# Required form for integral of exponentially weighted cosine distribution across $xy$ plane

I am currently trying to find the required form of an integral for my code. In this case, I have an $$xy$$ plane ranging from $$-1$$ to $$1$$ in both dimensions and fixed at $$z = 0$$. At each point, I am assuming that I have a $$\exp(-(x^2 + y^2))\cos(\theta_{emiss})$$ distribution where $$x$$ and $$y$$ are the aforementioned coordinates on the plane and $$\theta_{emiss}$$ is an angle defined relative to the normal vector of the plane in the $$+z$$ direction. I am also assuming this distribution is azimuthally symmetric.

What I would like to do is to see what would the integrated distribution look like as a function of $$\theta$$ and $$\phi$$ at a distance $$d$$ away from the origin. So for that I would need to set up some integral, but I am unsure what the required form would be in order to do this. Does anyone have any ideas?

Edit: For a physical context, this problem represents a Gaussian temperature distribution on a finite plane. Within this problem, I am also assuming that emission occurs from the surface with a emission distribution proportional to cos($$\theta$$), as defined above, and the emission strength is proportional to the value of the Gaussian at the coordinates ($$x$$,$$y$$). This emission also has a 1/$$r^2$$ dependence where $$r$$ is the distance between an observation point ($$x_0,y_0,z_0$$) and a point on the plane $$(x,y,0)$$

Now assuming I have a hemispherical shell with a radius, $$d$$, from the origin of the plane $$(0,0,0)$$ (where $$d$$ is much larger than the size of the plane), I would like to know if an integral expression can be found to calculate the distribution across the hemispherical screen as a function of polar angle, $$\theta_{screen}$$ and the azimuthal angle, $$\phi_{screen}$$.

I am also assuming that emission occurs in the $$+z$$ direction and that the screen spans from $$0 - 2\pi$$ in $$\phi_{screen}$$ and $$0-\pi/2$$ in $$\theta_{screen}$$.

• It is not entirely clear what you're trying to achieve with that gaussian distribution in a finite domain. What is the setup of the physical problem? What are the random variables here? Sep 13 at 14:08
• @DinosaurEgg I have added the physical context to the question. Sep 14 at 15:05

Here's a model for the problem after the edit. We break down the plane in infinitesimal sized black bodies. The power emitted per unit solid angle of the plane per unit solid angle can be described by

$$dP(x,y,0)=\sigma (T(x,y))^4\frac{\cos\theta}{\pi}d\Omega d xdy$$

Here $$d\Omega=\sin \theta d\theta d\phi$$. Now consider an infinitesimal part of the hemispherical shell detector located at $$(\theta_s,\phi_s)$$, which subtends a solid angle determined by

$$d\Omega=\hat{n}\cdot \frac{(\mathbf{r}-\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|^3}dS$$

where $$\hat{n}=(\sin\theta_s\cos\phi_s, \sin\theta_s\sin\phi_s, \cos\theta_s)$$ is the outward normal vector to the spherical surface, $$dS=R^2\sin\theta_s d\theta_s d\phi_s$$, $$\mathbf{r}=R\hat{n}$$ is the position of the spherical shell infinitesimal and $$\mathbf{r'}=(x',y',0)$$ denotes an arbitrary surface point. To obtain the distribution of luminosity per unit area of the semispherical detector we need to integrate over the whole luminous plane surface to take into account all the infinitesimal contributions:

\begin{align}&dP_{\text{received}}(\theta_s,\phi_s)=dS \int\int dx'dy' dP_{\hat{n}, \mathbf{r}}(x',y',0)\\\Rightarrow &\frac{dP_\text{received}}{dS}(\theta_s,\phi_s)=\frac{\sigma}{\pi }\int\int dx'dy'T^4(x',y')\frac{[\hat{n}\cdot(\mathbf{r}-\mathbf{r'})][\hat{z}\cdot(\mathbf{r}-\mathbf{r'})]}{|\mathbf{r}-\mathbf{r'}|^4}\end{align}

For large detector radius $$R$$, one requires a multipole expansion of the kernel $$\frac{[\hat{n}\cdot(r-r')][\hat{z}\cdot(r-r')]}{|r-r'|^4}$$ to obtain a reasonable approximation for large observer distances.