Gaussian Integral: Converting from Cartesian to Polar

[Adoniram]

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... :(

 

[RUber]

@Zondrina gave a thorough explanation of the conversion in this recent post.
https://www.physicsforums.com/threads/cartesian-to-polar-cordinates.828554/

 

[Ray Vickson]

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... :(

We can write
$$I^2 = \int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = \int_{R^2} e^{-x^2-y^2} \, dA,$$
where $dA = dx dy$ is the two-dimensional "area element". When we switch to polar coordinates, the area element becomes $dA = r \, dr \, d\theta$. Basically, we have $I^2 \approx \sum_i \Delta A_i e^{-r^2_i}$, where the $\Delta A_1, \Delta A_2, \ldots$ are finite but small areas of little regions and the $r_i$ are the $\sqrt{x_i^2 + y_i^2}$ values at some point $(x_i,y_i)$ inside the regions $i = 1,2, \ldots$. We can either take the little regions to be rectangular, with horizontal and vertical sides $\Delta x_i$ and $\Delta y_i$, or else we can take them to be slices of radial wedges with sides along radial lines at angles $\theta_i$ and $\theta_i + \Delta \theta_i$ and inner-outer radii at $r_i$ and $r_i + \Delta r_i$. In rectangular coordinates the area element is $\Delta A_i = \Delta x_i \, \Delta y_i$, while in polar coordinates it is $\Delta A_i = r_i \, \Delta r_i \, \Delta \theta_i$.

All this was supposed to have been covered thoroughly in Calculus II, where things such as Jacobians and the like were introduced and motivated.