# Rate of change of surface area of sphere

**Problem**

Gas is escaping from a spherical balloon at the rate of 2 cm^{3}/min. Find the rate at which the surface area is decreasing, in cm^{2}/min, when the radius is 8 cm..

**Solution**

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

$\dfrac{dV}{dt} = 4\pi r^2 \, \dfrac{dr}{dt}$

$2 = 4\pi r^2 \, \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \dfrac{1}{2\pi r^2}$

When r = 8 cm

$\dfrac{dr}{dt} = \dfrac{1}{2\pi (8^2)}$

$\dfrac{dr}{dt} = \dfrac{1}{128\pi}$

Surface area of the sphere:

$A = 4\pi r^2$

$\dfrac{dA}{dt} = 8\pi r \, \dfrac{dr}{dt}$

When r = 8 cm

$\dfrac{dA}{dt} = 8\pi(8)\left(\dfrac{1}{128\pi}\right)$

$\dfrac{dA}{dt} = \frac{1}{2} ~ \text{cm}^2\text{/min}$ *answer*

**Another Solution**

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

$V = \frac{1}{3}(4\pi r^2)r$

$V = \frac{1}{3}Ar$

$\dfrac{dV}{dt} = \frac{1}{3}\left( A \dfrac{dr}{dt} + r \dfrac{dA}{dt} \right)$

From Fluid Mechanics, discharge = area × velocity:

$Q = \frac{1}{3}\left( Q + r \dfrac{dA}{dt} \right)$

When r = 8 cm

$2 = \frac{1}{3}\left( 2 + 8 \cdot \dfrac{dA}{dt} \right)$

$\dfrac{dA}{dt} = \dfrac{2(3) - 2}{8}$

$\dfrac{dA}{dt} = \frac{1}{2} ~ \text{cm}^2\text{/min}$ *answer*