91Ó°ÊÓ

From the given information, the integral setup for M_x, M_y, and A will be: \(M_x = \frac{1}{2} \int_{0}^{\pi} \int_{0}^{2} (r\cos(\theta))^2 r dr d\theta\) \(M_y = \frac{1}{2} \int_{0}^{\pi} \int_{0}^{2} (r\sin(\theta))^2 r dr d\theta\) \(A = \int_{0}^{\pi} \int_{0}^{2} r dr d\theta\)

Now, evaluate the integrals: \(M_x = \frac{1}{2} \int_{0}^{\pi} \int_{0}^{2} (r^3\cos^2(\theta)) dr d\theta = \frac{1}{2} \int_{0}^{\pi} (\frac{r^4}{4})|_0^{2} \cos^2(\theta) d\theta = \int_{0}^{\pi} 4\cos^2(\theta) d\theta\) \(M_y = \frac{1}{2} \int_{0}^{\pi} \int_{0}^{2} (r^3\sin^2(\theta)) dr d\theta = \frac{1}{2} \int_{0}^{\pi} (\frac{r^4}{4})|_0^{2} \sin^2(\theta) d\theta = \int_{0}^{\pi} 4\sin^2(\theta) d\theta\) \(A = \int_{0}^{\pi} \int_{0}^{2} r dr d\theta = \int_{0}^{\pi} (\frac{r^2}{2})|_0^{2} d\theta = \int_{0}^{\pi} 2 d\theta\) Using trigonometric identities and evaluating the integrals, we find: \(\int_{0}^{\pi} \cos^2(\theta) d\theta = \int_{0}^{\pi} \frac{1+\cos(2\theta)}{2} d\theta = \frac{\pi}{2}\) \(\int_{0}^{\pi} \sin^2(\theta) d\theta = \int_{0}^{\pi} \frac{1-\cos(2\theta)}{2} d\theta = \frac{\pi}{2}\) \(\int_{0}^{\pi} d\theta = \pi\) Therefore, \(M_x = 4\cdot\frac{\pi}{2} = 2\pi\), \(M_y = 4\cdot\frac{\pi}{2} = 2\pi\), and \(A = \pi \cdot 2 = 2\pi\).

Finally, calculate the centroid (x, y) coordinates: \(x = \frac{M_x}{A} = \frac{2\pi}{2\pi} = 1\) \(y = \frac{M_y}{A} = \frac{2\pi}{2\pi} = 1\) So, the centroid of the semicircular disk is \((1, 1)\).