Divergence Theorem I'm looking for an alternate proof of a result. Let $A,B,C\in\mathbb{S}^d$. Let $T$ be the intersection of $\mathbb{S}^d$ with the cone generated by $A,B,C$. Then, $T$ is a spherical triangle. The vertices $A,B,C$ are opposite arcs with lengths $a,b,c$. Call $[T]$ the area of $T$. The centroid of $T$ is
$$g:=\frac{1}{[T]}\int_T xd\mu$$
where $\mu$ is uniform on the sphere. JE Brock found the centroid of a spherical triangle $T=\triangle ABC$ to be (paraphrased, but it checks out for me)
$$g=\frac{1}{2[T]}\left(\frac{A\times B}{|A\times B|}c+\frac{B\times C}{|B\times C|}a+\frac{C\times A}{|C\times A|}b\right)$$
Thinking of $A\times B/|A\times B|$ as the unit vector perpendicular to side $c$, we can write this as
$$\int_T xd\mu=g*[T]=\frac{1}{2}\int_{\partial T} \vec{n} ds$$
where $\vec{n}$ is the inward pointing unit vector.
This formulation looks like the divergence theorem. So, my question is how to prove it with the divergence or Stokes Theorem.
$$\int_T\mathrm{div}F=\int_{\partial T} F\cdot\vec{n}$$
I've tried separating components setting the vector field $F$ to $(x_1^2,0,\dots,0)$ to get a linear term in the divergence, but that doesn't match the right side where I expect $F=(1,0,\dots,0)$ to get just the normal around the boundary.
So, I suspect I need to use Stoke's Theorem on manifolds. The form can be the unit tangent so that the derivative points toward the center of the sphere. That would fit the form I'm looking for, I just need some details.
Does that seem like the right course? Thanks!
Happy Holidays!
 A: You can prove this centroid formula using Stokes' theorem of vector calculus.
First note that using standard identities (e.g. from Wikipedia) we can find the relation
$$
\nabla \times \left(\frac{1}{2} \vec c \times \vec r\right) = \vec c,
$$
for a constant vector $\vec c$ and $\vec r = (x,y,z)$.
Inserting this particular vector field into Stokes' theorem yields
$$
\vec c \cdot \int d\vec a = \int \nabla \times \left(\frac{1}{2} \vec c \times \vec r\right) \cdot d\vec a = \frac{1}{2}\int ( \vec c \times \vec r)\cdot d\vec l = \frac{1}{2}\vec c \cdot \int \vec r \times d\vec l,
$$
and as $\vec c$ is arbitrary, we can conclude
$$
\int d\vec a = \frac{1}{2}\int \vec r \times d\vec l.
$$
The integral on the LHS is the quantity you are interested in computing as $d\vec a = \hat n da$. Thus we can equivalently compute it by integrating the RHS over the three side of the triangle. This is easy as $\vec r$ has unit length on the sphere and is perpendicular to $d\vec l$. That is the integrand is constant and we find
$$
\int_{A\rightarrow B} \vec r \times d\vec l = \frac{\vec A\times \vec B}{|\vec A\times \vec B|} \int _{A\rightarrow B} dl,
$$
where the integral in your notation equal $c$. Adding up the three sides gives the final answer.
