Area of the intersection of two cones I would like to find an expression for the area of the intersection of the two cones shown in the following figure:

The axes of both cones are situated on the plane YZ, and the angles are respectively $\alpha_1$ and $\alpha_2$, while the angle of the intersection is $\theta$, as indicated in the picture. Obviously, I'm interested in the case when $\alpha_1+\alpha_2<\theta$, since otherwise there would be no intersection between the cones. 
My idea is to do it by integration on the sphere, but how can I find the limits I need for the integral? 
 A: Your question can be answered with spherical geometry. In the notation of the question, and referring to the diagram, the circle of angular radius $\alpha_{1}$ centered at $p_{1}$ and the circle of angular radius $\alpha_{2}$ centered at $p_{2}$ meet at $p_{3}$ and $p_{4}$. The segments shown are great circle arcs.



*

*The area sought is the area of the spherical sector $p_{1} p_{3} p_{4}$ plus the area of the sector $p_{2} p_{4} p_{3}$ minus the area of the spherical quadrilateral $p_{1} p_{3} p_{2} p_{4}$.

*If $\phi_{1} = \angle p_{4} p_{1} p_{3}$ is the "apex" angle of the spherical triangle $\triangle p_{1} p_{3} p_{4}$, the area of the spherical sector $p_{1} p_{3} p_{4}$ is $\phi_{1}(1 - \cos \alpha_{1})$ by Archimedes' hat box theorem.
Similarly, if $\phi_{2} = \angle p_{3} p_{2} p_{4}$, the area of the spherical sector $p_{2} p_{4} p_{3}$ is $\phi_{2}(1 - \cos \alpha_{2})$.

*If $\psi_{1} = \angle p_{1} p_{3} p_{4}$ and $\psi_{2} = \angle p_{2} p_{4} p_{3}$ are the "base" angles of the spherical triangles, the spherical quadrilateral $p_{1} p_{3} p_{2} p_{4}$ has area $\phi_{1} + \phi_{2} + 2(\psi_{1} + \psi_{2}) - 2\pi$.
Putting everything together, your "digon" has area
\begin{multline*}
  \phi_{1}(1 - \cos\alpha_{1}) + \phi_{2}(1 - \cos\alpha_{2}) - \bigl[\phi_{1} + \phi_{2} + 2(\psi_{1} + \psi_{2}) - 2\pi\bigr] \\
  = 2\pi - 2(\psi_{1} + \psi_{2}) - \phi_{1} \cos\alpha_{1} - \phi_{2} \cos\alpha_{2}.
\end{multline*}
If $A$, $B$, $C$ are points on the unit sphere, the normalized cross products
$$
n_{B} = \frac{A \times (B - A)}{\|A \times (B - A)\|},\qquad
n_{C} = \frac{A \times (C - A)}{\|A \times (C - A)\|}
$$
are perpendicular to the planes through $O$, $A$, $B$ and $O$, $A$, $C$ respectively, so
$$
\angle CAB = \arccos(n_{B} \cdot n_{C}).
\tag{1}
$$
To express this completely in terms of $\alpha_{1}$, $\alpha_{2}$, and $\theta$, use Cartesian coordinates with
\begin{align*}
  p_{1} &= (0, 0, 1), \\
  p_{2} &= (\sin\theta, 0, \cos\theta), \\
  p_{3} &= (a, -b, \cos\alpha_{1}), \\
  p_{4} &= (a, \phantom{-}b, \cos\alpha_{1}).
\end{align*}
The condition $p_{2} \cdot p_{3} = \cos\alpha_{2}$ gives
$$
a = \frac{\cos\alpha_{2} - \cos\alpha_{1} \cos\theta}{\sin\theta},
$$
and then
$$
b = \sqrt{\sin^{2}\alpha_{1} - a^{2}}.
$$
Equation (1) now gives the angles $\phi_{i} = \angle p_{3} p_{i} p_{4}$ and $\psi_{i} = p_{3} p_{4} p_{i}$ in terms of $\alpha_{1}$, $\alpha_{2}$, and $\theta$.
(I haven't tried to substitute everything and simplify.)
