Find the volume bounded above the sphere $r=2a\cos\theta$ and below the cone $\phi=\alpha$, where $0<\alpha<\frac{\pi}{2}.$

I'm supposed to use triple integrals in spherical coordinates to solve it.

The image I have in my mind is as such

Image of cone and sphere mentioned in the question

After some thinking I thought that the volume of the cone may be expressed as such (please note that the horizontal line represents the x-y plane, I forgot to include it in the picture):

$$\int^\frac{\pi}{2}_b \int^{2a}_0\int^{2\pi}_0 r^2\sin\theta\, d\phi\, dr \,d\theta$$ after transformation in spherical coordinates. Is it correct? I'm not confident in this answer.

The most troublesome part is the spherical cap on top. I know there's a formula for it in Cartesian coordinates, but I have absolutely no clue on how to find its volume via integration in spherical coordinates.

[FYI: The answer is $4\pi a^3(1-\cos^4\alpha)$]

Also, I have tried plain old geometry, ie, finding the volume of the sphere, then subtracting off the volume of the cone and spherical camp through their usual formulas, but I get some weird expression like $\frac{4}{3}\pi a^3 (2\sin^2\alpha(\sin^2\alpha-\cos^2\alpha)+1-6\sin^5\alpha \cos\alpha).$ If the diagram is wrong, please do let me know.

  • $\begingroup$ I do not think that you need to calculate the volume of the spherical cap. The original integral could be easily represented in spherical coordinates. So is the spherical cap plus the cone. $\endgroup$ – xbh Sep 8 '18 at 14:19
  • $\begingroup$ Sorry, I don't understand what you mean when you say the original integral being represented in spherical coordinates $\endgroup$ – Yip Jung Hon Sep 8 '18 at 14:20
  • $\begingroup$ How comes the answer is independent of $\theta$? $\endgroup$ – Bernard Sep 8 '18 at 14:21
  • $\begingroup$ @YipJungHon Sorry, not the original integral. It should be the original region. $\endgroup$ – xbh Sep 8 '18 at 14:22
  • $\begingroup$ @Bernard I assume you mean $\alpha$ right? There's an alpha hidden in the answer, the a in the coefficient and the alpha in the cosine are 2 different things. $\endgroup$ – Yip Jung Hon Sep 8 '18 at 14:26

I did the following plot:

enter image description here

It seems that, it was not an intersection between an sphere and a cone as you had thought. @xbh last comment refered to this. I think $$0\leq \rho\leq 2 \alpha\cos(\theta), \phi\in(\alpha, \pi), \theta\in[0,2\pi]$$

  • $\begingroup$ Hi, actually I've gotten that as well. But actually the question explicitly mentions that it is a sphere. As such, I had no choice but to assume that the theta mentioned in the expression refers to the $\theta=arctan(x^2=y^2)^\frac{1}{2}/x$, rather than the $arctan=\frac{y}{x}$ one. $\endgroup$ – Yip Jung Hon Sep 8 '18 at 23:28
  • $\begingroup$ I don't know why did it comes from? why did the book call it an sphere? I hope you can get the right asap. :-) $\endgroup$ – mrs Sep 9 '18 at 6:41


I would use cylindical coordinates with

  • $x=r\cos \theta$
  • $y=r\sin \theta$
  • $z=z$


  • $0\le \theta \le 2\pi$
  • $2a \cos^2 \alpha\le z\le 2a $
  • $0\le r \le 2a \cos \alpha\sin \alpha$
  • $dV=r\,dz\, dr\, d\theta$

that is

$$V=\int_0^{2\pi} d\theta\int^{2}_{2a \cos^2 \alpha}\, dz\int^{2a \cos \alpha\sin \alpha}_0 r\,dr$$


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