Given a polygon $P$, with geodesic edges, on the surface of the unit sphere in $\mathbb R^3$, what is the integral of the unit normal vector $\hat n$ over the polygon's area? (The normal vector is simply the position vector, or its opposite.)

$$\iint_P \hat n\,dA$$

I know that any polygon can be broken up into triangles, and any triangle can be split into 2 right triangles. So it suffices to integrate the normal vector over an arbitrary right triangle.

I've calculated the integral over a half-lune $H$, which is a triangle with 2 right angles. If the sphere is parametrized by $\vec x = (\cos\theta\sin\phi)\hat e_1 + (\sin\theta\sin\phi)\hat e_2 + (\cos\phi)\hat e_3$, and two vertices are at the equator ($\phi = \frac\pi 2$) and the other at the north pole ($\phi = 0$), then

$$\iint_H \hat n\,dA = \frac\pi 4(\sin\theta_2 - \sin\theta_1)\hat e_1 - \frac\pi 4(\cos\theta_2 - \cos\theta_1)\hat e_2 + \frac12(\theta_2 - \theta_1)\hat e_3$$

I'm still trying to figure out the geometric meaning of this, and I don't know what to do when the triangle has only 1 right angle.


Aha! I see that the half-lune's integral is half the sum of products of edge lengths with the unit vector that is tangent to the sphere and normal to the edge, pointing into the triangle.

$$\iint_H \hat n\,dA = \frac12(L_1\hat N_1 + L_2\hat N_2 + L_3\hat N_3)$$

I don't know if this generalizes, but it looks like it should.

  • $\begingroup$ do you mean polyhedron $\endgroup$ Feb 25, 2018 at 21:33
  • 1
    $\begingroup$ @phdmba7of12 -- No, a polygon. A spherical polyhedron would just be the whole sphere, with lines drawn on it. The average normal vector over a sphere is zero. $\endgroup$
    – mr_e_man
    Feb 25, 2018 at 21:36

1 Answer 1


This is related to the Fundamental Theorem of Geometric Calculus!

Hestenes' Tutorial (page 11) describes it as

$$\int_M (d^m\mathbf x\,\partial_\mathbf x F) = \pm\int_{\partial M} (d^{m-1}\mathbf x\,F)$$

where $M$ is a manifold, $m$ is its dimension ($2$ for a surface), $\partial M$ is its boundary, $F$ is some multivector function, $\partial_\mathbf x$ is the vector derivative on the manifold, and $d^m\mathbf x$ is a directed area/volume element. (I put the $\pm$ sign there; I think it should be negative for the special case of Stokes' Theorem.)

For a surface $S$, define $\vec H = H\hat n$ as the mean curvature vector, and $i$ as the surface's unit tangent bivector (which is orthogonal to $\hat n$). It can be shown that

$$i\,\partial_\mathbf xi = 2\vec H$$

which Hestenes calls the "spur". The directed area element is $i\,dA$, so, using the Fundamental Theorem,

$$\int_S 2\vec H\,dA = \int_S (i\,dA\,\partial_\mathbf x i) = \pm\int_{\partial S} (d\mathbf x\,i) = \pm\int_{\partial S} (\hat T\,i)ds$$

where $\frac{d\mathbf x}{ds} = \hat T$ is the unit tangent vector for the boundary curve. The bivector $i$ is the product of the tangent and the normal to the curve (which are both tangent to the surface): $i = \hat T\hat N$, so

$$\hat T i = \hat N$$ $$\int_S 2\vec H\,dA = \pm\int_{\partial S} \hat N\,ds$$

This is quite a general result, comparable to the Gauss-Bonnet Theorem!

$$\int_S K\,dA = 2\pi\chi - \int_{\partial S} \left(\frac{d^2\mathbb x}{ds^2}\cdot\hat N\right)ds$$

We can apply this result to our polygon problem: a unit sphere's mean curvature is $|H| = 1$, $\vec H = -\hat n$, so

$$\int_P \hat n\,dA = -\frac12\int_P 2\vec H\,dA = \mp\frac12\int_{\partial P} \hat N\,ds$$

The normal vector is constant on each edge of the polygon (it's parallel to the great circle's axis), and the edge's length is $L = \int ds$, so this becomes

$$\int_P \hat n\,dA = \mp\frac12\sum_j L_j\hat N_j$$

as suspected! (The lower sign should be taken, to agree with the OP special case.)


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