# Liquid in a spherical container

## Liquid in a spherical container

If the sphere is three quarters filled with liquid, what is the depth of the liquid?

This seems like it should be a pretty simple question, but I think exceeds the requirements of A-Level.

The easiest way to approach this question is to treat it as a volume of revolution question, using the equation of a sphere and then impose a condition on the limits.

$$V = \pi\int_{-r}^a y^2dx$$

As the volume of a sphere can be generated by rotating a circle, with radius $r$, then we can say $y^2 = r^2 - x^2$.

Substitution into the integral yields

$$\begin{align} V &= \pi\int_{-r}^a (r^2 - x^2)\\ &= \pi\left[r^2x - \frac{1}{3}x^3\right]_{-r}^a\\ &= \pi\left(r^2a - \frac{1}{3}a^3 + \frac{2}{3}r^3 \right)\end{align}$$

We know that the sphere, has a volume of $V_S = \frac{4}{3}\pi r^3$. If the container is $\frac{3}{4}$ full, then we get the equation

$$\begin{align}\frac{3}{4}V_S &= \pi\left(r^2a - \frac{1}{3}a^3 + \frac{2}{3}r^3 \right) \\ \pi r^3 &= \pi\left(r^2a - \frac{1}{3}a^3 + \frac{2}{3}r^3 \right) \\ r^3 &= \left(r^2a - \frac{1}{3}a^3 + \frac{2}{3}r^3 \right) \\ \frac{1}{3}r^3 -r^2a +\frac{1}{3}a^3 &= 0 \end{align}$$

This yields an equation

$$r^3 -3r^2a +a^3 =0$$

Solution of this, for $a$ requires advanced techniques for solving cubics - in fact this is a perfect cubic for applying the Cardano method to, (Example 2) which after considerable trigonometry, gives $a = 2\sin\left(\frac{\pi}{18}\right)r$. Though, for speed, numerical methods give $a = 0.347296r$

The depth of the liquid is therefore $1.347296r$.