# Setting up bounds of integral in polar coordinates

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 $$\theta$$?

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$$\int_0^{\pi/2}\int_0^1re^{-r^2}\,\mathrm dr\,\mathrm d\theta.$$
You are integrating over the first quarter of the unit circle, so the bounds converts to $$0\leq r\leq 1$$ and $$0\leq \theta\leq \frac{\pi}{2}$$