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 I`m 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 I`m 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 $\vec{n}(t)$.


1 Answer 1


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 $\Sigma.$

This concept also holds for differentiable curves, which can be shown by means of calculus. We only have to find out a formula for the "felt total curvature".

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.

  • $\begingroup$ Thank you so much, I'm totally rust on multivariable calculus and parametrization but I'll follow your links to follow each step until the demonstration feels natural to me. $\endgroup$
    – PedroDM
    Commented Apr 25, 2020 at 12:57

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