91Ó°ÊÓ

The given equation is \(x^{2}+y^{2}=16\). This is the equation for a circle centered at the origin with a radius of 4. Since we only want the region in the first quadrant, we are looking at the quarter-circle in quadrant I.

Let the density of the plate be \(\rho\). The mass of a thin plate with constant density is given by the formula \(M = \rho \cdot A\), where A is the area of the plate. The area of the quarter-circle in the first quadrant is given by \(A = \frac{1}{4}\pi (4)^{2} = 4\pi\). Therefore, the mass of the plate is \(M = \rho \cdot 4\pi\).

The centroid (center of mass) of the plate is given by the coordinates \((\bar{x}, \bar{y})\). By symmetry, we can determine that the centroid will lie on the line \(y = x\) in the first quadrant. The centroid can be found using the following formulas: \(\bar{x} = \frac{1}{A} \iint_R x \,dA\) and \(\bar{y} = \frac{1}{A} \iint_R y \,dA\), where R is the quarter-circle region. Since the region is in polar coordinates, we'll convert the integrals into polar form. The Jacobian of the polar coordinates is \(r\), so we need to multiply the integrals by \(r\). The limits of integration for polar coordinates are: \(r\) will go from 0 to 4, and \(\theta\) will go from 0 to \(\frac{\pi}{2}\). Thus, the integrals become: $$ \bar{x} = \frac{1}{4\pi} \int_0^{\frac{\pi}{2}} \int_0^4 r \cos{\theta} \cdot r \, dr \, d\theta \\ \bar{y} = \frac{1}{4\pi} \int_0^{\frac{\pi}{2}} \int_0^4 r \sin{\theta} \cdot r \, dr \, d\theta $$ Calculating these integrals, we get: $$ \bar{x} = \frac{4}{\pi} \\ \bar{y} = \frac{4}{\pi} $$ So the centroid of the quarter-circle in the first quadrant is \((\frac{4}{\pi}, \frac{4}{\pi})\).

For this step, draw the quarter-circle within the first quadrant bounded by the equation \(x^{2}+y^{2}=16\). Mark the center of the circle (0,0) and the end-points on the x and y-axes (4,0) and (0,4). Finally, mark the centroid of the plate as a point \((\frac{4}{\pi}, \frac{4}{\pi})\), which lies along the line \(y=x\).