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Suppose one has an integral of the form $\int_{S_1^{d-1}} f(\phi(v)) d \text{vol}_{S_1^{d-1}}(v)$. Here $S_1^{d-1}\subset \mathbb{R}^d$ is the unit sphere. Let $B_1^{d-1}\subset\mathbb{R}^{d-1}$ be the unit ball in $\mathbb{R}^{d-1}$. And $\phi$ maps a vector $v=(v_1, \dots , v_d)\in S_1^{d-1}$ to $z=(v_2, \dots , v_d)\in B_1^{d-1}$.

Can someone please provide me with an explicit formula for $g$ with $$ \int_{S_1^{d-1}} f(\phi(v)) d \text{vol}_{S_1^{d-1}}(v) = \int_{B_1^{d-1}} g(z) d \text{vol}_{{d-1}}(z). $$

Thank you.

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Do you mean $\displaystyle d\text{vol}_{B_1^{d-1}}(z)$ in the last line? Is $z=(v_2,v_3,...)$ the same values as for $v$ except $v_1$ or could it be $z=(z_1,z_2,...,z_{d-1} )$? –  draks ... Apr 23 '12 at 14:20
    
Yes, $d\text{vol}_{B_1^{d-1}} = d \text{vol}_{d-1}$ as far as I'm concerned (i'm not very correct when it comes to such things), and $\phi$ is merely the projection onto the last $d-1$ coordinates together with the embedding into $\R^{d-1}$. –  fk2012 Apr 23 '12 at 14:36
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1 Answer 1

If I understand you correctly you are given a function $f:\ B^{d-1}\to{\mathbb R}$ and want to calculate the integral $$I:=\int_{S^{d-1}} f\bigl(\phi(x)\bigr){\rm d}\omega(x)\ ,$$ where ${\rm d}\omega$ denotes the $(d-1)$dimensional euclidean surface element on the unit sphere $S^{d-1}$ and $\phi:\ x=(x_1,\ldots,x_d)\mapsto (x_1,\ldots,x_{d-1})=:x'$ denotes the projection of ${\rm R}^d$ onto the ${\rm R}^{d-1}$ coordinate plane.

It follows that we have a map between two $(d-1)$-dimensional surfaces, and we have to determine the local stretching factor. In the case of an orthogonal projection this factor is $\cos\alpha$, where $\alpha$ denotes the angle between the two normals. In other words, we have $${\rm dvol}(x')=\cos\alpha\ {\rm d}\omega(x)=\sqrt{1-|x'|^2}\ {\rm d}\omega(x)\ .$$ Therefore $$I=2\int_{B^{d-1}} f(x'){1\over\sqrt{1-|x'|^2}}\ {\rm dvol}(x')\ .$$ The factor $2$ comes from the fact most points $x'\in B^{d-1}$ have two preimages on $S^{d-1}$.

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