# Evaluate a certain double integral over a specific region.

Evaluate: $$\int \int_R e^{-(x^2 + y^2)} \,dA$$ where R: $x^2 + y^2 = 4 \qquad x \geq 0 \qquad y \geq 0$

What I have done so far: By graphing the region I note that it resembles the first quadrant of a circle. Recognizing that $x^2 + y^2 = r^2$. I obtain the following integral: $$\int_0^{\pi/2} \int_0^2 re^{-(r^2)} \, dr d\theta$$ I get stuck here. I also checked this question on mathstackexchange: Is there really no way to integrate $e^{-x^2}$?

• It doesn't resemble the first quadrant of a circle, but rather is one. =) – Pedro Tamaroff Oct 18 '15 at 18:41
• I hope you mean "$x^2+y^2 \leqslant 4$": otherwise, the region of integration has measure zero, so the integral vanishes... – Chappers Oct 18 '15 at 18:45
• You’re not trying to integrate $e^{-x^2}$, though. You’ve got $xe^{-x^2}$, which is a different matter. – amd Oct 18 '15 at 21:17

You can integrate that using a simple substitution. $u=r^2$ will work fine.
And to answer your other question you can't find $$\int e^{-x^2}~dx$$ in terms of elementary functions but you can calculate it using a little trick which produces and integral similar to the one you have now. You switch into polars and it becomes easy to calculate.
$$\int_0^{\pi/2} \int_0^2 re^{-(r^2)} \, dr d\theta$$ $$=\int_0^{\pi/2} \frac{1}{2}\int_0^2 2re^{-(r^2)} \, dr d\theta$$ $$=\int_0^{\pi/2} \frac{1}{2}\int_0^2 e^{-(r^2)} \, d(r^2) d\theta$$ $$=\frac{1}{2}\int_0^4 e^{-(u)} \, d(u) \int_0^{\pi/2}d\theta$$ where $u=r^2$ as pointed out by Chris $$=\frac{1}{2}[-e^{-(u)}]_0^4 \frac{\pi}{2}$$ $$=\frac{\pi}{4}[1-e^{-4}]$$