Multiple integrals in polar coordinates

So, I was practicing some problems and found this:

$$\int_0^2 \int_0^{\sqrt{4-x^2}} e^{x^2+y^2} dydx.$$

While converting from Cartesian to polar coordinates, $$\theta$$ limits will be from $$0$$ to $$\pi /2$$ but what about the $$r$$ limits? I am stuck on this. Is it $$0$$ to $$2cosec θ$$?

• Thanks @Robert Z – Korra Feb 19 at 11:04

The integration region corresponds to the part of a circle with radius $$2$$ centered at $$(0,0)$$ that lies in the first quadrant. So the distance to the origin, $$r$$, is between $$0$$ and $$2$$. the integral becomes $$\int_0^{\pi/2} \int_0^2 r e^{r^2} dr d \theta = \frac{\pi}{2} [\frac 12 e^{r^2}]_0^2 = \frac{\pi}{4}(e^4-1)$$