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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!

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  • $\begingroup$ It’s considered bad form here to repost the same question only an hour or so later. If you have things to add to your original question, you should edit it instead of posting what’s virtually a duplicate of it. $\endgroup$ – amd Apr 10 at 19:55

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