Answer
$(e^R-1) \pi$
Work Step by Step
We write the region in the polar co-ordinates as: $0 \leq r \leq \sqrt R$ and $0 \leq \theta \leq 2\pi$
$\iint_{D} f(x,y) \ dA=\int_0^{2\pi} \int_0^{\sqrt R} (e^{r^2}) (r) \ dr \ d\theta$
Now, we have:
$\int_0^{2\pi} \int_0^{\sqrt R} (e^{r^2} (r) \ dr \ d\theta =\int_0^{2\pi} [\dfrac{e^{r^2}}{2}]_0^{\sqrt R } d\theta$
or, $=\dfrac{1}{2} \times \int_0^{\pi} [e^R-1] d\theta$
or. $=(1/9) \times[-\cos \theta]_0^{\pi}$
or. $=\dfrac{1}{2}[e^R-1] \times 2 \pi$
or. $=(e^R-1) \pi$
