If the volume of a sphere doubles, what is the ratio of the surface area of the new, larger sphere to the old?

Feb 23, 2016

The ratio of the surface area of the new, larger sphere to the old is
$\sqrt[3]{4}$

Explanation:

Let's start with two formulas - for surface area of a sphere $S$ and for its volume $V$, assuming the radius of a sphere is $R$:
$S = 4 \pi {R}^{2}$
$V = \frac{4}{3} \pi {R}^{3}$

To double the volume, we have to increase the radius by multiplying it by $\sqrt[3]{2}$.
Indeed, let ${R}_{1} = R \sqrt[3]{2}$
Then the volume of a sphere with radius ${R}_{1}$ will be
${V}_{1} = \frac{4}{3} \pi {R}_{1}^{3} = \frac{4}{3} \pi {\left(R \sqrt[3]{2}\right)}^{3} = \frac{8}{3} \pi {R}^{3}$ - twice larger than original volume.

With radius ${R}_{1} = R \sqrt[3]{2}$ the surface area of a new sphere will be
$4 \pi {R}_{1}^{2} = 4 \pi {R}^{2} {\left(\sqrt[3]{2}\right)}^{2} = 4 \sqrt[3]{4} \pi {R}^{2}$

The ratio of the new surface area to the old one equals to
$\frac{4 \sqrt[3]{4} \pi {R}^{2}}{4 \pi {R}^{2}} = \sqrt[3]{4}$