I am working in a problem of fluxes of vector maps, and after having applied the divergence theorem I have obtained the following integral:

$\Phi = \int_{x=-R}^{x=R} \int_{y=-\sqrt{R^2-x^2}}^{y=\sqrt{R^2-x^2}} \exp\left(-\frac{\sqrt{(x-x_i)^2+(y-y_i)^2}}{\delta}\right)\left(\frac{1}{\sqrt{(x-x_i)^2+(y-y_i)^2}}-\frac{1}{\delta}\right) dydx$

This integral is the surface integral over a circle of radius $R$ centered at $(0,0)$ of the function:

$f(x,y)= \exp\left(-\frac{\sqrt{(x-x_i)^2+(y-y_i)^2}}{\delta}\right)\left(\frac{1}{\sqrt{(x-x_i)^2+(y-y_i)^2}}-\frac{1}{\delta}\right)$

This function have rotation symmetry but is centered at $(x_i,y_i)$. I have tried to different approaches to solve this integral:


using the changes of variables $x = \bar{x}+x_i$ and $y = \bar{y}+y_i$ we obtain the integral \begin{equation} \Phi = \int_{\bar{x}=-R-x_i}^{\bar{x}=R-x_i} \int_{\bar{y}=-\sqrt{R^2-(\bar{x}+x_i)^2}-y_i}^{\bar{y}=\sqrt{R^2-(\bar{x}+x_i)^2}-y_i} \exp\left(-\frac{\sqrt{\bar{x}^2+\bar{y}^2}}{\delta}\right)\left(\frac{1}{\sqrt{\bar{x}^2+\bar{y}^2}}-\frac{1}{\delta}\right) d\bar{y}d\bar{x} \end{equation}

now let's use polar coordinates $\bar{x} = r\cos\theta$ and $\bar{y} = r\sin\theta$

\begin{equation} \Phi= \int_{\bar{x}=-R-x_i}^{\bar{x}=R-x_i} \int_{\bar{y}=-\sqrt{R^2-(\bar{x}+x_i)^2}-y_i}^{\bar{y}=\sqrt{R^2-(\bar{x}+x_i)^2}-y_i} \exp\left(-\frac{r}{\delta}\right)\left(\frac{1}{r}-\frac{1}{\delta}\right) rd\theta dr \end{equation}

but now I don't know how to change the integration limits and I'm stuck here.


using directly the change to polar coordinates $x = r\cos\theta$ and $y = r\sin\theta$

$ \begin{aligned} \Phi = \int_{r=0}^{r=R} \int_{\theta=0}^{\theta=2\pi} \exp&\left(-\frac{\sqrt{r^2+r_i^2-2r(x_i\cos\theta+y_i\sin\theta)}}{\delta}\right)\\ &\left(\frac{1}{\sqrt{r^2+r_i^2-2r(x_i\cos\theta+y_i\sin\theta)}}-\frac{1}{\delta}\right)rdrd\theta \end{aligned} $

where I have defined $r_i = \sqrt{x_i^2+y_i^2}$. In this case I am also stuck and I don't know how to continue.

So my question is: which is the correct method to compute this integral, 1 or 2, and how I have to continue the computations? In the case that nor method 1 nor method 2 are correct, which is the correct way to do it?

Thanks for your attention and your help!

PD: I have also looked for this integral in the Gradshteyn bu I have not found it.

  • $\begingroup$ I think Method 1 is how I would do it. Looks like your bounds of integration are going to depend on if $(x_i, y_i)$ is in the circle of radius $R$. Maybe look at the bounds for $r$ first. It should be independent of $\theta$. Although getting an expression of $\theta$ in terms of $r$ might be tricky. $\endgroup$ – Trevor Norton Jul 19 '17 at 12:52

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