# Area inside a curve on the surface of a sphere

Suppose there is a surface $$\Sigma$$ on the surface of a sphere with unit radius, the surface is bounded by a curve $$\Gamma$$.

The curve is closed and has no wiggles (sorry, im a physicist I forgot the correct term, is it simply connected?, idk).

The curve is parametrized by a vector $$\vec{n}(t)$$ with the parameter $$t$$ (define the range as you want).

The grey area is what Im interested

How can I calculate the area inside the curve on the surface of the sphere using the parametrization of the vector $$\vec{n}(t)$$?

What Im basically asking is the area of $$\Sigma$$ but in terms of the parametrization of the curve that bounds this surface, knowing that $$\Sigma$$ resides in the surface of a sphere of unit radius, the information is enough to find a formula for this situation.

I`m looking for the formula and its complete and detailed demonstration. The answer must be in terms of the $$\vec{n}(t)$$.

First imagine that $$\Sigma$$ is a spherical polygon. The area oft this polygon is given by $$A = \left(\sum\limits_{n=1}^{N}\alpha_n\right)-(N-2)\pi$$ where $$\alpha_n$$ are the interior angles of the polygon, see Spherical trigonometry
Let $$\beta_n$$ be the angle that you have to turn in the $$n$$-th vertex of the polygon, if you are a 2D creature crawling on the surface of the sphere along the boundary of the polygon. Then $$\alpha_n = \pi-\beta_n.$$ The area $$A$$ can be rewritten as $$A = \left(\sum\limits_{n=1}^{N}(\pi-\beta_n)\right)-(N-2)\pi \;=\; 2\pi \;-\; \left(\sum\limits_{n=1}^{N}\beta_n\right)$$ So the area of $$\Sigma$$ only depends on the "felt" total curvature of the polygon, which is the accumulated rotation a 2D creature would think it has accomplished after a complete walk on the boundary of $$\Sigma.$$
In $$\mathbb{R}^2,$$ the total curvature is simply the integral of the signed curvature $$k(t).$$ For a curve $$\gamma : [a,b] \rightarrow \mathbb{R}^2$$, we have $$k(t) = \det\big(\dot{\gamma}(t),\, \ddot{\gamma}(t)\big)$$ if the curve is parameterized by its length, see curvature
On the surface of the sphere, we have to project $$\dot{\vec{n}}(t)$$ and $$\ddot{\vec{n}}(t)$$ into the plane that is tangent to the sphere in the current point of the curve, which is the plane that is perpendicular to $$\vec{n}(t).$$ Therefore, we get $$k(t) = \det\big(\vec{n}(t),\,\dot{\vec{n}}(t),\,\ddot{\vec{n}}(t)\big)$$ given that $$\vec{n}$$ is parameterized by its length. So the overall solution is $$A = 2\pi - \int\limits_a^b \det\big(\vec{n}(t),\,\dot{\vec{n}}(t),\,\ddot{\vec{n}}(t)\big) \,dt$$ where $$\vec{n}$$ is parameterized by its length.