# About the shape of a gaussian function on a cylinder

In a three-dimensional space, given a cylinder of radius $R$ and height $L$ with the axis forming a couple of angles $\{\phi,\theta\}$ with respect to the $z$ axis of the $\{x,y,z\}$ reference frame, a gaussian function: $$G=\frac{1}{2\pi\sigma^2}\exp\left(-\frac{1}{2}\frac{x^2+y^2}{\sigma^2}\right)$$ intersects the surface of the cylinder. What is the equation of the shape of the figure formed by this intersection? We can assume the center of the axis of the cylinder to be coincident with the origin of the cartesian coordinates. Thanks.

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Do you mean the surface $z=G(x,y)$? – joriki Oct 10 '12 at 11:19
@joriki: yes, $z=G(x,y)$ – Riccardo.Alestra Oct 10 '12 at 11:22
You can write the cylindrical surface as $(\vec a\cdot\vec r)^2+(\vec b\cdot\vec r)^2=R^2$ and substitute $z=G(x,y)$ to get a relationship between $x$ and $y$ on the curve, but since that results in a transcendental equation I doubt you'll find a closed-form solution. – joriki Oct 10 '12 at 11:30