When is $\int_{\vec{x} \in X} \langle \vec{x}, \vec{v} \rangle$ where $X$ is a closed path in the unit sphere zero?
I was about thinking equations for great circles in different coordinate systems and came across a nice characterization in $\mathbb{R}^3$.
Let $S$ be the unit sphere.
Suppose $X$ is a closed path in $S$.
I claim that $X$ is a great circle if and only if there exists some $\vec{v}$ such that $\langle \vec{x}, \vec{v} \rangle = 0$ for all $\vec{x}$ in $X$.
As proof, a great circle is an intersection of a plane with $S$. If $\vec{n}$ is the normal vector associated with the plane, then the line generated by $\vec{n}$ intersects $S$ in exactly two places.
When $X$ is a great circle, then $\int_{\vec{x} \in X} \langle \vec{x}, \vec{v} \rangle$ would be $0$ because the value of each component is zero. (Also, if people don't like this integral, I think I can paraphrase it away using an arc-length parameterization of $X$.)
I'm curious under what other circumstances the value of that integral can be zero.
By linearity of integration one has \begin{equation} \int_{\vec{x} \in X} \langle \vec{x}, \vec{v} \rangle = \left \langle \vec{v}, \int_{\vec{x} \in X} \vec{x} \right \rangle = \langle \vec{v}, \vec{x}_I\rangle \end{equation} where $\vec{x}_I = \int_{\vec{x} \in X} \vec{x}$. From the last expression we see that the integral is $0$ if and only if $\vec{x}_I = \vec{0}$ or is orthogonal to $\vec{v}$.
If you think of the sphere in $\mathbb{R}^3$, and choose $v$ to be pointing to the north pole, then all of the rings of latitude will also integrate to 0.
There are a lot of other things where the integral is 0, but it's not clear to me that there are any nicer conditions than saying the integrated vector needs to be 0 or orthogonal to $\vec{v}$.