Volume of a bead with square hole- Program approach?

[Albert Chan]

Hi, DM48

Dimensional analysis shows the formula is definitely wrong, with Length^6 for first term.
Yes, a = radius of sphere, but b was half the length of square.

I corrected your formula, and rewrite as volume of sphere - square hole

\(\frac{4}{3}\pi a^3 - \frac{8}{3}
\left (b^2\sqrt{a^2-2b^2}
-a^3\;arctan \left( \frac{b^2}{a\sqrt{a^2-2b^2}} \right)
+b\left(3a^2-b^2\right )arctan\left(\frac{b}{\sqrt{a^2-2b^2}} \right ) \right )\)

Note, formula for square hole, except for a factor of 8, is same as my code.
My formula doubled up all dimensions, diameter instead of radius, and length, width for rectangle.