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I need help solving this integral ($\hat{z}$ denotes the polar axis):$$\int_V\dfrac{\vec{r}\cdot(\vec{r}-c\hat{z})}{|\vec{r}|^3|\vec{r}-c\hat{z}|^3} dV$$ Where $V$ denotes all space.

Attempt: $$2\pi \int_0^\infty \int_0^{\pi}\dfrac{\vec{r}\cdot(\vec{r}-c\hat{z})}{|\vec{r}|^3|\vec{r}-c\hat{z}|^3} \sin \theta d\theta \;|\vec{r}|^2dr $$

let $r = |\vec{r}|$, as $\hat{r}\cdot \hat{z} = \cos \theta $ $$2\pi \int_0^\infty \int_0^{\pi} \dfrac{r - c\cos \theta}{|\vec{r}-c\hat{z}|^3} \sin \theta d\theta dr$$

$$2\pi \int_0^\infty \int_0^{\pi} \dfrac{r - c\cos \theta}{(r^2+c^2-2 rc \cos \theta)^{3/2}}\sin \theta d\theta dr$$

Let $x = \cos \theta$

$$2\pi \int_0^\infty \int_{-1}^{1} \dfrac{r-cx}{(r^2+c^2 - 2 r c x)^{3/2}}dx dr$$

How do I reduce this further?

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1 Answer 1

up vote 2 down vote accepted

This answers the original version of the question:

Use $ \vert \vec{r} - c \hat{z} \vert^2 = r^2 + c^2 - 2 r c \cos \theta$. Now integration with respect to $\theta$ can be carried out, change $t = \cos\theta$.

$$ \begin{eqnarray} 2 \pi \int_0^ \pi \frac{r(1-\cos \theta)}{ \left( r^2 + c^2 - 2 \, c \cdot r \cdot \cos \theta \right)^{3/2}} \sin \theta \, \mathrm{d} \theta &=& 2 \pi \int_{-1}^1 \frac{r (1 -t) }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } \mathrm{d} t \\ &=& 2 \pi \frac{ c^2 + r^2 - \vert c^2 - r^2 \vert }{c^2 r ( c+ r)} \end{eqnarray} $$

Integration with respect to $r$ is now trivial:

$$ \begin{eqnarray} \int_0^\infty 2 \pi \frac{ c^2 + r^2 - \vert c^2 - r^2 \vert }{c^2 r ( c+ r)} \mathrm{d} r &=& \int_0^c 2 \pi \frac{ 2 r^2 }{c^2 r ( c+ r)} \mathrm{d} r + \int_c^\infty 2 \pi \frac{ 2 c^2 }{c^2 r ( c+ r)} \mathrm{d} r \\ & = & 2 \pi \frac{2 - 2 \log 2}{c} + 2 \pi \frac{2 \log 2}{c} = \frac{4 \pi}{c} \end{eqnarray} $$

It now remains to show steps to integrate with respect to $t$:

$$ \begin{eqnarray} \frac{r (1 -t) }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } &=& \frac{1}{c} \frac{ (r^2 + c^2 - 2 r c t) - (r-c)^2 }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } \\ &=& \frac{1}{c} \frac{1}{\left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{1/2}} - \frac{1}{c} \frac{(r-c)^2 }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } \end{eqnarray} $$ Now integration with respect to $t$ can be done trivially using $\int ( a-b t)^{-c} \mathrm{d} t =\frac{ (a - b t)^{1-c}}{b (c-1)}$.

Modified version

$$ \begin{eqnarray} 2 \pi \int_0^ \pi \frac{(r-c \cos \theta)}{ \left( r^2 + c^2 - 2 \, c \cdot r \cdot \cos \theta \right)^{3/2}} \sin \theta \, \mathrm{d} \theta &=& 2 \pi \int_{-1}^1 \frac{(r -c t) }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } \mathrm{d} t \\ &=& 2 \pi \frac{ 1 - \operatorname{sign}(c - r ) }{r^2} \end{eqnarray} $$

Integrating with respect to $r$:

$$ \int_0^\infty 2 \pi \frac{ 1 - \operatorname{sign}(c - r ) }{r^2} \mathrm{d} r = \int_c^\infty 2 \pi \frac{ 2 }{r^2} \mathrm{d} r = \frac{4\pi}{c} $$

The logic of carrying out integration with respect to $t$ is the same.

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The numerator in the first integtrand is $r-c\cos \theta$ –  kuch nahi Sep 20 '11 at 18:59
    
This simplified things, I will update the post in a second. –  Sasha Sep 20 '11 at 19:01
    
Thanks for answering, how did you get: $\int_{-1}^1 \frac{(r -c t) }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } \mathrm{d} t = \frac{ 1 - \operatorname{sign}(c - r ) }{r^2}$? This is where I was stuck. –  kuch nahi Sep 20 '11 at 20:21
    
Use $\int r (r^2+c^2-2 c r t)^{-3/2} \mathrm{d} t = \left( c \sqrt{ r^2+c^2-2 c r t} \right)^{-1}$ and $\int c t (r^2+c^2-2 c r t)^{-3/2} \mathrm{d} t = \frac{1}{c r^2} \frac{ c^2 + r^2 - c r t}{\sqrt{ r^2+c^2-2 c r t}}$, then use the fundamental theorem of calculus. –  Sasha Sep 20 '11 at 20:26

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