Loss of height rate in a half sphere?

Question

The question: A flask with a shape of half of a sphere with 5 meters as its radius is said to be fully filled, the flask has a hole at the bottom spot. If the hole were to be opened with the volume loss rate of $0.125m^3/minute$ How much is the height loss rate when the water is 4 meters from below?

From what I see in the problem, $\frac{dV}{dt} = 0.125m^3/minute$ and what we are trying to find is $\frac{dh}{dt}$ which means we need to find $\frac{dV}{dh}$ so we can multiply it like this $\frac{dh}{dV} * \frac{dV}{dt} = \frac{dh}{dt}$ but then I can't seem to find a way to derivate the half sphere volume formula ($\frac{2}{3}πr^3$) against h to make $\frac{dV}{dh}$

Any ideas?

Answer 1

Hint: At any stage, $R^2 = r^2+ (R-h)^2$ where $R$ is the initial radius, $r$ is the new radius and $h$ is the height of the water from the hole.

Answer 2

Hint: As far as the velocity of the surface is concerned, at any fixed point in time it is irrelevant that the walls of the container are slanted. It only matters how large the surface is.

Answer 3

Hint:

2D problem.

Consider a circle radius $r$ , centre $(0,r)$.

$x^2+(y-r)^2=r^2$;

$y=r \pm \sqrt{r^2-x^2}$;

Let $x \ge 0$.

Choose $y=r-\sqrt{r^2-x^2}.$

Volume of revolution when rotated about $y-$axis.

$dV = πx^2dy;$

Integrate from $0$ to $h$, which gives $V(h)$, the cap.

Try to finish.