# 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?

• It is not clear if the area is on the sphere or on the surface of the cones. From the figure, I understand it is on the sphere. From the text I understand it is on the cones. Mar 10, 2017 at 10:14
• The area is on the cones. Sorry, English is not my first language and I probably didn't explain it well. I know how to compute the area that each of the cones delimites on the sphere ($4\alpha_1$ and $4\alpha_2$ if we consider the sphere to be the unit sphere, right?), but I don't know how to compute the area of the intersection, on the surface of the sphere. Mar 10, 2017 at 10:37
• @toliveira, sorry, I forgot to tag you in my previous comment, and I can't figure out how to edit it. Mar 10, 2017 at 10:53
• @WildFeather, I would try to: 1) express the blue circle in spherical coordinates; 2) express the red circle in spherical coordinates; 3) find an expression for the two lines delimiting the surface on the sphere. I haven't found out yet how to do (2). Mar 10, 2017 at 14:49
• @WildFeather, the equation of the blue circle in spherical coordinates is (R,θ,α1), where R is the radius of the sphere, θ∈[0,2π]. Note that by θ in this comment is a dummy variable and not the one you have drawn. Mar 11, 2017 at 1:19

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. 1. 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 $p_{1} p_{3} p_{2} p_{4}$.

2. 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 $\phi_{2}(1 - \cos \alpha_{2})$.

3. 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.)

• Ah, yes, you did say the cone axes were in the $(y, z)$-plane. Swapping the first two coordinates reconciles things with your notation (though of course the area is the same either way). ;) Mar 10, 2017 at 20:51
• @toliveira: I used ePiX. Mar 11, 2017 at 11:56
• @WildFeather: If $\alpha_{1} = 0$, the spherical cap has area zero, not $2\pi$, and if $\alpha > \pi/2$, the area is positive. ;) In my previous comment, I probably should have written that one nappe of a cone of half-angle $\alpha$ cuts out a spherical cap of area $2\pi(1 - \cos\alpha)$. (This cap is technically a "zone", but "zone" might suggest the complement of the cap.) In your calculation, if you replace $1 - \cos\alpha$ with $\cos\alpha$, the area of the sector comes out correctly. Mar 11, 2017 at 12:03
• The claim about the area of a quadrilateral comes from a fundamental property of spherical (geodesic) triangles, itself a special case of the Gauss-Bonnet theorem (but admitting an elementary proof along the lines of this post): The area is proportional to their angular defect. Dividing the quadrilateral into two triangles gives the stated property. Mar 11, 2017 at 16:19
• The "other" case turns out not to be when the axis of one cone is inside the other, but when one or both sectors subtend an angle greater than $\pi$. In this event, more-or-less the same formula holds, aside from signs on square roots. (It may be easiest to express the area of a "segment" of a spherical cap cut by a chord, and to view your digon as a union of two such segments.) Mar 11, 2017 at 21:28