A problem about how the rate of the water's volume changes in a hemisphere.

Question

Suppose there's a pool in the shape of a hemisphere whose radius is 5 meters, and at some time the level of the water is 2 meters from the bottom of the hemishere. If the rate at which the water level decreases is 1/3 meter per minute, then what's the rate at which the water's volume changes?

My attempt: Say the volume is $V$, the time is $t$, the radius of the circle formed by the water level is $r$, and the distance between the water level and the bottom is $x$. I'm aware that $r^2=25-x^2$ and I need to find the relation of this equation with $V$ so that I can differentiate them with respect to $t$ and find out $dV/dt$. But I cannot find the relation.

Answer 1

Denote the changing water level by $z$. At the instant in question we have $z=-5+2=-3$. The water surface at this level is a circle of radius $r=\sqrt{25-z^2}=4$ meters. Under an infinitesimal change $dz$ (in meters) of the water level we have a volume change $$dV=\pi r^2\>dz=16\pi\>dz$$ cubic meters. It follows that $${dV\over dt}= 16\pi{dz\over dt}={16\pi\over 3}$$ cubic meters per second.

Answer 2

We know that the volume produced with rotating a curve $y=f(x)$ about $x$-axis from $a$ to $b$ is $${\bf V}=\pi\int_a^by^2dx$$ for a pool with a hemisphere shape we see $y=\sqrt{r^2-x^2}$ $${\bf V}= \pi\int_{r-h}^{r}y^2dx=\pi\int_{r-h}^{r}r^2-x^2\,dx=\pi(xr^2-\frac13x^3)\Big|_{r-h}^{r}=\pi(r-\frac13r^3)-\pi(r-h-\frac13(r-h)^3)=\frac{\pi}{3}\left(3rh^2-h^3\right)=\pi h^2(r-\frac{h}{3})$$