Find depth of a half-filled parabolic cross-section Given a cross-section of an object that is parabolic in shape, how do you find the depth of the object when it is "half full".
A full example given in an exam:

A long trough whose cross-section is parabolic is $1\frac{1}{2}$ metres wide at the top and $2$ metres deep. Find the depth of water when it is half-full.


Model the cross-section as a parabola centred at the origin. Then we know that:
$$
y = ax^2
$$
for some constant $a$. But from the given dimensions, we also know that the parabola contains the point $(3/4, 2)$. Substituting, we find that:
$$
2 = a(3/4)^2 \iff 2 = \frac{9a}{16} \iff a = \frac{32}{9}
$$
Now the area of the cross-section when full is given by:
$$
A = 2\int_0^{3/4} (2 - \tfrac{32}{9}x^2) \, dx = 2 \left[2x - \frac{32}{27}x^3 \right]_0^{3/4} = 2\left[\frac{3}{2} - \frac{1}{2} \right] = 2
$$
Now suppose that the half-width of the cross-section when half-full is $x = k$ so that its depth when half-full is $\frac{32}{9}k^2$. Then it follows that:
\begin{align*}
\frac{A}{2} &= 2\int_0^{k} (\tfrac{32}{9}k^2 - \tfrac{32}{9}x^2) \, dx \\
\frac{9A}{128} &= \int_0^{k} (k^2 - x^2) \, dx \\
\frac{9(2)}{128} &= \left[k^2x - \frac{x^3}{3} \right]_0^k \\
\frac{9}{64} &= \frac{2k^3}{3} - 0 \\
\frac{27}{128} &= k^3 \\
k &= \frac{3}{4\sqrt[3]{2}}
\end{align*}
Hence, we conclude that the desired depth is:
$$
\frac{32}{9}k^2 = \frac{32}{9}\cdot \frac{9}{16\sqrt[3]{4}} = \frac{2}{\sqrt[3]{4}} \cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}} = \sqrt[3]{2}
$$
