Let $K:=\bar{{B_{r_0}^3}}$ be the ball in $\Bbb R^3$ around zero with radius $r_0$. Let $\rho: K \to \Bbb R_+^0$ be symmetric under rotation, i. e. $\rho (x) =f(\mid x\mid)$. Let $y \in \Bbb R^3 \backslash K $.
I have to show that $$U(y):=\int _K{\frac{\rho(x)}{\mid x-y \mid }dx}=\frac{\int_K {\rho(x)dx}}{\mid y\mid}$$
I had the idea of transforming into sherical coordinates, to make the expression $\mid x \mid$ easier to integrate but i don't know how to handle the $\mid x-y \mid$, meaning i don't know how to pull out the $y$ from the integral.
Any tipps or ideas? Thanks in advance!
Show: $U(y):=\int {\frac{f(\mid x \mid)}{\mid x-y \mid }dx}=\frac{\int {f(\mid x \mid)dx}}{\mid y\mid}$
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real-analysis
integration
change-of-variable
1 Answers
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We can recognize the identity, it's related to the well known fact in physics that the potential due to a spherically symmetric distribution of mass in a point out of the distribution it's the same as if the total mass were in the center of the distribution (Newton Proposition LXXI for the totally equivalent statment about forces). We interpret $f(\mid x\mid)$ as being the spherically symmetric density (mass per unit of volume) with center at the origin and $\mid x-y\mid$ as the distance from the point at $y$ to the center of the distribution. So $\frac{f(\mid x\mid)}{\mid x-y\mid}$ is the potential at $y$ due to the infinitesimal mass at $x$ and $\int_K f(\mid x\mid)$ is the total mass of the distribution. Anyway, ti's nice to try to prove the identity "as is".
To prove it the idea is, as you think, to write the expressions in spherical coordinates. First of all, thanks to the problem's symmetry, without loss of generallity we can consider $y$ placed somewhere along the only ray with all its points having coordinate $\theta=0$: if the identity holds for a point in that line, it holds for any point at the same distance from the origin than $y$. Symbolically, we choose: $y=(0,0,R)$ with $R>r_0$
Here I assume that $dx=dx_1dx_2dx_3$ is the volume element. In spherical coordinates, $r^2sin\theta d\theta d\phi dr$
It's easy to see with a simple drawing that $\mid x-y \mid=\sqrt{r^2+R^2-2rR\cos\theta}$
$$I=\int _K{\frac{f(\mid x \mid)}{\mid x-y \mid }dx}=\int_0^{r_0}\int_{0}^{2\pi}\int_{R-r}^{R+r}\frac{f(r)}{\sqrt{r^2+R^2-2rR\cos\theta}}r^2\sin\theta d\theta d\phi dr$$
Making the change of variable $s^2=r^2+R^2-2rR\cos\theta$
We have: $2sds=2rR\sin\theta d\theta$ (we are integrating for $\theta$ with r constant)
$\theta =0 \implies$ $s=R-r$ and $\theta=\pi\implies s=R+r$
$$I=\int_0^{r_0}\int_{0}^{2\pi}\int_{R-r}^{R+r}\frac{f(r)}{sR}rsdsd\phi dr=$$
$$=\int_0^{r_0}\int_{0}^{2\pi}\int_{R-r}^{R+r}\frac{f(r)r}{R}dsd\phi dr=$$ $$=\int_0^{r_0}\frac{f(r)r}{R}\int_{0}^{2\pi}\int_{R-r}^{R+r}dsd\phi dr=$$
$$=\int_0^{r_0}\frac{f(r)r}{R}\int_0^{2\pi}[s]_{R-r}^{R+r}d\phi dr=\int_0^{r_0}\frac{f(r)r}{R}\int_0^{2\pi}2rd\phi dr=$$ $$=\int_0^{r_0}\frac{f(r)r}{R}2r2\pi dr=\frac{4\pi}{R}\int_0^{r_0}f(r)r^2dr$$
Now, we have: $4\pi=\int_0^{2\pi}\int_0^{\pi}\sin \theta d\theta d\phi$
$$I=\frac{1}{R}\int_0^{r_0}\int_0^{2\pi}\int_0^{\pi}f(r)r^2\sin\theta d\theta d\phi dr=$$
$$=\frac{\int_K f(\mid x \mid)dx}{R}$$
And due to symmetry:
$$I=\frac{\int_K f(\mid x \mid)}{\mid y\mid}$$