\).

Volume of the sphere = \(\dfrac{4}{3}\pi r^3\), volume of the cube = a³.

\(\dfrac{V_{sphere}}{V_{cube}} = \dfrac{\frac{4}{3}\pi r^3}{a^3} = \dfrac{4\pi}{3} \left(\dfrac{3}{2\pi}\right)^{3/2}\)

= \(\dfrac{4\pi}{3} \times \dfrac{3}{2\pi} \times \sqrt{\dfrac{3}{2\pi}} = 2\sqrt{\dfrac{3}{2\pi}} = \sqrt{\dfrac{12}{2\pi}} = \sqrt{\dfrac{6}{\pi}}\)

∴ \(V_{sphere} : V_{cube} = \sqrt{6} : \sqrt{\pi}\).