Answer
a. $ \pi ah^2-\frac{\pi}{3}h^3$
b. $\frac{1}{120\pi}\ m/sec$
Work Step by Step
a. Draw a diagram as shown in the figure where the equation of the bowl can be written as $x^2+(y-a)^2=a^2$ with $0\leq y\leq a$. Using disk approximation, we have the volume of water at level $h$ as
$V=\int_0^h \pi (x^2) dy=\int_0^h \pi (a^a-(y-a)^2) dy= \pi\int_0^h (y(2a-y)) dy= \pi\int_0^h (2ay-y^2) dy=\pi (ay^2-\frac{1}{3}y^3)|_0^h= \pi ah^2-\frac{\pi}{3}h^3$
b. Given $\frac{dV}{dt}=0.2\ m^3/sec$, we need to find $\frac{dh}{dt}$ at $a=5m, h=4m$. From the integral above, we have $\frac{dV}{dh}= \pi (2ah-h^2) $, and thus we have
$\frac{dV}{dt}= \pi (2ah-h^2)\frac{dh}{dt}= \pi (2(5)(4)-4^2)\frac{dh}{dt}=24\pi \frac{dh}{dt}=0.2\ m^3/sec$
which gives $\frac{dh}{dt}=\frac{1}{120\pi}\ m/sec$
