Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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 $S^{d-1}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.