(41) Bulk Cylindrical Tank

[Thomas Klemm]

(10-12-2018 10:41 PM)Thomas Klemm Wrote:  Instead this formula can be used:

\(V(x)=\left [ x(\pi-\cos^{-1}x)+\tfrac{1}{3}\sqrt{1-x^2}(2+x^2) \right ]Hr^2\)

For \(x=0\) this leads to \(V(0)=\tfrac{2}{3}Hr^2\) which surprised me because it doesn't contain \(\pi\).
This means the slanted part of the cylinder is exactly \(\frac{1}{6}\) of the square box that envelops it.

Kind regards
Thomas