Calculate the area on a sphere of the intersection of two spherical caps Given a sphere of radius $r$ with two spherical caps on it defined by the radii ($a_1$ and $a_2$) of the bases of the spherical caps, given a separation of the two spherical caps by angle $\theta$, how do you calculate the surface area of that intersection? 
To clarify, the area is that of the curved surface on the sphere defined by the intersection. At the extreme where both $a_1,a_2 = r$, we would be describing a spherical lune. 
Alternatively define the spherical caps by the angles 
$\Phi_1 = \arcsin(a_1/r)$ and $\Phi_2 = \arcsin(a_2/r)$.
 A: This problem can be solved by elementary geometry, resp., spherical trigonometry.
We may assume $r=1$. Let $\rho$ and $\rho'$ be the spherical radii of the two caps.
When the caps are disjoint there is nothing to compute.
When the smaller cap (of radius $\rho$) is a subset of the larger then the area $A_\cap$ of the intersection is given by $$A_\cap=4\pi\sin^2{\rho\over2}\ .$$
Now the interesting case: Assume that  the two boundary circles intersect in two points $P$ and $Q$. Connect $P$ and $Q$ by an arc of a great circle; furthermore connect the centers $C$, $C'$ of the two caps with both $P$ and $Q$. The two arcs emanating from $C$ have length $\rho$ and enclose an angle $\alpha>0$, and similarly there is an angle $\alpha'$ at $C'$. The two angles $\alpha$ and $\alpha'$ showing up at $C$ and $C'$ can be computed from the given data by spherical trigonometry. I leave that to the OP.

You will then see that the intersection of the two caps (also in cases not covered by the above figure) can be written as an algebraic (i.e. $\pm$) sum  of shapes of the following kind:
(a) Sectors $S$ of spherical caps of radius $\rho$, resp. $\rho'$, and central angle $\alpha>0$, resp. $\alpha'>0$. The area of such a sector is given by $$A_S=2\alpha\sin^2{\rho\over2}\ .$$
(b) Isosceles spherical triangles $\Delta$  with two sides equal to $\rho$ (or $\rho'$) and central angle $\alpha>0$. The base angle $\beta$ of such a triangle can be computed from $$\tan\beta={\cot(\alpha/2)\over\cos\rho}\ ;$$
then its  area $A_\Delta$  is given by
$$A_\Delta=\alpha+2\beta-\pi\ .$$
A: Here's a simplified formula as a function of your 3 variables, $a_1$, $a_2$, and $\theta$:
$$
2\cos(a_2)\arccos \left ( \frac{-\cos(a_1) + \cos(\theta)\cos(a_2)}{\sin(\theta)\sin(a_2)} \right ) \\
-2\cos(a_1)\arccos \left ( \frac{\cos(a_2) - \cos(\theta)\cos(a_1)}{\sin(\theta)\sin(a_1)} \right ) \\
-2\arccos \left ( \frac{-\cos(\theta) + \cos(a_1)\cos(a_2)}{\sin(a_1)\sin(a_2)} \right ) \\
-2\pi\cos(a_2)
$$
As previously stated, the caps must intersect and one cannot entirely contain the other.
This solution is copied from a graphics presentation by AMD, and is originally from a biochemistry paper. I have no proof that it works, so take it for what it's worth.
TOVCHIGRECHKO, A. AND VAKSER, I.A. 2001. How
common is the funnel-like energy landscape in protein-protein
interactions? Protein Sci. 10:1572-1583
A: You probably don't use the site anymore since the question was asked a year ago, but I'll post an answer anyway because it was a good problem and there's an ever so slight chance someone else will take a look at this.
Before I solve this, let me get some nomenclature/assumptions of my own out of the way:


*

*Instead of $\theta$ being the separating angle for the two spherical caps, let it be $\alpha$.

*When I refer to the central axis of a spherical cap (SC) I imply the axis which intersects the center of the SC's base and the center of our sphere.

*I'll assume none of the SCs are larger than hemispheres.

*I'll assume the SCs intersect.


Therefore, the farthest any point from the intersection of the two SCs from the center of the SC can only be $\Delta\phi = \pi/2$, using the spherical coordinate system.
The problem states that there is are two SCs with base radii $a_1$ and $a_2$ on a sphere of radius $r$ with an angle of $\alpha$ between the central axes of the two SCs.
As you mentioned, the angle between the SC's central axis and a radius from the center of the sphere to the circumference of the SC's base are $\Phi_1 = \arcsin(a_1/r)$ and $\Phi_2 = \arcsin(a_2/r)$.
Align the sphere such that SC-1's central axis and the $x$-axis are parallel (in the $\mathbb{R^3}$ coordinate system).  Translate the sphere to the origin.  Rotate the sphere such that the central axis of SC-2 is on the $xz$-plane.  Angle $\alpha$ now rises from the $x$-axis towards the $z$-axis.  Angle $\Phi_2$ is contained in $\alpha$ in the same plane, rising in the same direction (but not starting from the $x$-axis).  These actions are legal since they don't distort the shape or its surface area.
Now that somewhat rigidly defined our problem we can approach it.
Let the space curve which defines the intersection of the SC-2's base with the sphere be $\vec s_2(t)$. After some (read: a lot of) vector arithmetic and trigonometry, we arrive at $\vec s_2(t)$.
$$
\vec s_2(t) = \left\{\sqrt{r^2-a_2^2}\cos\alpha+r\cos(\alpha-\Phi_2)\cos t, a_2\sin t, z_2(t)\right\}
$$
for $0\le t\le 2\pi$.  This expression of $\vec s_2(t)$ allows us to project onto the $xy$-plane.  The projection of this base-sphere intersection looks is an ellipse.
The parametrization of the space curve $\vec s_1(t)$ is a bit easier because of the way we set our axes up.
$$
\vec s_1(t)=\left\{\sqrt{r^2-a_1^2},a_1\sin t, a_1\cos t\right\}
$$
When projected onto the $xy$-axis this looks like a line segment.  Using the two projected space curves, I can find the angular limits for $\theta$ in the spherical coordinate system for 3D by finding the intersection of the ellipse and the line segment.
$$
\theta_0=\arccos\frac{\sqrt{r^2-a_1^2}}{r^2-a_1^2+(a_2\sin\arccos\frac{\sqrt{r^2-a_1^2}-\sqrt{r^2-a_2^2}\cos\alpha}{r\cos(\alpha-\Phi_2)})^2}
$$
It's messy, but it's a constant at least!  We know that for the surface area we are interested in is for values of $\theta$ such that $-\theta_0\le \theta\le\theta_0$
Now we need to find the angular limits for $\phi$ in our spherical coordinate system.  These we know to change over $\theta$ since you can imagine that as we rotate our sphere along the $z$-axis the area of interest's width changes (the angle from the $z$-axis to the position vectors of the surface area we're looking at changes).
For the top limit of $\phi$, we look at the bottom spherical cap's surface along the sphere.  We use the parametrization I provided earlier.  Since $z_1(t) =  a_1cos(t)$ and $t$ is analogous to $\theta$ in my parametrization, I can use $\cos \phi = \frac{z}{r}$ to find that 
$$
\phi_1(\theta) = \arccos\frac{a_1\cos\theta}{r}
$$
Similarly, using 
$$
z_2(t)=r\sin(\alpha-\Phi_2)+a_2(1-\cos t)\sin(\arcsin(\frac{r}{a_2}\sin\Phi_2)+\alpha)
$$
which I didn't post at $\vec s_2(t)$ because of spatial concerns, I can find
$$
\phi_1(\theta)  = \arccos\frac{z_2(\theta)}{r}
$$
Thus, for the surface area $A$ of the intersection of two spherical caps $\Sigma$, 
$$
A=\int\int_\Sigma dS=\int_{-\theta_0}^{\theta_0}\int_{\phi_2(\theta)}^{\phi_1(\theta)}(r^2 \sin\phi) d\phi d\theta
$$
I'll be first to admit this integral is probably only numerically solvable, but I couldn't find any elegant geometric approaches to this, so I went with the "brute force" method.
EDIT
Thanks to Christian Blatter's answer to this question, I can answer in a more concise manner.
If I take our sphere situated on the origin as described before, and rotate it in the positive direction about $\hat j$ until $\alpha$ is centered on the $z$-axis, then I say:


*

*Project space curves $s_1(t)$ and $s_2(t)$ onto the $xy$-plane.

*Said curves must be ellipses or circles, since the extreme case $\alpha=\pi$ is the only one which yields straight lines upon projection and the answer to that scenario is that the surface area $A=0$.


If an ellipse is contained within another, then the surface area can be found using the formula for finding the external surface area of a spherical cap (see Christian's answer).
If the ellipses intersect at two points (they cannot intersect in this case in three or four), then using the answer to the aforementioned question, we can set up a surface integral for the calculation of $A$, given that the intersection of the ellipses on the $xy$-plane can be described by the two inequalities $a\le u\le b$ and $g(u)\le v\le f(u)$ such that $\left\{a,b:a,b\in\mathbb{R}\land a<b\right\}$, $f:\mathbb{R}\rightarrow\mathbb{R}$, $g:\mathbb{R}\rightarrow\mathbb{R}$, and $f(u)>g(u)$ over $u:[a, b]$.
Let $x=\frac{u}{\sqrt{r_x}}+c_x$ and $y=v$, for $r_x$ being the semi-diameter in the $x$ direction and $c_x$ being the horizontal shift from the origin to the center of the ellipse from the projection of $s_2(t)$ onto the $xy$-plane.
Let $\vec F=\{x, y,\sqrt{r^2-x^2-y^2}\}$, the position vector.
$$
A=\int\int_\Sigma dS=\int_{u=a}^b\int_{v=f(u)}^{g(u)}\left\|\frac{\partial\vec F}{\partial v}\times\frac{\partial\vec F}{\partial u}\right\|dvdu
$$
A: The next to last negative sign in the above formula should be a plus. The formula in the paper of Tovchigrechko and Vakser is right. The typo with the sign here comes from the presentation quoted above.
A: The formula for the area is complicated (see other answers), but it has a simple Legendre expansion
$$
A(\theta) = \sum_{l=0}^{\infty} a_l P_l(\cos\theta)
$$
with coefficients given by the associated Legendre functions,
$$
a_l = \pi \, (2l+1) \sin(\alpha_1) \sin(\alpha_2) \, P_l^{-1}(\cos\alpha_1) \, P_l^{-1}(\cos\alpha_2)
$$
which may be helpful.
