Problem source

\begin{center}
    \begin{tikzpicture}[scale=0.5]
        \draw[domain = -2.5:2.5, samples=180, variable = \x]  plot ({\x},{1+\x*\x});
        \draw[domain = 0:5, samples=180, variable = \x]  plot ({\x+8},{2*(ln(\x + sqrt(\x^2+1)))});
        \draw[domain = 0:5, samples=180, variable = \x]  plot ({-\x+8},{2*(ln(\x + sqrt(\x^2+1)))});
        \draw[domain = 173:353, samples=180, variable = \x]  plot ({4+sqrt(65/4)*cos(\x)},{0.5+sqrt(65/4)*sin(\x)});
        \draw[dashed] (-2.5, 1) -- (2.5, 1);
        \draw[dashed] (-2.5, 0) -- (12, 0);
        \def\h{3.63689};
        \draw ({-3+8}, \h) -- ({3+8}, \h);
        \draw[dashed] ({-3+8}, \h) -- ({sqrt(\h-1)}, \h);
        \draw ({-sqrt(\h-1)}, \h) -- ({sqrt(\h-1)}, \h);
        \node at (0,5) {$A$};
        \node at (8,5) {$B$};
        \draw[->] (1, 0.5) -- (1, 1);
        \draw[->] (1, 0.5) -- (1, 0);
        \node at (1, 0.5) [right] {$1$ unit};
        \draw[->] (4, {\h/2}) -- (4, \h);
        \draw[->] (4, {\h/2}) -- (4, 0);
        \node at (4, {\h/2}) [right] {$h$};
    \end{tikzpicture}
\end{center}
Two funnels $A$ and $B$ have surfaces formed by rotating the curves $y=x^{2}$ and $y=2\sinh^{-1}x$ $(x>0)$ above the $y$-axis. The bottom of $B$ is one unit lower than the bottom of $A$ and they are connected by a thin rubber tube with a tap in it. The tap is closed and $A$ is filled with water to a depth of 4 units. The tap is then opened. When the water comes to rest, both surfaces are at a height $h$ above the bottom of $B$, as shown in the diagram. Show that $h$ satisfies the equation 
\[
h^{2}-3h+\sinh h=15.
\]

Solution source

The initial volume of water in $A$ is:

\begin{align*}
\pi \int_0^4 x^2 \, \d y &= \pi \int_0^4 y \d y \\
&= \pi [ \frac{y^2}{2}]_0^4 \\
&= 8\pi
\end{align*}

We assume that no water is in the tube as it is `thin'.

Therefore we must have:

\begin{align*}
&& 8\pi &= \pi \int_0^{h-1} x^2 \d y +\pi \int_0^{h} x^2 \d y \\
&&&= \pi \int_0^{h-1} y \d y +\pi \int_0^{h} \l \sinh \frac{x}{2}\r^2 \d y \\
&&&= \pi \left [\frac{y^2}{2} \right]_0^{h-1} + \pi \int_0^h \frac{-1+\cosh y}{2}\d y \\
&&&= \pi \frac{(h-1)^2}{2} + \pi \left [ -\frac{y}{2} +\frac{\sinh y}{2}\right]_0^h \\
&&&= \pi \frac{(h-1)^2}{2} -\pi \frac{h}{2} + \pi \frac{\sinh h}{2}  \\
\Rightarrow && 0 &= h^2-2h+1-h+\sinh h -16 \\
&&&= h^2 -3h+\sinh h - 15 \\
\Rightarrow && 15 &= h^2 -3h+\sinh h
\end{align*}