- #1

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## Homework Statement

I'm encountering these integrals a lot lately, and I can solve them because I know the "trick" but I'd like to know actually how the cartesian to polar conversion works:

##\int_{-\infty}^{\infty}e^{-x^2}dx##

## Homework Equations

##\int_{-\infty}^{\infty} e^{-x^2} = I##

##I^2=\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy##

##=\int_{-\infty}^{\infty} e^{-r^2}r dr dθ = π##

## The Attempt at a Solution

So, if I look at ##r=\sqrt{x^2+y^2}##, it's easy to see that ##dr=(1/r)(x dx+y dy)##

Which leads me to believe that ##dθ=(x dx+y dy)## ... ???

If ##θ=ArcTan(y/x)##, then how does ##dθ=(x dx+y dy)##?

Say ##y=Tan(θ)x##, then taking ##dy## and simplifying, I can get:

##x dy - y dx=(x^2+y^2) dθ##

So unless ##x/(x^2+y^2) = y## and ##y/(x^2+y^2) = -x##, I am at a loss... :(