Find \( M_{x^{\prime}} M_{y^{\prime}} \) and \( (\bar{x}, \bar{y}) \) for the lamina of uniform density \( \rho \) bounded by the graphs of the equation: \[ y=\frac{1}{8} x, y=0, x=8 \] \( M_{x}= \) \( M_{y}= \) \( (\bar{x}, \bar{y})= \)

To find \( M_{x^{\prime}} M_{y^{\prime}} \) and \( (\bar{x}, \bar{y}) \) for the lamina bounded by the given equations, let's first interpret the region of integration. 1. The lines given are \( y = \frac{1}{8} x \), \( y = 0 \) (the x-axis), and the vertical line \( x = 8 \). The intersection of these lines forms a triangular region with vertices at \( (0, 0) \), \( (8, 0) \), and \( (8, 1) \) where \( y = \frac{1}{8} \times 8 = 1 \). 2. Next, we calculate the mass \( M \) for uniform density \( \rho \): \[ M = \int_A \rho \, dA = \rho \cdot \text{Area} \] The area \( A \) of the triangular region can be calculated as: \[ \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 8 \cdot 1 = 4 \] Thus, \[ M = \rho \cdot 4 \] 3. The moments \( M_x \) and \( M_y \): - For \( M_y \): \[ M_y = \int_A x \cdot \rho \, dA = \rho \int_0^8 \int_0^{\frac{1}{8} x} x \, dy \, dx \] Computing the inner integral: \[ \int_0^{\frac{1}{8} x} x \, dy = x \cdot \frac{1}{8} x = \frac{1}{8} x^2 \] Now, evaluating the outer integral: \[ M_y = \rho \int_0^8 \frac{1}{8} x^2 \, dx = \rho \cdot \frac{1}{8} \cdot \left[\frac{x^3}{3}\right]_0^8 = \rho \cdot \frac{1}{8} \cdot \frac{512}{3} = \frac{64}{3} \rho \] - For \( M_x \): \[ M_x = \int_A y \cdot \rho \, dA = \rho \int_0^8 \int_0^{\frac{1}{8} x} y \, dy \, dx \] Calculating the inner integral: \[ \int_0^{\frac{1}{8} x} y \, dy = \left[\frac{y^2}{2}\right]_0^{\frac{1}{8} x} = \frac{1}{2} \left(\frac{1}{8} x\right)^2 = \frac{x^2}{128} \] Thus, the outer integral: \[ M_x = \rho \int_0^8 \frac{x^2}{128} \, dx = \rho \cdot \frac{1}{128} \cdot \left[\frac{x^3}{3}\right]_0^8 = \rho \cdot \frac{1}{128} \cdot \frac{512}{3} = \frac{4}{3} \rho \] 4. Finally, we compute the center of mass \( (\bar{x}, \bar{y}) \): \[ \bar{x} = \frac{M_y}{M} = \frac{\frac{64}{3} \rho}{4 \rho} = \frac{16}{3} \] \[ \bar{y} = \frac{M_x}{M} = \frac{\frac{4}{3} \rho}{4 \rho} = \frac{1}{3} \] Thus, the results are: \[ M_{x} = \frac{4}{3} \rho, \quad M_{y} = \frac{64}{3} \rho, \quad (\bar{x}, \bar{y}) = \left(\frac{16}{3}, \frac{1}{3}\right) \]