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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 $z=(z_1,z_2,...,z_{d-1} )$?2012-04-23
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    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 $\R^{d-1}$.2012-04-23

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