1. The problem asks to find the correct integral to evaluate the area of the shaded region, which is the bottom half of a circle described in polar coordinates. 2. The formula for the area enclosed by a polar curve $r(\theta)$ from $\theta=a$ to $\theta=b$ is: $$\text{Area} = \frac{1}{2} \int_a^b r(\theta)^2 \, d\theta$$ 3. Here, the radius function is given as $r(\theta) = 2 + 2 \sin \theta$. 4. The shaded region corresponds to the bottom half of the circle, which spans from $\theta = \pi$ to $\theta = 2\pi$. 5. Therefore, the correct integral to find the area is: $$\text{Area} = \frac{1}{2} \int_{\pi}^{2\pi} (2 + 2 \sin \theta)^2 \, d\theta$$ 6. Among the options, only option (b) and (d) have the $\frac{1}{2}$ factor, but (b) integrates from $\pi$ to $\frac{3\pi}{2}$, which is only half of the bottom semicircle. 7. Option (d) integrates from $\frac{3\pi}{2}$ to $2\pi$, which is only the right quarter of the bottom semicircle. 8. The full bottom half-circle integral is from $\pi$ to $2\pi$, so the correct integral is: $$\boxed{\text{Area} = \frac{1}{2} \int_{\pi}^{2\pi} (2 + 2 \sin \theta)^2 \, d\theta}$$ This matches none of the options exactly, but the closest correct form is option (b) if the interval is extended to $2\pi$. Hence, the correct integral is option (b) with the interval $[\pi, 2\pi]$. Final answer: $$\text{Area} = \frac{1}{2} \int_{\pi}^{2\pi} (2 + 2 \sin \theta)^2 \, d\theta$$