1. The problem asks to find the correct integral to evaluate the area of the shaded region in a polar graph where the radius is given by $r = 2 + 2 \sin \theta$. 2. The formula for the area $A$ enclosed by a polar curve $r(\theta)$ from angle $\alpha$ to $\beta$ is: $$A = \frac{1}{2} \int_{\alpha}^{\beta} r(\theta)^2 \, d\theta$$ This formula calculates the area by integrating the square of the radius over the given angle interval, multiplied by $\frac{1}{2}$. 3. The shaded region is described as the lower half part between angles $\pi$ and $2\pi$. 4. Therefore, the limits of integration should be from $\pi$ to $2\pi$. 5. The radius function is $r(\theta) = 2 + 2 \sin \theta$, so the integrand is $(2 + 2 \sin \theta)^2$. 6. Applying the formula, the correct integral for the area is: $$A = \frac{1}{2} \int_{\pi}^{2\pi} (2 + 2 \sin \theta)^2 \, d\theta$$ 7. Among the options given, option (a) integrates from $\frac{3\pi}{2}$ to $2\pi$, which is only part of the lower half. 8. Option (b) integrates from $\pi$ to $\frac{3\pi}{2}$, which is the other part of the lower half. 9. Option (c) lacks the $\frac{1}{2}$ factor, so it is incorrect. 10. Option (d) integrates from $\pi$ to $2\pi$ but also lacks the $\frac{1}{2}$ factor. 11. Since the shaded region covers the entire lower half from $\pi$ to $2\pi$, the total area is the sum of (a) and (b), or equivalently: $$A = \frac{1}{2} \int_{\pi}^{2\pi} (2 + 2 \sin \theta)^2 \, d\theta$$ 12. Therefore, the correct integral is option (a) plus option (b) combined, which matches the integral in option (a) and (b) together, or equivalently the integral in option (a) plus option (b). 13. Since the question asks for the correct integral to evaluate the area of the shaded region, the best single integral expression is: $$A = \frac{1}{2} \int_{\pi}^{2\pi} (2 + 2 \sin \theta)^2 \, d\theta$$ which corresponds to the sum of (a) and (b). 14. Hence, the correct choice is the integral with limits from $\pi$ to $2\pi$ and the $\frac{1}{2}$ factor, which is not explicitly listed but is the sum of (a) and (b). Since the question asks to find the correct integral from the options, the correct integral is the sum of (a) and (b), or equivalently: $$\boxed{\text{Area} = \frac{1}{2} \int_{\pi}^{2\pi} (2 + 2 \sin \theta)^2 \, d\theta}$$