Setting up bounds of integral in polar coordinates

Question

I need to solve the integral in polar coordinates $$\int_0^1\int_0^\sqrt{1-x^2}e^{-x^2-y^2}dydx$$

What I've tried:

$$x= r\cos(\theta), y = r\sin(\theta) \Rightarrow e^{-r^2} $$ $$|J| = r$$

Therefore,

$$\int_{a}^{b} \int_{c}^{d} e^{-r^2}rdrd\theta$$

But I don't know what to do with the lower and upper limits of the integral $a, b, c, d$. I've tried doing the following: $0 = r\sin(\theta), r = c = 0$ and $y = \sqrt{1-x^2} \iff r\sin(\theta) = \sqrt{1-r^2\cos^2(\theta)} \iff r = 1$ so $d$ would be $d = 1$.

But I'm not sure I'm setting these bounds correctly. How would it go for the limits of $\theta$?

Answer 1

You have $0\leqslant x\leqslant1$ and, for each such $x$, you have $0\leqslant y\leqslant\sqrt{1-x^2}$. This latter assertion is equivalent to $0\leqslant y^2\leqslant1-x^2$. So, $x^2+y^2\leqslant1$. Since $x,y\geqslant0$, you get $\theta$ can take any value from $0$ to $\frac\pi2$ and that $r$ can take any value from $0$ to $1$. Therefore, in polar coordinates your integral becomes$$\int_0^{\pi/2}\int_0^1re^{-r^2}\,\mathrm dr\,\mathrm d\theta.$$

Answer 2

You are integrating over the first quarter of the unit circle, so the bounds converts to $0\leq r\leq 1$ and $0\leq \theta\leq \frac{\pi}{2}$