Evaluate the given double integral by converting to polar coordinates

Question

Evaluate the given double integral by converting to polar coordinates

$$\int_0^2 \int_0^{\sqrt{4-x^2}} e^{x^2+y^2} \, dy\,dx$$

my work is that

$$x^2 +y^2 =r^2$$

$$x= r\cos(\theta)$$

$$y=r\sin(\theta)$$

$$x^2 +y^2= r^2$$

$$y=\sqrt{4-x^2}$$

$$y^2+x^2 =4$$

then $r = 2$

$$0\le r\le2$$

$$\int_{0}^{2}\int_{r=0}^{r=2} e^{r^2}r \,dr\,dx$$

let $$u= r^2$$ then $$ du=2r\,dr$$

$$\int_0^2 \int_{r=0}^{r=2} e^u \, du\,dx$$

I got $$e^4/2$$ is that correct ?????

stuck here

Answer 1

You are in the first quadrant, so $I = \displaystyle \int_{0}^{\frac{\pi}{2}} \int_{0}^2 re^{r^2}drd\theta$