# Use polar coordinates to evaluate the integral

use polar coordinates to evaluate the following integral $$\int^{a}_{-a} \int^{\sqrt{a^2-x^2}}_0 e^{-(x^2+y^2)}dydx$$

i did in my paper was my limit are after converting it to polar coordinates $$\int^{\pi}_0 \int^{{a}}_0 e^{-r^2}rdrd\theta$$

Are the limits in the above integral true?

• $x^2+ y^2= r^2$, not $r$. Your integral should have $e^{-r^2}$ rather than $e^{-r}$. Then use the substitution $u= r^2$. Feb 21, 2019 at 12:22
• Shouldnot the value be $x^2+y^2=a^2$? Feb 21, 2019 at 14:08

The domain is a half-disk of radius $$a$$, on the side of the positive $$y$$.

In Cartesian coordinates, $$x$$ runs from $$-a$$ to $$a$$ and $$y$$ is between $$0$$ and $$\sqrt{a^2-x^2}$$ (where $$x^2+y^2=a^2$$).

In polar coordinates, $$\theta$$ runs from $$0$$ to $$\pi$$ and $$r$$ from $$0$$ to $$a$$.

In both cases,

$$x^2+y^2\le a^2\land y\ge0.$$

Indeed $$x^2\le a^2\implies x^2+(a-x^2)\le a^2,$$

and

$$r^2\le a^2\land \theta\in[0,\pi]\implies\sin\theta\ge0.$$

• yes thanks i exatly did it like this Feb 21, 2019 at 12:58