In the figure, a metal ball is placed at the bottom of a spherical container with water. The radii of the container and the metal ball are $8$ cm and $4$ cm respectively. Let $h$ cm be the depth of water.
(a) It is given that the volume of a portion of a sphere cut off by a horizontal plane is $\pi \left(rH^2 - \frac{H^3} 3 \right) \ \text{cm}^3$, where $r$ cm is the radius of the sphere and $H$ cm is the height of the portion. Show that the volume of water in the container is
(i) $\frac{\pi}{3}(-h^3 + 24h^2 - 256) \ \text{cm}^3$ for $8 \leq h \leq 16$,
(ii) $4\pi h^2 \ \text{cm}^3$ for $0 \leq h \lt 8$
(b) When the container is fully filled up with water, its bottom cracks suddenly, and the water leaks at a rate of $\pi \ \text{cm}^3/s$. Find the rate of decrease of the depth of water after the container has been leaking for 6 minutes. (Give your answer correct to 3 significant figures.)
I have done (a) part but got the wrong answer in (b). Here is how I do it:
Let V denote the volume of the sphere, $v(h)$ denotes the volume of water and $h$ denotes the depth of water,
After $6$ minutes, the volume of water is
$$\frac{4}{3} \pi 8^3 - 360 \pi = \frac{968 \pi}{3} \lt \frac{1}{2}V$$
Then find $h_0$ which is the final depth of water,
\begin{align}
4 \pi h_0^2 &= \frac{968 \pi}{3} \\
h_0 &= \sqrt{\frac{242}{3}}
\end{align}
Then,
\begin{align}
\frac{dv}{dt} &= \frac{dv}{dh} \frac{dh}{dt} \\
-\pi &= 8 \pi h_0 \frac{dh}{dt} \\
\frac{dh}{dt} = -0.139
\end{align}
But my answer was wrong, the suggested one was $-0.0162 \ \text{cm/s}$, where did I go wrong? Thanks in advance.
After 6 minutes, the volume of water is actually:
$$\frac{4}{3} \pi (8)^3 - 6(60) \pi - \frac{4}{3}\pi(4)^3= \frac{712 \pi}{3} \lt \frac{1}{2}V$$
because you have forgotten to subtract the volume of the steel ball.
Then repeating your steps (which are correct), $h_0 = \frac{\sqrt{534}}{3}$, and so $\frac{dh}{dt} = -\frac{1}{8h_0} = -0.0162 \ \text{cm}/\text{s}$.