Assume that $r$= radius of cylinder = radius of cone = height of cylinder = height of cone.
Hence we can calulate the following:
- Volume of cylinder = volume of water = $\pi r^3$
- Volume of cone = $\frac13 \pi r^3$
- Volume of container = total volumes of cylinder and conee = $\frac 43 \pi r^3$
- Volume of air in container = $\frac 13 \pi r^3$
We know that
- (from above) the volume of water is more than half the volume of container
- container is symmetrical about its longitudinal axis
hence when container is lying on its side, the level of water must be greater than the midheight, $r$, of the container (on its side).
Let the level of water in the container on its side be $r+a$ where $0<a<1$.
(a) Cylinder
Volume of air is the area of a circular segment multiplied by its height $r$, i.e.
$$\begin{align}
V_1&=\frac 12 r^3 \ (2\alpha - \sin 2\alpha)\cdot r\qquad
\text{where $\cos\alpha=\frac ar$}\\
&=\frac 12 r^3\ (2\alpha-2\sin\alpha\cos\alpha)\\
&=r^3\left[\cos^{-1}\left(\frac ar\right) -\left(\frac ar\right)\sqrt{1-\left(\frac ar\right)^2}\ \ \right]\\
&=r^3(\cos^{-1}\lambda-\lambda\sqrt{1-\lambda^2})\qquad
\text{where $\lambda=\frac ar$}\\
&=\frac{r^3}3\left[3\cos^{-1}\lambda-3\lambda\sqrt{1-\lambda^2}\right]
\end{align}$$
(b) Cone
Calculate the the volume of air in the cone by integrating circle segments of infinitesimally small thickness and of different radius $u$, as $u$ varies from $r$ (the part adjacent to the cylinder) to $a$ (edge of container).
Area of circle segment with radius $u$ is given by
$$\frac 12 u^2 \ (2\phi - \sin 2\phi)
= u^2\ \left[ \cos^{-1} \left(\frac au\right) - \frac au \sqrt{1-\left(\frac au \right)^2} \;\right]\qquad
\text{where $\cos\phi=\frac au$}\\$$
Volume of air in cone is given by integrating this from $u=r$ to $u=a$, i.e.
$$\begin{align}
V_2&=\int_r^a
u^2\ \left[ \cos^{-1} \left(\frac au\right) - \frac au \sqrt{1-\left(\frac au \right)^2}\;\right] du\\
&=\int_r^a u^2\cos^{-1} \left(\frac au\right) -a\sqrt{u^2-a^2} \; du\\
&=a^2 \int_r^a \left(\frac ua \right)^2\cos^{-1} \left(\frac au\right) -\sqrt{\left(\frac ua\right)^2-1} \; \; du\
\end{align}$$
Put
$$\begin{align}
\frac au = \cos\phi \Rightarrow
u&=a\sec\phi\quad \Rightarrow \sec\phi=\frac ua\\
du&=a\sec\phi\tan\phi \; d\phi\\
u=r \Rightarrow \phi&=\cos^{-1}\frac ar=\cos^{-1}\lambda\\
u=a \Rightarrow \phi&=\cos^{-1}1=0
\end{align}$$
This gives
$$\begin{align}
V_2&=a^2\int_{\cos^{-1}\lambda}^{0}\left[(\sec^2\phi) \cdot \phi-\sqrt{\sec^2\phi-1}\right]\ a\sec\phi\tan\phi \; d\phi\\
&=a^3\int_{\cos^{-1}\lambda}^0 \phi\sec^3\phi\tan\phi-\sec\phi\tan^2\phi \; d\phi\\
&=a^3\int_{\cos^{-1}\lambda}^0 \phi\sec^3\phi\tan\phi-\sec^3\phi +\sec\phi \; d\phi\\
&=\cdots \text{(and after some tedious integration)}\cdots\\
&=\frac {a^3}3 \left[\phi\sec^3\phi-2\sec\phi\sqrt{\sec^2\phi-1}+\ln(\sec\phi+\ sqrt{\sec^2\phi-1})\right]_0^{\cos^{-1}\lambda}\\
&=\frac {a^3}3\left[ \left(\cos^{-1}\frac au\right)\left(\frac ua\right)^3-2\frac ua\sqrt{\left(\frac ua\right)^2-1}+\ln\left(\frac ua+\sqrt{\left(\frac ua\right)^2-1}\right)\right]_{u=a}^{u=r}\\
&=\frac {a^3}3 \left[ \left(\cos^{-1} \frac ar\right)\left(\frac ra\right)^3-2\frac ra \sqrt{\left(\frac ra\right)^2-1}+\ln\left(\frac ra+\sqrt{\left(\frac ra\right)^2-1}\right)\right]\\
&=\frac {a^3}3 \left[ \frac 1{\lambda^3} \cos^{-1}\lambda -\frac 2\lambda \sqrt{\frac 1{\lambda^2}-1}+\ln\left(\frac 1\lambda+\sqrt{\frac 1{\lambda^2}-1}\right) \right]\\
&=\frac {r^3}3 \left[ \cos^{-1} \lambda-2\lambda\sqrt{1-\lambda^2}+\lambda^3\ln\left(\frac 1\lambda +\sqrt{\frac 1{\lambda^2}-1}\right)\right]
\end{align}$$
(c) Total
Total volume of air in container is
$$\begin{align}
V_1+V_2&=\frac {\pi r^3}3\\
\frac{r^3}3\left[4\cos^{-1}\lambda-5\lambda\sqrt{1-\lambda^2}+\lambda^3\ln\left(\frac 1\lambda (1 + \sqrt{1-\lambda^2})\right)\right]&=\frac {\pi r^3}3\\
4\cos^{-1}\lambda-5\lambda\sqrt{1-\lambda^2}+\lambda^3\ln\left(\frac 1\lambda (1 + \sqrt{1-\lambda^2})\right)-\pi&=0\\
\lambda &\approx 0.36876\\
a=\lambda r &\approx 0.36876 r
\end{align}$$
Hence, height of water in container on its side is
$$r+a=(1+\lambda)r\approx 1.36876r \qquad \blacksquare$$
NB:
1. See this for a useful guide to the messy integration and the volume of a vertical cut of a cone.
2. See this for a useful guide for integrating $\sec^3\theta$.
So we know the following:
Rate at which water is entering cone, lets call that $\dfrac{df}{dt}$.
Rate at which the height of the water in the cone is rising, $\dfrac{dh}{dt}$.
We need to find the leak rate, call it $\dfrac{dk}{dt}$.
My hint was that the change in volume of water in the tank, $\dfrac{dv}{dt}$, satisfies
$\dfrac{dv}{dt}=\dfrac{df}{dt}-\dfrac{dk}{dt}$
We have only one of the things we need, but we can find $\dfrac{dv}{dt}$ using our other given (this is the trickiest part). Note that the cone has fixed proportions (the relationship between $r$ and $h$ is a constant fixed ratio). Hence we can rewrite $v=\dfrac{1}{3}\pi r^2h$ entirely in terms of $h$. This uses the dimensions of the tank given in the problem. Hence $r=\dfrac{1}{4}h$. Substituting for $r$, we obtain
$v=\dfrac{1}{3}\pi \left( \dfrac{1}{4}h \right)^2h=\dfrac{1}{48}h^3$
Differentiate implicitly with respect to time
$\dfrac{dv}{dt}=\dfrac{3}{48}h^2\dfrac{dh}{dt}$
But we know what $h$ and $\dfrac{dh}{dt}$ are from the givens in the problem (you found $h$ yourself).
Thus you can find $\dfrac{dv}{dt}$ and use that and then given for $\dfrac{df}{dt}$ to solve for $\dfrac{dk}{dt}$.
Best Answer
Does this side view of your cone help?
I answered your question "why."