263. A reservoir in the form of a hemisphere with an internal radius of $R(m)$ is being filled with water at a rate of $Q(l)$ per second. Determine the rate of increase of the water level in the reservoir at the moment when it is equal to $0.5 R$.
Solution. Let $h$ be the water level in $m$ and $v$ its volume in $\mu^{3}$. We will find the dependence between the variables $h$ and $v$ using the formula for the volume of a spherical cap:

$$
v=\pi h^{2}\left(R-\frac{h}{3}\right)
$$

Differentiating this equality with respect to time $t$, we will find the dependence between the rates of change of the variables $h$ and $v$:

$$
\frac{d v}{d t}=\frac{d v}{d h} \cdot \frac{d h}{d t}=\pi\left[2 h\left(R-\frac{h}{3}\right)-\frac{1}{3} h^{2}\right] \frac{d h}{d t}=\pi\left(2 R h-h^{2}\right) \frac{d h}{d t}
$$

Assuming, according to the condition, $\frac{d v}{d t}=0.001 Q\left(\frac{\mu^{3}}{\text { sec }}\right)$, we get $\frac{d h}{d t}=\frac{0.001 Q}{\pi h(2 R-h)}\left(\frac{M}{\text { sec }}\right)$.

When $h=\frac{R}{2}$, we get $\frac{d h}{d t}=\frac{0.004 Q}{3 \pi R^{2}}\left(\frac{M}{\text { sec }}\right)$.