Area enclosed by loop on $S^2$ Let $l:S^1\to S^2$ be a simple closed loop on $S^2$. How do you calculate the area enclosed by this curve (Up to exchange of which "cap"* you choose?)


*

*I wrote "cap" because the cap defined by the loop may not be a proper cap, because the loop may not necessarily be a circle on the sphere.


EDIT
To make it easier, let's say you have a closed simple curve in $\mathbb{R}^2$, then instead of doing $Area=\int\int_{inside\,curve} dx dy$ you would want something like $\int y(x) dx$ where you assume you can express the curve as a function $y(x)$ (if that's not possible perhaps one needs to make several divisions of horizontal lines). 
Now I'm looking for the analogue of that, moving from $$ \int_{\varphi,\theta\,in\,curve} \sin(\theta)d\theta d\varphi $$ to $$ \int_{\varphi=0}^{\varphi=2\pi} (1-\cos(\theta_l(\varphi))) d\varphi $$ where we assume that $l:S^1->S^2$ defines a function $\theta_l(\varphi)$ which is not always true if the curve bends... but perhaps for starters we can assume that $l:S^1->S^2$ indeed defines a function $\theta_l(\varphi)$.
 A: Because the curvature of the sphere is constant, provided your function is smooth, the area is given by the Gauss-Bonnet theorem:
$$ A(M) = 2\pi \chi(M) - \int_{\partial M} k_g \, ds $$
(notice the Wikipedia article mentions that the Euler characteristic of the closed disc is 1; I assume this generalises to non-simple curves). So essentially what you have to work out is $k_g$ for a curve on the sphere, which is given by the projection of the tangent vector onto the normal vector, most simply expressed by this formula:
$$ k_g = \frac{\gamma''(t) \cdot (n \times \gamma'(t))}{\lVert \gamma'(t) \rVert^3}, $$
which you can compute explicitly using the usual coordinates on the sphere. You also have
$$ ds = \frac{dt}{\lVert \gamma'(t) \rVert} $$
if you're not using a unit parameter, of course.

Edit:
As a trivial example/check, a great circle has $k_g=0$ (being a geodesic, for example), and so the cap has area $2\pi$, as expected from it being half the sphere. For a simple closed curve, the opposite cap has area
$$ 4\pi - A = 2\pi + \int_{\partial M} k_g \, ds, $$
so the side of the curve you choose is basically dependent on which sign the curvature has, i.e. which way your normal vector points.
More explicitly, we have, suppressing the explicit $t$s everywhere
$$ \gamma = ( \cos{\phi} \sin{\theta}, \sin{\phi}\sin{\theta}, \cos{\theta} ) = e_r, $$
and the relations between the vectors of the orthonormal basis $(e_r,e_{\theta},e_{\phi})$
$$ e_r' = \theta' e_{\theta} + \phi' \sin{\theta} \, e_{\phi} \\
e_{\theta}' = -\theta' e_r + \phi'\cos{\theta} \, e_{\phi} \\
e_{\phi}' = -\phi'\sin{\theta} \, e_r - \phi' \cos{\theta} \, e_{\theta}
$$
give
$$ \gamma' = e_r' = \theta' e_{\theta} + \phi' \sin{\theta} \,  e_{\phi}, $$
from which we have the familiar
$$ \lVert \gamma' \rVert^2 = \theta'^2 + \phi'^2 \sin^2{\theta}, $$
and taking another derivative gives
$$\begin{align*}
\gamma'' &= \theta'' e_{\theta} + \theta' e_{\theta}' + (\phi'' \sin{\theta}  +\phi' \theta' \cos{\theta}) e_{\phi} + \phi' \sin{\theta} \, e_{\phi}' \\
&= \theta'' e_{\theta} + \theta' ( -\theta' e_r + \phi'\cos{\theta} \, e_{\phi}) + (\phi'' \sin{\theta}  +\phi' \theta' \cos{\theta}) e_{\phi} + \phi' \sin{\theta} (-\phi'\sin{\theta} \, e_r - \phi' \cos{\theta} \, e_{\theta}) \\
&= -(\theta'^2 + \phi'^2\sin^2{\theta} ) e_r + (\theta''- \phi'^2 \sin{\theta} \cos{\theta} ) e_{\theta} + (\phi'' \sin{\theta} + 2\theta'\phi' \cos{\theta} ) e_{\phi}  
\end{align*}$$
Having done the nasty differentiation, we now do some dull vector algebra using the relations
$$ e_r \times e_{\theta} = e_{\phi}, \quad e_{\theta} \times e_{\phi} = e_r, \quad e_{\phi} \times e_r = e_{\theta}; $$
in our case, $n = e_r$, so
$$ n \times \gamma' = \theta' e_{\phi} - \phi' \sin{\theta} \, e_{\theta}, $$
and then dotting gives
$$ \gamma'' \cdot (n \times \gamma') = (\phi'' \sin{\theta} + 2\theta'\phi' \cos{\theta})\theta'-( \theta''- \phi'^2 \sin{\theta} \cos{\theta} )\phi'\sin{\theta} $$
It doesn't really simplify beyond this, but now we can check another example: the circle $\theta=a$. Parametrising it as $\phi=t$, $0<t<2\pi$ we have
$$ \lVert \gamma' \rVert^2 = 0 + \sin^2{a}, $$
and the only nonzero derivative is $\phi'=1$. Therefore the integral is
$$ \int_0^{2\pi} \frac{1^3 \cos{a} \sin^2{a}}{\sin^2{a}} dt = 2\pi \cos{a}, $$
so we obtain the result
$$ A = 2\pi(1-\cos{a}), $$
which we again recognise from more elementary results.
