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 equations. \[ y=\sqrt{x}+3, y=\frac{1}{4} x+3 \] \( M_{x}=\square \) \( M_{y}=\square \) \( (\bar{x}, \bar{y})=\square \)

To find the moments \( M_{x^{\prime}} \), \( M_{y^{\prime}} \) and the center of mass \( (\bar{x}, \bar{y}) \) for the lamina of uniform density \( \rho \) bounded by the curves \( y = \sqrt{x} + 3 \) and \( y = \frac{1}{4}x + 3 \), we will follow these steps: 1. **Find the points of intersection of the curves.** 2. **Set up the integrals for \( M_{x^{\prime}} \), \( M_{y^{\prime}} \), and the area \( A \).** 3. **Calculate \( M_{x^{\prime}} \), \( M_{y^{\prime}} \), and \( A \).** 4. **Calculate the coordinates of the center of mass \( (\bar{x}, \bar{y}) \).** ### Step 1: Find the points of intersection To find the points of intersection, we set the equations equal to each other: \[ \sqrt{x} + 3 = \frac{1}{4}x + 3 \] Subtracting 3 from both sides gives: \[ \sqrt{x} = \frac{1}{4}x \] Squaring both sides results in: \[ x = \frac{1}{16}x^2 \] Rearranging gives: \[ \frac{1}{16}x^2 - x = 0 \] Factoring out \( x \): \[ x\left(\frac{1}{16}x - 1\right) = 0 \] This gives \( x = 0 \) or \( x = 16 \). ### Step 2: Set up the integrals The area \( A \) between the curves from \( x = 0 \) to \( x = 16 \) is given by: \[ A = \int_{0}^{16} \left( \sqrt{x} + 3 - \left( \frac{1}{4}x + 3 \right) \right) \, dx \] This simplifies to: \[ A = \int_{0}^{16} \left( \sqrt{x} - \frac{1}{4}x \right) \, dx \] The moments \( M_{x^{\prime}} \) and \( M_{y^{\prime}} \) are given by: \[ M_{x^{\prime}} = \int_{0}^{16} \left( \sqrt{x} + 3 \right) \left( \sqrt{x} - \frac{1}{4}x \right) \, dx \] \[ M_{y^{\prime}} = \int_{0}^{16} \left( \sqrt{x} + 3 \right) \cdot x \, dx \] ### Step 3: Calculate \( A \), \( M_{x^{\prime}} \), and \( M_{y^{\prime}} \) Let's calculate these integrals. 1. **Calculate \( A \)**: \[ A = \int_{0}^{16} \left( \sqrt{x} - \frac{1}{4}x \right) \, dx = \int_{0}^{16} x^{1/2} \, dx - \frac{1}{4} \int_{0}^{16} x \, dx \] Calculating each integral: \[ \int x^{1/2} \, dx = \frac{2}{3}x^{3/2} \quad \text{and} \quad \int x \, dx = \frac{1}{2}x^2 \] Evaluating from 0 to 16: \[ \int_{0}^{16} x^{1/2} \, dx = \frac{2}{3}(16)^{3/2} = \frac{2}{3}(64) = \frac{128}{3} \] \[ \int_{0}^{16} x \, dx = \frac{1}{2}(16^2) = \frac{1}{2}(256) = 128 \] Thus, \[ A = \frac{128}{3} - \frac{1}{4}(128) = \frac{128}{3} - 32 = \frac{128 - 96}{3} = \frac{32}{3} \] 2. **Calculate \( M_{x^{\prime}} \)**: \[ M_{x^{\prime}} = \int_{0}^{16} \left( \sqrt{x} + 3 \right) \left( \sqrt{x} - \frac{1}{4}x \right) \, dx \] Expanding the integrand: \[ = \int_{0}^{16} \left( x - \frac{1}{4}x^{3/2} + 3\sqrt{x} - \frac{3}{4}x \right) \, dx \] Combining like terms: \[ = \int_{0}^{16} \left( -\frac{1}{4}x + 3\sqrt{x} - \frac{1}{4}x^{3/2} \right) \, dx \] Calculating each term: \[ \int_{0}^{16} x \, dx = 128, \quad \int_{0}^{16} \sqrt{x} \, dx = \frac{128}{3}, \quad \int_{0}^{16} x^{3/2} \, dx = \frac{2}{5}(16)^{5/2} = \frac{2}{5}(1024) = \frac{2048}{5} \] Thus, \[ M_{x^{\prime}} = -\frac{1}{4}(128) + 3\left(\frac{128}{3}\right) - \frac{3}{4}\left(\frac{2048}{5}\right) \] Calculating: \[ = -32 + 128 - \frac{1536}{20} = 96 - 76.8 = 19.2 \] 3. **Calculate \( M_{y^{\prime}} \)**: \[ M_{y^{\prime}} = \int_{0}^{16} \left( \sqrt{x} + 3 \right) x \, dx = \int_{0}^{16} \left( x^{3/2} + 3x \right) \, dx \] Calculating each term: \[ \int_{0}^{16} x^{3/2} \, dx = \frac{2048}{5}, \quad \int_{0}^{16} 3x \, dx = 3(128) = 384 \] Thus, \[ M_{y^{\prime}} = \frac{2048}{5} +