The surface areas of two spheres are in the ratio of 64 : 81. Find the ratio of their volumes, in the order given.

Solution

Let's assume the radius of two spheres is $$R_1$$ and $$R_2$$.

The surface areas of two spheres are in the ratio of 64 : 81.

$$\frac{Surface\ area\ of\ the\ first\ sphere}{Surface\ area\ of\ the\ second\ sphere}\ =\ \frac{64}{81}$$

$$\frac{4\times\ \pi\ \times\ \left(R_1\right)^2}{4\times\ \pi\ \times\ \left(R_2\right)^2}\ =\ \frac{64}{81}$$

$$\frac{\ \left(R_1\right)^2}{\ \left(R_2\right)^2}\ =\ \frac{64}{81}$$

$$\left(\frac{\ R_1}{\ R_2}\right)^2\ =\ \left(\frac{8}{9}\right)^2$$
$$\frac{\ R_1}{\ R_2}\ =\ \frac{8}{9}$$    Eq.(i)
Ratio of their volumes, in the order given = $$\frac{\ \frac{4}{3}\times\ \pi\ \times\ \left(R_1\right)^3}{\frac{4}{3}\times\ \pi\ \times\ \left(R_2\right)^3}$$
= $$\frac{\ \left(R_1\right)^3}{\ \left(R_2\right)^3}$$
= $$\frac{\ \left(8\right)^3}{\ \left(9\right)^3}$$
= $$\frac{\ 512}{729}$$
= 512 : 729