Finding $\iint_{G} e^{x^2 + y^2} \,dx\,dy$, $G=\{(x,y) : x^2 + y^2 ≤ R^2\}$

Question

I used polar coordinates and my answer is $\pi(e^{R^2} - 1)$ but in the answer sheet it's double of that somewhy, is that my mistake?

Answer 1

I get $\int\int_Ge^{({x^2+y^2})}dxdy=2\pi\int_0^{R}e^{r^2}rdr=\pi(e^{R^2}-1)$ It looks like the author forgot to divide by 2 to get the $r$ integral.

Answer 2

Actually, your statement of the problem doesn't make sense! "dxdy" means this is an integral over an area but "$\{(x, y)| x^2+ y^2= R^2$" is a curve. Did you mean "$\{(x, y)| x^2+ y^2\le R^2$, the disk rather than the circle?

Assuming that, $x^2+ y^2= r^2$ and $dxdy= r drd\theta$ so the integral is $\int_0^R\int_0^{2\pi} e^{r^2}drd\theta$. After integrating with respect to $\theta$ we have $2\pi\int_0^{R}e^{r^2}rdr$. Letting $u= r^2$, $du= 2rdr$ so the integral becomes $\pi\int_0^{R^2} e^u du= \pi(e^{R^2}- 1)$.

However, if the problem is to integrate around the circle, not over the disk, then $x^2+ y^2= R^2$, a constant, so the integral is $\int_0^{2\pi} e^{R^2}d\theta= 2\pi e^{R^2}$.