QUESTION
$$I =\int_0^\infty e^{-x^2}\mathrm{d}x$$
such that $$ \int_0^\infty \int _0^\infty e^{-(x^2+y^2)} \mathrm{d}x \ \mathrm{d}y = I^2 $$
Change the variable to polar coordinates in order to evaluate $$ \int_0^\infty \int _0^\infty e^{-(x^2+y^2)} \mathrm{d}x \mathrm{d}y = I^2 $$
SOLUTION:
Here is where my question lies in these steps $$\int _0^\frac{\pi}{2} \int_0^\infty\ e^{-r^2} r \mathrm{d}r\mathrm{d}\theta$$ Why $\frac{\pi}{2}$ not any other $\theta$ value in the integral? That's a solution I found in a textbook, regardless the following steps, (conducting a non proper integral and computing the definte integral's value), this is the point where I don't understand.
- And why do we consider the $e^{-a^2} =0$ in:
$$\frac{-1}{2} \left( \lim\limits_{a \to \infty}\ e^{-a^2}-1 \right)\int_0^\infty \mathrm{d}\theta=\frac{\pi}{4}$$
You're only interested in the first quadrant of $\Bbb R^2$ in the double integral. That is, the double integral is integrating over the region $(0,\infty)\times(0,\infty)$, which is precisely the first quadrant. If you think about the unit circle values for $\theta$ and determine which angles lie in the first quadrant, you'll see that the first quadrant corresponds to $\theta\in(0,\pi/2)$.
For the values of $r$, well, you want to cover the entire first quadrant. This means that along any ray with an angle of $\theta$ (as measured from the positive $x$-axis), you'll need to take $r$ from $0$ to $\infty$. Since you do this for every value of $\theta$, you end up with a double integral: one that integrates with respect to $\theta$ and another that integrates with respect to $r$.