Double integrals transforming into Polars This is my first post here.
I'm reading about double integrals and can't catch how to get the new limits of integration when converting to polar form.
$$\left(\int_{-\infty}^{\infty} e^{-x^2}dx\right)^2=\left(\int_{-\infty}^{\infty} e^{-x^2}dx\right)\left(\int_{-\infty}^{\infty} e^{-y^2}dy\right)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}dx\,dy$$
$x=r\cos(\phi), y=r\sin(\phi)$ we get the region of integration is the $(x,y)$ plane (how?) and
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}dxdy=\int_{0}^{2\pi}\int_{0}^{\infty} e^{-r^2}r\; dr\, d\phi=\pi$$
so 
$$\int_{-\infty}^{\infty} e^{-x^2}dx=\sqrt{\pi}$$
So here is the question, how did we go from infinities to $0$ to $2\pi$? And can I transform something easy as $\int_0^1 x dx$ into polar form? What would the limits be in this case? 
Thanks.
 A: When you had Cartesian coordinates with $-\infty < x < \infty$ and $-\infty < y < \infty$ in your first limits, they covered the whole plane.
Now you have polar coordinates. How do you fill the whole plane with polar coordinates? Well, if $0 \le r < \infty$ (and $\theta =0$) then you have the positive $x$-axis. Next start to rotate this around the origin by a whole turn: $0 \le \theta \le 2\pi$. You will sweep out the whole plane.
Imagine standing at the origin holding an infinity long broom handle along the positive $x$-axis. If you spin $360^{\circ}$ on the spot then the broom handle hits everything in the room.
You can't do anything with $\int_0^1 x \, \operatorname{d}\!x$ in terms of polar coordinates. Polar coordinates are two dimensional. Your integral is along a one dimensional line.
A: For your first question: how would you use polar substitution to evaluate the integral
$$
\int_{-\rho}^\rho\int_{-\sqrt{\rho^2-x^2}}^{\sqrt{\rho^2-x^2}}e^{x^2+y^2}dy\,dx\,?
$$
Now, think of
$$
\int_{-\infty}^\infty\int_{-\infty}^\infty e^{x^2+y^2}dy\,dx
$$
as the limit of the first integral as $\rho\to\infty$.
For your second question: not exactly. You can change a double-integral to polar form, but that integral is over a single variable. The only substitution you can do there is the familiar u-substitution.
