If a hemispherical bowl of radius 6cm contains water to a depth of h cm, the volume of the water is 1/3πh^2(18-h). Water is poured into the bowl at a rate 4 cm^3/s . Find the rate at ehich the water level is rising when the depth is 2 cm.

Answer 1

V = (1/3)πh^2 (18-h)
= 6πh^2 - (1/3)πh^3

dV/dt = 12πh dh/dt - πh^2 dh/dt

4 = dh/dt(12πh - πh^2)
when h = 2

4 = dh/dt(24π - 4π)
dh/dt = 4/(20π) = 1/(5π) cm/s