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Feb11-10, 02:33 PM
P: 85
Quote Quote by LCKurtz View Post
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]
Did you derive this triple integral?