Thread: Arc Circle Volume View Single Post

 Quote by LCKurtz It is a simple integral to set up in spherical coordinates. I will call the angle you denote y by $\alpha$ and the radius a: $$V = \int_0^{2\pi}\int_0^\alpha\int_0^a \rho^2\sin{(\phi)}d\rho\ d\phi\ d\theta = \frac{2\pi a^3}{3}(1 - cos{(\alpha)})$$
Did you derive this triple integral?