1. **Problem Statement:** Find the volume of the solid under the surface defined by the function $f(x,y) = 7$ and above the region $D$ bounded by the cardioid $r = 1 + \cos(\theta)$ for $0 \leq \theta \leq \pi$. 2. **Set up the iterated integral in polar coordinates:** The volume under the surface $f(x,y)$ over region $D$ is given by the double integral $$V = \iint_D f(x,y) \, dA$$ In polar coordinates, the area element $dA = r \, dr \, d\theta$. Given $f(x,y) = 7$, the integral becomes $$V = \int_0^{\pi} \int_0^{1 + \cos(\theta)} 7 \cdot r \, dr \, d\theta$$ 3. **Evaluate the inner integral:** $$\int_0^{1 + \cos(\theta)} 7r \, dr = 7 \int_0^{1 + \cos(\theta)} r \, dr = 7 \left[ \frac{r^2}{2} \right]_0^{1 + \cos(\theta)} = \frac{7}{2} (1 + \cos(\theta))^2$$ 4. **Evaluate the outer integral:** $$V = \int_0^{\pi} \frac{7}{2} (1 + \cos(\theta))^2 \, d\theta = \frac{7}{2} \int_0^{\pi} (1 + 2\cos(\theta) + \cos^2(\theta)) \, d\theta$$ 5. **Simplify the integral:** Recall the identity: $$\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$$ So, $$\int_0^{\pi} (1 + 2\cos(\theta) + \cos^2(\theta)) \, d\theta = \int_0^{\pi} \left(1 + 2\cos(\theta) + \frac{1 + \cos(2\theta)}{2}\right) d\theta$$ Combine terms: $$= \int_0^{\pi} \left(\frac{3}{2} + 2\cos(\theta) + \frac{\cos(2\theta)}{2}\right) d\theta$$ 6. **Integrate term-by-term:** $$\int_0^{\pi} \frac{3}{2} d\theta = \frac{3}{2} \pi$$ $$\int_0^{\pi} 2\cos(\theta) d\theta = 2 \sin(\theta) \Big|_0^{\pi} = 0$$ $$\int_0^{\pi} \frac{\cos(2\theta)}{2} d\theta = \frac{1}{2} \cdot \frac{\sin(2\theta)}{2} \Big|_0^{\pi} = 0$$ 7. **Sum the results:** $$\int_0^{\pi} (1 + 2\cos(\theta) + \cos^2(\theta)) d\theta = \frac{3}{2} \pi + 0 + 0 = \frac{3\pi}{2}$$ 8. **Calculate the volume:** $$V = \frac{7}{2} \times \frac{3\pi}{2} = \frac{21\pi}{4}$$ **Final answer:** $$\boxed{\frac{21\pi}{4}}$$