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)}$

However what now? Dont belive that you have a tan 1/2

Thank you!


Using spherical coordinates $$x = rcos\theta cos\phi$$ $$y = rsin\theta cos\phi$$ $$z = rsin\phi$$

The sphere becomes $r = 2$ and the cone becomes $tan\phi = 2$

The volume becomes $$\int_{\theta=0}^{2\pi}\int_{\phi=tan^{-1}2}^{\frac{\pi}{2}}\int_{r=0}^{2}r^{2}cos\phi dr d\phi d\theta$$ $$= \frac{16\pi}{3}(1 - \frac{2}{\sqrt{5}})$$

  • $\begingroup$ Thanks, im struggling with the pi/2 part of the integral, why pi/2? $\endgroup$ – Shaun Weinberg Apr 13 at 17:23
  • $\begingroup$ $\phi$ is the angle made with the xy plane and perpendicular to the xy plane. Since the volume is above the cone intersected with the sphere, so the angle should be from $\tan^{-1} 2$ to the z axis which is $\phi = \frac{\pi}{2}$ $\endgroup$ – KY Tang Apr 14 at 1:14

You have $tan(\phi)=1/2$, so $\phi=arctan(1/2)$ and the bounds for $\theta$ are also correct.

Going back to the definition of $\phi$, which is the angle $\rho$ makes between the positive z axis. We are considering the region above the cone, so the upper bound of $\phi=arctan(1/2)$ and the lower bound starts at the z axis, so $\phi\in[0,arctan(1/2)]$

The region lies inside the sphere, so $\rho$ cannot extend past the sphere, in other words the sphere $x^2+y^2+z^2=4$ describes the upper bound of $\rho$, which you have as 2. So, $\rho\in[0,2]$, now you just integrate over these bounds.

  • $\begingroup$ I'm with you now, ok thank you. $\endgroup$ – Shaun Weinberg Apr 13 at 17:27

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