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Let's suppose we have a sphere with radius $R>0$, and half a cone with an opening angle $\alpha \in (0, \pi/4)$. The vertex of the cone is in the surface of the sphere, and the center of the sphere is in the surface of the cone.

How can I find the region of integration of their intersection?

I have placed the cone so that its axis is in the direction of the Z axis, and its vertex is $(0, 0, 0)$. I think spherical coordinates will make things easier, but I don't know how to determine the region of integration.

Here's the region of integration, and how the cone and sphere are placed:

Cone and sphere

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If I understand the geometric description, the center of the sphere can be chosen to lie at $(R\sin\alpha, 0, R\cos\alpha)$ (i.e., in the positive $x$-direction in the $(x, z)$-plane). The ball bounded by the sphere is the solution set of $$ (x - R\sin\alpha)^{2} + y^{2} + (z - R\cos\alpha)^{2} \leq R^{2}, $$ or $$ x^{2} + y^{2} + z^{2} \leq 2xR\sin\alpha + 2zR\cos\alpha. $$ Using spherical coordinates $$ x = \rho \cos\theta \sin\phi,\qquad y = \rho \sin\theta \sin\phi,\qquad z = \rho \cos\phi, $$ in which $\theta$ denotes longitude and $\phi$ denotes colatitude, and canceling a common factor of $\rho$, the ball is described by $$ \rho \leq 2R(\cos\theta \sin\phi \sin\alpha + \cos\phi \cos\alpha). $$ The bounds on the solid region of intersection are therefore $$ 0 \leq \theta \leq 2\pi,\qquad 0 \leq \phi \leq \alpha,\qquad 0 \leq \rho \leq 2R(\cos\theta \sin\phi \sin\alpha + \cos\phi \cos\alpha). $$

If you really did mean "half a cone with vertex angle $\alpha$" (rather than a "cone with half-vertex angle $\alpha$", as I've assumed), this description will require modification (and more information: where is the center of the sphere on the half-cone?).

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  • $\begingroup$ Yes, that's what I meant. I've added an image in order to see the region of integration and what I meant. $\endgroup$
    – user139019
    Apr 1 '14 at 21:09

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