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Suppose we have positive density $q$ with "good" qualities (continuity, etc..). I need to calculate this integral: $$\int_B q(\textbf{z}) d \textbf{z},\ \textbf{z} \in \mathbb{R}^d,$$ where $B \subset \mathbb{R}^d$ is a Borel set that is bounded away from the origin. In order to do that it is necessary (in my case) to make a d-dimensional "polar coordinate" substitution, where $r=||\textbf{z}||$ ($||\cdot||$ is $L_2$ norm) and $\textbf{w}=\textbf{z}/r$, respectively are "radius" and "direction". $$\int_B q(r\textbf{w}) r^{d-1} drd\lambda(\textbf{w}),$$ where $\lambda(\textbf{w})$ is Hausdorff measure (I assume the "surface area" on d-dimensional hypersphere). The question is how do I prove that Jacobian is $r^{d-1}$? I found that this is the right Jacobian for this transform in this article p. 1818 and also intuitively the idea looks fine, but I could not find a mathematical proof anywhere. I know how to find Jacobian in the simple cases like d=2, or d=3, but I don't really know how to deal with this Hausdorff measure.

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    $\begingroup$ It's not really trivial. You may proceed by induction on the dimension. But the cleanest proof is by the co-area formula. $\endgroup$
    – Siminore
    Commented May 8, 2012 at 13:41
  • $\begingroup$ You can think as it being the Riemann integral since you q is continuous. If q is positive and symmetric radial you may use the distribution function $\omega(\alpha)=|{|q|>\alpha}|$ and integrate it from zero to infinity to obtain the result desired! $\endgroup$
    – checkmath
    Commented May 8, 2012 at 13:46

2 Answers 2

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What we have before us is an instance of "Fubini" in disguise.

I assume your $B$ is a ball of radius $R$ centered at ${\bf 0}\in{\mathbb R}^d$. Since the integrand $q(\cdot)$ is well behaved we can use an approximation by Riemann sums in order to arrive at the correct formula.

Partition the interval $[0,R]$ into $M$ subintervals $I_i:=[r_{i-1},r_i]$. Likewise, partition $S^{d-1}$ into $N$ essentially disjoint patches $\Omega_k$, and for each $k\in[N]$ choose a sampling point $\omega_k\in\Omega_k$.

In this way you obtain a partition of $B$ into $MN$ essentially disjoint pieces $$B_{ik}:=\{{\bf x}=r \omega\,|\, r\in I_i, \ \omega\in\Omega_k\}\ .$$ The $d$-dimensional volume of $B_{ik}$ is given by (this is elementary geometry) $$\eqalign{\mu(B_{ik})&={1\over d}(r_i)^{d-1}\lambda(\Omega_k)\cdot r_i -{1\over d}(r_{i-1})^{d-1}\lambda(\Omega_k)\cdot r_{i-1}\cr &={1\over d}(r_i^d-r_{i-1}^d)\lambda(\Omega_k)=\rho_i^{d-1}(r_i-r_{i-1})\lambda(\Omega_k)\ ,\cr}$$ where a suitable value $\rho_i\in I_i$ is guaranteed by the MVT of differential calculus. In particular $\rho_i\,\omega_k\in B_{ik}$ for all $i$ and $k$..

Now we approximate our integral as a Riemann sum: $$\eqalign{\int_B q({\bf x})\,{\rm d}({\bf x})&\doteq \sum_{i,\, k} q(\rho_i \omega_k)\ \mu(B_{ik})=\sum_{i,\, k} q(\rho_i \omega_k)\ \rho_i^{d-1}(r_i-r_{i-1})\lambda(\Omega_k)\cr &=\sum_k\left(\sum_i q(\rho_i \omega_k)\ \rho_i^{d-1}(r_i-r_{i-1})\right)\lambda(\Omega_k)\cr &\doteq\sum_k\left(\int_0^R q(r \omega_k) \, r^{d-1} dr\right)\lambda(\Omega_k)\cr &\doteq\int_{S^{d-1}} \int_0^R q(r\,\omega)\, r^{d-1}\ dr \ {\rm d}\lambda(\omega)\ .\cr}$$ Since the used approximations can be made as precise as desired, the integrals on the LHS and the RHS of this chain have to be of equal value.

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  • $\begingroup$ Although, I forgot to mention that B is a Borel set that is bounded away from the origin and not necessarily a ball set, I see the point in your proof. $\endgroup$
    – jem77bfp
    Commented Sep 3, 2012 at 11:56
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Easiest way is to use really simple idea - two dimensional polar coordinates. Vector $\mathbf{z}$ can be mapped to vector $(r \cos \phi, r \sin \phi, rw_3, rw_4, \dots, rw_d)$, where $\phi$ and $r$ are such that $w_1=r \cos \phi$ and $w_2=r \sin \phi$. Then the Jacobian is $$ \left| \begin{array}{ccccc} \frac{\partial r \cos \phi}{\partial r} & \frac{\partial r \cos \phi}{\partial \phi} & \frac{\partial r \cos \phi}{\partial w_3} & \cdots & \frac{\partial r \cos \phi}{\partial w_d} \\ \frac{\partial r \sin \phi}{\partial r} & \frac{\partial r \sin \phi}{\partial \phi} & \frac{\partial r \sin \phi}{\partial w_3} & \cdots & \frac{\partial r \sin \phi}{\partial w_d} \\ \frac{\partial r w_3}{\partial r} & \frac{\partial r w_3}{\partial \phi} & \frac{\partial r w_3}{\partial w_3} & \cdots & \frac{\partial r w_3}{\partial w_d} \\ \vdots & \vdots &\vdots & \ddots & \vdots \\ \frac{\partial r w_d}{\partial r} & \frac{\partial r w_d}{\partial \phi} & \frac{\partial r w_d}{\partial w_3} & \cdots & \frac{\partial r w_d}{\partial w_d} \end{array} \right| . $$ Then by taking derivatives and evaluating determinant we get that Jacobian equals $r^{d-1}$

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