A hemispherical bowl of radius R is filled with water

A hemispherical bowl of radius $R$ is filled with water at the constant rate of $a$ litres per second. Find the rate at which the depth increases at depth $h$, and show that it is inversely proportional to the area of the free surface.

Step 1: The cross-section at height $z$

Put the origin at the lowest point of the bowl, $z$ upward; the centre of the sphere is then at $z = R$, and the wall satisfies

\[ x^{2}+y^{2}+\bigl(z-R\bigr)^{2} = R^{2} \;\Longrightarrow\; x^{2}+y^{2} = 2Rz-z^{2}. \]

So the horizontal cross-section at height $z$ is a disc of area $\pi\bigl(2Rz-z^{2}\bigr)$.

Step 2: Volume filled to depth $h$

\[ V(h) = \int_{0}^{h}\pi\bigl(2Rz-z^{2}\bigr)dz = \pi\left[Rz^{2}-\frac{z^{3}}{3}\right]_{0}^{h} = \pi\left(Rh^{2}-\frac{h^{3}}{3}\right). \]

Step 3: Differentiate with respect to time

Both $V$ and $h$ change with $t$: \[\frac{dV}{dt} = \pi\bigl(2Rh-h^{2}\bigr)\frac{dh}{dt}. \tag{1}\]

The bracket is precisely the cross-sectional area at the top of the water — which is no accident, since the volume added in time $dt$ is a thin layer of that area.

Step 4: Solve for the rate of rise

The inflow is $\dfrac{dV}{dt} = a\cdot10^{-3}\ \text{m}^{3}\text{/s}$, so

\[ \frac{dh}{dt} = \frac{a\cdot10^{-3}}{\pi h\bigl(2R-h\bigr)}. \]

Step 5: The free surface area

At depth $h$ the water surface is a disc of radius $\sqrt{2Rh-h^{2}}$, of area

\[ S(h) = \pi\bigl(2Rh-h^{2}\bigr) = \pi h\bigl(2R-h\bigr), \]

so the answer takes the compact form below and the inverse proportionality is immediate.

Step 6: Verification

\[\text{Full bowl: } V(R) = \pi\left(R^{3}-\frac{R^{3}}{3}\right) = \frac{2\pi R^{3}}{3}, \text{ a hemisphere} \;\;\checkmark\]

\[h\to0:\ \frac{dh}{dt}\to\infty \text{ (the bottom is a point); } h = R:\ \frac{dh}{dt} = \frac{a\cdot10^{-3}}{\pi R^{2}}, \text{ the slowest rise} \;\;\checkmark\]

\[R = 0.5\ \text{m},\ a = 1,\ h = 0.25:\ \frac{dh}{dt} = \frac{10^{-3}}{\pi\cdot0.25\cdot0.75} \approx 1.70\ \frac{\text{mm}}{\text{s}} \;\;\checkmark\]