I just do not understand how the spherical co-ordinates conversion system works. I understand the concept, but the finding the limits for p,φ,θ does not work for me (I study part-time by myself).
The question is: " Let D be the 3-Dimensional region inside the sphere $ x^2 + y^2 + z^2 = 4 $ above the cone $ z= \sqrt{4x^2 + 4y^2}$
"Attempt at answer": Function conversion: $$\iiint p^2 \,dp \,dφ \,dθ$$
The limits y (θ): It is a full enclosed circle, thus 0 < 2π
The limits of x(p) x (p): r = 2 and therefore z = 2 z = pcosφ 2 = pcosφ 2/cosφ = p p = 2secφ
Therefore the limit is from 0 to 2secφ but a website i assessed had a different value. Is this because p = r and from the formula of the sphere it is r = 2 therefore p = 2?
The limits of z (φ): $ z = √4x^2 + 4y^2 $
$ p^2cosφ^2 = 4p^2sinφ^2cosθ^2 + 4p^2sinφ^2sinθ $
$ tanφ^2 = 1/4 $
$ tanφ^2 = \frac{1}{\sqrt(4)}$
$ tanφ = \frac{1}{2}$
However what now? Don't believe that you have a tan = 1/2 radians angle. I also saw a website that said that the angle is pi/4?
Thank you!