Why did I convert the outer integeral from $ \int_0^\infty $ into $ \int _0^\frac{\pi}{2} $ not any other value of theta?

Question

QUESTION: $$I =\int_0^\infty e^{-x^2}\ dx$$

such that $$ \int_0^\infty \int _0^\infty e^{-x^2+y^2} \ dx \ dy = I^2 $$

Change the variable to polar coordinates in order to evaluate $$ \int_0^\infty \int _0^\infty e^{-x^2+y^2} \ dx \ dy = I^2 $$

SOLUTION:

Here is where my question lies in these steps $$ \int _0^\frac{\pi}{2} \int_0^\infty \ e^{-r^2} r \ dr\ d\theta $$ why $\frac{\pi}{2}$ not any other $\theta$ value in the integeral? that's a solution I found in a textbook , regardless the following steps , (conducting a non proper integeral and computing the definte integeral's value ) , this is the point where I don't understand.

And one more question related to this workout why do we consider the $e^a =0$ where

$\frac{-1}{2} ( \lim\limits_{a \to \infty}\ e^{-a^2}-1 )\int_0^\infty d\theta$=$\frac{\pi}{4}$

Answer 1

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 $r$.