# Liquid in a sphere

Solution
This problem can be solved using geometry alone, but the quickest method is to use the volume of revolution formulae $$V_{Liquid} = \frac{3}{4}V_{Sphere} = \frac{3}{4}\frac{4}{3}\pi r^3= \pi r^3$$ We now require that the volume of revolution between $-r$ and some unknown position, $a$ is $\pi r^3$, that is, $$\pi \int_{-r}^a y^2 dx = \pi r^3$$ For a sphere, centre $O$, the equation of the sphere is given by $x^2 + y^2 = r^2$, this gives $y^2 = r^2 - x^2$. $$V_{Liquid} = \pi\int_{-r}^a r^2 - x^2dx = \pi\left[ r^2x - \frac{1}{3}x^3 \right]_{-r}^a$$ Substitution yields $$V_{Liquid} =\pi\left(\left(r^2a - \frac{1}{3}a^3 \right) - \left( -r^3 +\frac{1}{3}r^3 \right)\right)$$ Let us now apply $r=1$ then we get $$V_{Liquid} = \pi = \pi\left(-\frac{1}{3}a^3 + a + \frac{2}{3} \right)$$ dividing by $\pi$ and rearranging yields $$a^3 - 3a + 1 = 0$$ The exact solutions of the cubic are unhelpfully ugly, however, we get a decimal approximation of $a\approx 0.3473$ The depth of the liquid is therefore $h = 1.3473 (4\mbox{dp})$