The equation I used is cos (phi)*sin(phi)*rho^2 d(rho)d(phi)d(theta)
I misspelled it on the first message.
Evaluate, in spherical coordinates, the triple integral of f(rho,theta,phi)=cos (phi) , over the region 3<rho<7, 0<theta<2pi, 0<phi<pi/3.
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.
3 answers
Hmmm.
∫∫∫ 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 :-)
∫∫∫ 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 :-)
oops. That'd be 79π on that last step.
1 answer