Chapter 8, Problem 8P

The volume V of liquid in a hollow horizontal cylinder of radius r and length L is related to the depth of the liquid h by

$V=\left[r^2{\text{cos}}^{-1}\left(\frac{r-h}{r}\right)-\left(r-h\right)\sqrt{2rh-h^2}\right]L$

Determine h given $r=2\text{ m, }L=5\text{ m, and }V=8{\text{ m}}^3$. Note that if you are using a programming language or software tool that is not rich in trigonometric functions, the arc cosine can be computed with

${\text{cos}}^{-1}x=\frac{\pi }{2}-{\text{ tan}}^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$