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## Main Question or Discussion Point

Given the equation for a Gaussian as: ##z = f(x,y) = Ae^{[(x-x0)^2 + (y-y0)^2] /2pi*σ^2 }## , how would I go about converting this into cylindrical coordinates? The mean is non-zero, and this seems to be the biggest hurdle. I believe I read earlier that the answer is ~ ##z = f(r,θ) = Ae^{(r-r0)^2 / σ^2}## but this doesnt quite make sense since it completely neglects the angle (i.e. theta). Would this be the correct solution?

## x = rcosθ##;

##y = rsinθ##;

##x0 = r0cosθ0##;

##y0 = r0sinθ0##; where ##r0## and ##θ0## and ##σ## are constants

##z = f(r,θ) = Ae^{(r^2 + r0^2 - 2rcosθr0cosθ0 - 2rsinθr0sinθ0 / σ^2 }##

This is just a direct substitution, though. Shouldn't there be a change of variable and thus making the solution:

##z = f(r,θ) = rAe^{(r^2 + r0^2 - 2rcosθr0cosθ0 - 2rsinθr0sinθ0 /σ^2 }##

Any advice?

## x = rcosθ##;

##y = rsinθ##;

##x0 = r0cosθ0##;

##y0 = r0sinθ0##; where ##r0## and ##θ0## and ##σ## are constants

##z = f(r,θ) = Ae^{(r^2 + r0^2 - 2rcosθr0cosθ0 - 2rsinθr0sinθ0 / σ^2 }##

This is just a direct substitution, though. Shouldn't there be a change of variable and thus making the solution:

##z = f(r,θ) = rAe^{(r^2 + r0^2 - 2rcosθr0cosθ0 - 2rsinθr0sinθ0 /σ^2 }##

Any advice?

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