1. **State the problem:** We want to evaluate the double integral $$\iint_D x \, dA$$ where $D$ is the region in the first quadrant between the circles $$x^2 + y^2 = 4$$ and $$x^2 + y^2 = 2x$$. 2. **Rewrite the region:** The circle $$x^2 + y^2 = 2x$$ can be rewritten as $$x^2 - 2x + y^2 = 0$$ or $$(x-1)^2 + y^2 = 1$$, a circle centered at $(1,0)$ with radius $1$. 3. **Describe the region $D$:** It lies between the larger circle centered at the origin with radius $2$ and the smaller circle centered at $(1,0)$ with radius $1$, restricted to the first quadrant ($x \geq 0$, $y \geq 0$). 4. **Use polar coordinates:** Let $x = r \cos \theta$, $y = r \sin \theta$. - The circle $x^2 + y^2 = 4$ becomes $r = 2$. - The circle $(x-1)^2 + y^2 = 1$ becomes $r^2 - 2r \cos \theta + 1 = 1$ or $r = 2 \cos \theta$. 5. **Determine bounds:** Since $D$ is in the first quadrant, $\theta$ ranges from $0$ to $\frac{\pi}{2}$. For each $\theta$, $r$ ranges from the smaller circle $r = 2 \cos \theta$ to the larger circle $r = 2$. 6. **Set up the integral:** The integral becomes $$\int_0^{\frac{\pi}{2}} \int_{2 \cos \theta}^2 (r \cos \theta) \cdot r \, dr \, d\theta$$ Note: $x = r \cos \theta$ and $dA = r \, dr \, d\theta$. 7. **Simplify the integrand:** $$x \, dA = r \cos \theta \cdot r \, dr \, d\theta = r^2 \cos \theta \, dr \, d\theta$$ 8. **Evaluate the inner integral:** $$\int_{2 \cos \theta}^2 r^2 \cos \theta \, dr = \cos \theta \int_{2 \cos \theta}^2 r^2 \, dr = \cos \theta \left[ \frac{r^3}{3} \right]_{2 \cos \theta}^2 = \cos \theta \left( \frac{8}{3} - \frac{8 \cos^3 \theta}{3} \right) = \frac{8}{3} \cos \theta (1 - \cos^3 \theta)$$ 9. **Evaluate the outer integral:** $$\int_0^{\frac{\pi}{2}} \frac{8}{3} \cos \theta (1 - \cos^3 \theta) \, d\theta = \frac{8}{3} \int_0^{\frac{\pi}{2}} \cos \theta \, d\theta - \frac{8}{3} \int_0^{\frac{\pi}{2}} \cos^4 \theta \, d\theta$$ 10. **Calculate each integral:** - $$\int_0^{\frac{\pi}{2}} \cos \theta \, d\theta = \sin \theta \Big|_0^{\frac{\pi}{2}} = 1$$ - $$\int_0^{\frac{\pi}{2}} \cos^4 \theta \, d\theta = \frac{3\pi}{16}$$ (using reduction formulas or known integrals) 11. **Substitute back:** $$\frac{8}{3} (1) - \frac{8}{3} \cdot \frac{3\pi}{16} = \frac{8}{3} - \frac{8}{3} \cdot \frac{3\pi}{16} = \frac{8}{3} - \frac{8 \pi}{16} = \frac{8}{3} - \frac{\pi}{2}$$ 12. **Final answer:** $$\boxed{\frac{8}{3} - \frac{\pi}{2}}$$