Question: Let $\mathcal S^2:=\{x\in \mathbb R^3: \;\vert \vert x \vert \vert=1 \}$ and assume $f$ is a function of one variable, i.e $f(x)=f(x\_1)$. I'm trying to understand the following formula:

> 
> $\int\_{\mathcal S^2} f(x) \;d\mathcal H^2(x)=2\pi \int\_{-1}^1
>  f(x\_1)\:dx\_1$
> 
> 
> 

It seems to me that polar coordinates are used here, however we are in a 3-dimensional space. 

Is been a really long time since I encountered this kind of calculus and I feel I'm forgetting something essential here. I would appreciate any help in explaining why this formula holds true.

Thanks in advance

Answer: 
This result makes use of the interesting fact that in $3$ dimensions (and in $3$ dimensions only) slices of a sphere of equal width have equal surface area. So if you integrate a function of one Cartesian coordinate over the surface of the sphere, you can just integrate it over that coordinate, since slices of equal infinitesimal width $\mathrm dx\_1$ have equal infinitesimal surface area (which here seems to be denoted by $\mathrm d\mathcal H^2$). You can then derive the factor $2\pi$ from the fact that integrating over $f(x\_1)\equiv1$ must yield the surface area of the sphere, $4\pi$.
Set the $x\_3$ axis to be the variable that $f$ depends on. We can use the known Jacobian from parametrising a sphere in angular [spherical coordinates](https://en.wikipedia.org/wiki/Spherical_coordinate_system) (setting the radial coordinate to $1$). We get the integral

$$\int\_0^\pi \int\_0^{2\pi}f(x\_3)\sin\theta \:d\phi \:d\theta = \int\_0^\pi \int\_0^{2\pi}f(\cos\theta)\sin\theta \:d\phi \:d\theta = 2\pi\int\_0^\pi f(\cos\theta)\sin\theta \:d\theta$$

Then use the substitution $x=\cos\theta$ to get the final expression:

$$2\pi\int\_{-1}^1 f(x)\:dx$$

This works because we chose to rotate our coordinate system such that the function was not dependent on the $2\pi$ angular variable.