1. **Problem Statement:** Evaluate the integral $$\iint_R e^{x^2 + y^2} \, dy \, dx$$ where $R$ is the semicircular region bounded by the $x$-axis and the curve $y = \sqrt{1 - x^2}$, i.e., the upper half of the unit circle. 2. **Formula and Setup:** The integral is difficult in Cartesian coordinates because $e^{x^2 + y^2}$ has no elementary antiderivative. We use polar coordinates where $x = r \cos \theta$, $y = r \sin \theta$, and the area element $dy \, dx = r \, dr \, d\theta$. 3. **Region in Polar Coordinates:** The semicircle corresponds to $0 \leq r \leq 1$ and $0 \leq \theta \leq \pi$. 4. **Rewrite the Integral:** $$ \iint_R e^{x^2 + y^2} \, dy \, dx = \int_0^\pi \int_0^1 e^{r^2} r \, dr \, d\theta $$ 5. **Integrate with respect to $r$:** Use substitution $u = r^2$, $du = 2r \, dr$, so $r \, dr = \frac{du}{2}$. $$ \int_0^1 e^{r^2} r \, dr = \frac{1}{2} \int_0^1 e^u \, du = \frac{1}{2} [e^u]_0^1 = \frac{1}{2} (e - 1) $$ 6. **Integrate with respect to $\theta$:** $$ \int_0^\pi \frac{1}{2} (e - 1) \, d\theta = \frac{1}{2} (e - 1) \times \pi = \frac{\pi}{2} (e - 1) $$ 7. **Final Answer:** $$ \boxed{\frac{\pi}{2} (e - 1)} $$ This completes the evaluation of the integral over the semicircular region. --- **Note:** The user requested solutions to all questions, but per instructions, only the first problem is solved here.