91Ó°ÊÓ

The given region is bounded by two curves: \(y = \sin x\) and \(y = 1 - \sin x\). First, plot these functions over the given interval \([\frac{\pi}{4}, \frac{3\pi}{4}]\), and then draw the region they define. Notice that the region is symmetric with respect to the line \(x=\frac{\pi}{2}\).

Since this plate has a constant density, we will assume a density \(\delta\). Now we need to find the mass of the plate by integrating over the region and multiplying by density. The mass of the plate, \(M\), can be found as follows: $$M = \delta \int_{\pi/4}^{3\pi/4} [(1 - \sin x) - (\sin x)] dx$$ Now, we can find the moments \(M_x\) and \(M_y\). The centroid of the plate will have coordinates \((\bar{x},\bar{y})\) given by \(\bar{x} = \frac{M_x}{M}\) and \(\bar{y} = \frac{M_y}{M}\). To compute the moments, we evaluate the following integrals: $$M_x = \delta \int_{\pi/4}^{3\pi/4} x[(1 - \sin x) - (\sin x)] dx$$ $$M_y = \delta \int_{\pi/4}^{3\pi/4} \frac{1}{2}[(1 - \sin x) + (\sin x)] [(1 - \sin x) - (\sin x)] dx$$

Observing that the region is symmetric with respect to the line \(x=\frac{\pi}{2}\), we can use this property to simplify the work. Based on symmetry, we know that \(\bar{x} = \frac{\pi}{2}\). Now we only need to compute \(\bar{y}\). To find \(M\): $$M = \delta \int_{\pi/4}^{3\pi/4} (1 - 2\sin x) dx = \delta \left[x - 2\int_{\pi/4}^{3\pi/4} \sin x dx\right]_{\pi/4}^{3\pi/4} = \delta\left(\frac{\pi}{2} - 2\left[\cos x\right]_{\pi/4}^{3\pi/4}\right) = \delta \pi$$ Now, we find \(M_y\): $$M_y = \delta \int_{\pi/4}^{3\pi/4} \frac{1}{2}(1 - 2\sin x)(1-\sin x) dx = \frac{\delta}{2} \int_{\pi/4}^{3\pi/4} (1 - \sin x)^2 dx$$ To solve this integral, use u-substitution with \(u = 1 - \sin x\), giving \(du = -\cos x dx\), and changing the bounds to correspond to \(u\) values. $$M_y = -\frac{\delta}{2} \int_{\sqrt{2}/2}^{1-\sqrt{2}/2} u^2 du = -\frac{\delta}{6}\left[u^3\right]_{\sqrt{2}/2}^{1-\sqrt{2}/2} = \frac{\delta}{6}\left(\frac{\sqrt{2}}{2} - \frac{1-\sqrt{2}}{2}\right) = \frac{\delta\pi}{6}$$ Now we can find the coordinates of the centroid: $$\bar{x} = \frac{M_x}{M} = \frac{\pi}{2}$$ $$\bar{y} = \frac{M_y}{M} = \frac{\pi}{6}$$ The mass and centroid of the thin plate are \(M = \delta\pi\) and \((\bar{x},\bar{y}) = \left(\frac{\pi}{2},\frac{\pi}{6}\right)\), respectively.