∬_d x da, where d is the region in the first - quadrant that lies between the circles x² + y² = 4 and x² +…

Explanation:

Step1: Convert to polar coordinates

In polar coordinates, $x = r\cos\theta$, $dA=r\ dr\ d\theta$, $x^{2}+y^{2}=r^{2}$, and $x^{2}+y^{2}=2x$ becomes $r^{2}=2r\cos\theta$ or $r = 2\cos\theta$, and $x^{2}+y^{2}=4$ gives $r = 2$. For the first - quadrant, $0\leqslant\theta\leqslant\frac{\pi}{2}$. The region $D$ is described by $2\cos\theta\leqslant r\leqslant2$ and $0\leqslant\theta\leqslant\frac{\pi}{2}$.

Step2: Set up the double - integral

The double - integral $\iint_{D}x\ dA$ in polar coordinates is $\int_{0}^{\frac{\pi}{2}}\int_{2\cos\theta}^{2}(r\cos\theta)r\ dr\ d\theta=\int_{0}^{\frac{\pi}{2}}\cos\theta\left[\frac{r^{3}}{3}\right]_{r = 2\cos\theta}^{r = 2}d\theta$.

Step3: Evaluate the inner - integral

$\int_{0}^{\frac{\pi}{2}}\cos\theta\left(\frac{8}{3}-\frac{8\cos^{3}\theta}{3}\right)d\theta=\frac{8}{3}\int_{0}^{\frac{\pi}{2}}\cos\theta(1 - \cos^{3}\theta)d\theta=\frac{8}{3}\int_{0}^{\frac{\pi}{2}}(\cos\theta-\cos^{4}\theta)d\theta$.

Step4: Use integral formulas

We know that $\int\cos\theta d\theta=\sin\theta+C$ and $\int\cos^{n}\theta d\theta=\frac{1}{n}\cos^{n - 1}\theta\sin\theta+\frac{n - 1}{n}\int\cos^{n - 2}\theta d\theta$. For $n = 4$, $\int\cos^{4}\theta d\theta=\frac{1}{4}\cos^{3}\theta\sin\theta+\frac{3}{4}\int\cos^{2}\theta d\theta$, and $\int\cos^{2}\theta d\theta=\frac{1}{2}(\theta+\frac{1}{2}\sin2\theta)+C$. Then $\int_{0}^{\frac{\pi}{2}}\cos\theta d\theta=[\sin\theta]{0}^{\frac{\pi}{2}} = 1$ and $\int{0}^{\frac{\pi}{2}}\cos^{4}\theta d\theta=\frac{3}{4}\times\frac{\pi}{4}=\frac{3\pi}{16}$.

Step5: Calculate the result

$\frac{8}{3}\left(1-\frac{3\pi}{16}\right)=\frac{8}{3}-\frac{\pi}{2}$.

Answer:

$\frac{8}{3}-\frac{\pi}{2}$