Question

Could somebody show I how to do this triple integral? Integrate [0,2π]∫ [0,π/2]∫ [1/2 sec(φ),1]∫ (ρ^2 sin(φ))dpdφdθ.
These are what I did:
[0,2π]∫ [0,π/2]∫ (ρ^3)/3 sin(φ) [1/2sec(φ),1]dφdθ=>
[0,2π]∫ [0,π/2]∫ {((1/2sec(φ))^3)/3 sin(φ)}- {(1^3)/3sin(φ)}dφdθ=>
[0,2π]∫ [0,π/2]∫ (sec^3(φ))/24 sin(φ)- 1/3sin(φ)dφdθ=> I stuck here. Show me some steps to finish the integration.

Answer 1

sec(φ)^3 = (cosφ)^-3
So, using u=cosφ,
∫ {((1/2sec(φ))^3)/3 sin(φ)} dφ
= -1/24∫ u^-3 du
= 1/48 u^-2 = 1/48 cos^2φ
and now apply the limits for your integrals