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

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

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

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 $$r$$.
• @Mayar: You'd have to explain further what you mean. Here, the region is $(0,\infty)\times(0,\infty)$ which corresponds to the first quadrant; in other words, it's the same region as $(0,\pi/2)\times(0,\infty)$ in polar coordinates. If our region were $(-\infty,\infty)\times(0,\infty)\subset\Bbb R^2$, then $\theta$ would range between $0$ and $\pi$ since this would encompass the upper half plane. Oct 18, 2018 at 22:23