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October 2, 2014

October 2, 2014

Posted by **Mackenzie** on Wednesday, November 28, 2012 at 2:28pm.

I used the equation cos (phi)*sin(phi)*rho d(rho)d(phi)d(theta) with the given boundaries.

I got -632*pi*sqrt(3)/12

But that answer is wrong.

- Calculus -
**Mackenzie**, Wednesday, November 28, 2012 at 2:29pmThe equation I used is cos (phi)*sin(phi)*rho^2 d(rho)d(phi)d(theta)

I misspelled it on the first message.

- Calculus -
**Steve**, Wednesday, November 28, 2012 at 2:48pmHmmm.

∫∫∫ f(p,θ,φ) dv

=∫[0,π/3]∫[0,2π]∫[3,7] p^2 sinφ cosφ dp dθ dφ

= ∫[0,π/3]∫[0,2π] 316/3 * 1/2 sin2φ dθ dφ

= 316π/3 ∫[0,π/3] sin2φ dφ

= -158π/3 (-1/2 - 1)

= 316π

better double-check my math :-)

- Calculus -
**Steve**, Wednesday, November 28, 2012 at 2:56pmoops. That'd be 79π on that last step.

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