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

Question

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

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.

Answer 1

HINT

I would use cylindical coordinates with

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

and

• $0\le \theta \le 2\pi$
• $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

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

Answer 2

I did the following plot:

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 $$0\leq \rho\leq 2 \alpha\cos(\theta), \phi\in(\alpha, \pi), \theta\in[0,2\pi]$$