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.


The way to do this is to find the area of the parabola as a function of $x$. Let's assume for simplicity that function is: $x = c*y^2$. This is equivalent to $y = \sqrt{\frac{x}{c}}$ Since the parabola is symmetric, we can double this to get the total width at any point $x$.
$$ w(x) = 2 \sqrt{\frac{x}{c}}$$
Now we can integrate with respect to $x$ from $0$ to $2$. But first, we need to find the constant $c$ that gives us the particular parabola that we have in this case. To do this, at the point $(0.75,2)$ solve for $c$.
$$ y= \sqrt{\frac{x}{c}} = 2 = \sqrt{\frac{3/4}{c}} \rightarrow 4 = \frac{3}{4c} \rightarrow c = \frac{3}{16} $$
Now we can do the actual integration that will give the area.
$$A(u) = \int_0^u w(x) dx = \int_0^u 2 \sqrt {\frac{16x}{3}} dx = \frac{8}{\sqrt 3} \int_0^u \sqrt x dx = \frac{8}{\sqrt 3} [\frac{2}{3} x^{\frac{3}{2}}]_0 ^ u = \frac{16}{3 \sqrt 3} u^{\frac{3}{2}}$$.
Thus, you now have the area as a function of integration distance along the length. Just plug in $2$ to get the total area, and then solve it for $1/2$ the total area.
$$A_t = \frac{16}{3 \sqrt 3} 2^{\frac{3}{2}} = \frac{32}{3}\sqrt{\frac{2}{3}}$$
Solve:
$$\frac{\frac{32}{3}\sqrt{\frac{2}{3}}}{2} = \frac{16}{3 \sqrt 3} u^{\frac{3}{2}}$$
$$\frac{16}{3}\sqrt{\frac{2}{3}} = \frac{16}{3 \sqrt 3} u^{\frac{3}{2}}$$
$$\sqrt{\frac{2}{3}} = \frac{1}{\sqrt 3} u^{\frac{3}{2}}$$
$$\sqrt{2} =  u^{\frac{3}{2}}$$
$$\sqrt{2} =  u^{\frac{3}{2}}$$
$$u = \sqrt[3]{2}$$
