Let $S$ be an $n$--sphere $\mathcal{S}^n(0,R)\subset\mathbb{R}^{n+1}$, $\Theta\subset\mathbb{R}^n$ an open subset and let $\phi:\Theta\subset\mathbb{R}^n\longrightarrow S\subset\mathbb{R}^{n+1}$ be a diffeomorphism onto "most of the sphere", i.e. $S\backslash\phi(\Theta)$ is of spherical measure zero. I'm talking about measure described in this article


Let's also assume that this spherical measure is scaled in such a way that measure of $S$ is equal to its "surface area" or an $n$-dimensional volume.

We can integrate a real function $f:S\longrightarrow \mathbb{R}$ over $\phi(\Theta)$ which is an $n$-dimensional manifold using a formula

$$\int_{\phi(\Theta)}f=\int_{\Theta}f(\phi(\theta))\sqrt{\big|\det D\phi^TD\phi\big|}\mathrm{d}l^n(\theta)$$

where $l^n$ is a Lebesgue measure in $\mathbb{R}^n$. (Is this correct? The square root of the determinant of Gram matrix of $D\phi$ serves the same role as the usual Jacobian determinant in case when the codomain is of higher dimension, I think.) I'd like to know if the integral of $f$ over $S$ with respect to spherical measure is equal to the above integral, that is if


where $s^n$ is the spherical measure (appropriately scaled, like we assumed) and the rightmost integral is the integral defined above (the integral over a manifold).


"Yes, they are the same up to overall scale." If $\phi$ is a regular parametrization of the sphere $S^{n}$, the round metric on the sphere pulls back to $$ g(X, Y)_{p} = D\phi(p)X \cdot D\phi(p)Y = X \cdot D\phi(p)^{T} D\phi(p)Y, $$ whose volume form is, as you say, $\sqrt{\left|\det D\phi^{T} D\phi\right|}$ times Lebesgue measure in Euclidean $n$-space. If the image of $\phi$ has measure-zero complement, the integral of a measurable function $f$ over the image of $\phi$ is equal to the integral over $S^{n}$.

As you note, you need to take care with overall scale: Most mathematicians take "the sphere" to be the unit sphere $S^{n}$ centered at the origin of Cartesian $(n+1)$-dimensional space, whose $n$-dimensional volume is $$ \frac{2\pi^{\frac{n+1}{2}}}{\Gamma(\frac{n+1}{2})}. $$


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