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The polar coordinates of point $x \in \mathbb{R} \setminus \{0\}$ are pairs $(r,\gamma)$, where $0 < r < \infty$ and $\gamma \in S^{d-1} = \{x \in \mathbb{R}^{d}\mid |x| = 1\}$. These are determined by $$r = |x|, \quad\gamma = x/|x|,$$ and reciprocally by $x = r\gamma$.

Then we have: $$\int_{\mathbb{R}^{d}}f(x)dx = \int_{S^{d-1}} \left( \int_{0}^{\infty}f(r\gamma)r^{d-1}dr \right) d\sigma(\gamma).$$

The proof of this formula using the Fubini's theorem. However, I can't understand that the relationship between those measure spaces. And I was so confused by this equation: $$\mu_{1}(E) = \int_{E}r^{d-1}dr.$$

Besides, I also want to know about the integral formula for the general case. Is there a universal steps to construct the integral formula with respect to the Jacobi in the Reimann integral.

Siminore
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Eric
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  • What is $E$? What is $\mu_1$? – Siminore May 23 '14 at 12:11
  • A measure space $(X_{1}, M_{1}, \mu_{1})$, where $X_{1}=(0,\infty)$. $M_{1}$ is the collection of Lebesgue measure sets in (0,$\infty$). E is a set belongs to $M_{1}$.@Siminore – Eric May 23 '14 at 12:19
  • It seems you are missing the point: first you define the measure $\sigma$ on the sphere, then you prove the integration formula. – Siminore May 23 '14 at 12:35
  • Yes, you are right. First define two measure spaces: (X1, M1, $\mu$1) and (X2, M2, $\mu$2). My problem is that why can we calculate the measure: $$d\mu_{1}(r) = r^{d-1}dr.$$ – Eric May 23 '14 at 13:52
  • Roughly speaking, the volume element of $\mathbb{R}^d$ becomes the product of the surface element of the sphere times the height $dr$. Locally, the surface element behaves like a circle in dimension $d-1$, so the (infinitesimal) volume in polar coordinates become $r^{d-1}dr d\sigma$. Of course this can be made rigorous by the co-area formula. – Siminore May 23 '14 at 15:53

1 Answers1

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I recently discovered a very nice treatment of spherical integration in the paper Integration over spheres and the Divergence Theorem by John A. Baker (American mathematical monthly 104 (1997), 36-47). The story goes as follows. Let $n \geq 2$ and $g \colon S^{n-1} \to \mathbb{R}$ be continuous. Define $\hat{g} \colon \mathbb{R}^n \to \mathbb{R}$ by $$ \hat{g}(x)= \begin{cases} g(|x|^{-1}x) &\hbox{if $x \neq 0$} \\ 0 &\hbox{if $x=0$}. \end{cases} $$ Now define $$ \int_{S^{n-1}} g\, d\sigma_{n-1} = n\int_{B(0,n)} \hat{g}(x)\, dx. $$ The following result can be proved ($B(a,b)$ is the spherical shell with radii $a$ and $b$).

Theorem. Suppose $0 \leq a < b$ and $f \colon B(a,b) \to \mathbb{R}$ is continuous. Then $$ \int_{a \leq |x| \leq b} f(x)\, dx = \int_a^b r^{n-1} \left( \int_{S^{n-1}} f(rs)\, d\sigma_{n-1}(s)\right)dr. $$

The proof is very nice, and uses the differentiability properties of the map $$ \varphi(r) = \int_{a \leq |x| \leq r} f(x)\, dx, $$ since it turns out that $$ \frac{d\varphi}{dr} = r^{n-1} \int_{S^{n-1}} f(rs)\, d\sigma_{n-1}(s). $$

Siminore
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  • Yes, I got your proof. It's very beautiful. – Eric May 24 '14 at 02:24
  • I think we can get the theory about steps for construct the integral formula for any coordinates. Actually, we have the Jacobi in the Riemann integral, but if the x is not differential for the new variables, we may need the Fubini's theorem to help us. – Eric May 24 '14 at 02:40