2D Gaussian integration over arbitrary eccentric circle. Analytical solution? How can I find the solution for the integral of an axisymmetric Gaussian distribution over a circular surface? (A circular surface eccentric to the centre of the Gaussian distribution).
I am trying to create a theoretical framework for working with Gaussian jets entering circular pipes. I have found the solution for the problem when the Gaussian profile and the circle are aligned, but the problem now relies on the solution when this Gaussian distribution (of velocity) is not aligned with the centre of the pipe. 


*

*The Gaussian distribution of velocity has of the following form:


$u(r) = U e^{-\frac{r^2}{K^2}}$
Where $U$ and $K$ are constants and $r$ is the radial distance.


*My area (the inlet of the pipe) is a circle, with radius $R$ and eccentric to the centre of Gaussian distribution by a distance $d$ and with an angle $\theta$. Any arbitrary circle:


$(x-a)^2 + (y-b)^2 = R^2$ 
Where: $d = \sqrt(a^2+b^2)$, 
$\theta = tan^{-1}(\frac{b}{a})$
I am not sure whether this problem can be solved analytically or not. I have done a schematic of the integration problem that can be seen online.
$\\$
The goal is to solve:
$U_{avg} = \frac{1}{A} \int_{A} u(r) dA$
Where:


*

*$u(r)$ is the axisymetric Gaussian Distribution

*A is the area of an arbitrary circle


Is it possible an analytical solution? If so, how can the problem be solved?
Any help or suggestion would be extremely welcome.
Daniel
 A: As Lovsovs says, your integral appears not to be elementary. If the center of the disk of integration is sufficiently far from the center of the Gaussian, the integral will be roughly
$$
\pi R^{2} e^{-a^{2}/b^{2}},
$$
the value at the center multiplied by the area of the disk.
In case it's helpful: Since the Gaussian is radially symmetric, you may as well assume the center of your disk is at $(a, 0)$, with $0 < R < a$. Converting the Cartesian equation $(x - a)^{2} + y^{2} = R^{2}$ to polar gives
$$
r^{2} - 2ar\cos\theta + a^{2} - R^{2} = 0,
$$
or (using the quadratic formula) the polar equations
$$
r = a\cos\theta \pm \sqrt{a^{2} \cos^{2}\theta + R^{2} - a^{2}},\qquad
|\theta| \leq \arccos \tfrac{a^{2} - R^{2}}{a^{2}}.
$$
Setting $\theta_{0} = \arccos \frac{a^{2} - R^{2}}{a^{2}}$ and
$$
r_{1}(\theta) = a\cos\theta - \sqrt{a^{2} \cos^{2}\theta + R^{2} - a^{2}},\qquad
r_{2}(\theta) = a\cos\theta + \sqrt{a^{2} \cos^{2}\theta + R^{2} - a^{2}},
$$
the exact integral is
$$
2\int_{0}^{\theta_{0}} \int_{r_{1}(\theta)}^{r_{2}(\theta)} e^{-r^{2}/b^{2}}\, r\, dr\, d\theta
= b^{2} \int_{0}^{\theta_{0}} \bigl[e^{-r_{1}(\theta)^{2}/b^{2}} - e^{-r_{2}(\theta)^{2}/b^{2}}\bigr]\, d\theta.
$$
This can be formally simplified a bit by writing
\begin{align*}
r_{1}(\theta)^{2} &= \Bigl(2a^{2} \cos^{2}\theta + R^{2} - a^{2}\Bigr) - \Bigl(2a\cos\theta \sqrt{a^{2} \cos^{2}\theta + R^{2} - a^{2}}\Bigr) = b^{2}\bigl[f(\theta) - g(\theta)\bigr], \\
r_{2}(\theta)^{2} &= \Bigl(2a^{2} \cos^{2}\theta + R^{2} - a^{2}\Bigr) + \Bigl(2a\cos\theta \sqrt{a^{2} \cos^{2}\theta + R^{2} - a^{2}}\Bigr) = b^{2}\bigl[f(\theta) + g(\theta)\bigr],
\end{align*}
so that
\begin{align*}
  e^{-r_{1}(\theta)^{2}/b^{2}} - e^{-r_{2}(\theta)^{2}/b^{2}}
  &= e^{-(f(\theta) - g(\theta))} - e^{-(f(\theta) + g(\theta))} \\
  &= e^{-f(\theta)}\bigl[e^{g(\theta)} - e^{-g(\theta)}\bigr]
  = 2e^{-f(\theta)} \sinh g(\theta) \\
  &= 2e^{-(2a^{2} \cos^{2}\theta + R^{2} - a^{2})/b^{2}} \sinh \tfrac{1}{b^{2}}\Bigl(2a\cos\theta \sqrt{a^{2} \cos^{2}\theta + R^{2} - a^{2}}\Bigr).
\end{align*}
