 Quote by LCKurtz
It is a simple integral to set up in spherical coordinates. I will call the angle you denote y by [itex]\alpha[/itex] and the radius a:
[tex] 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)})[/tex]
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Did you derive this triple integral?