91Ó°ÊÓ

We need to find the mass and center of mass of a lamina. The lamina is shaped by the region bounded by the curves \(y = e^x\), \(y = 0\), \(x = 0\), and \(x = 1\). The density function of the lamina is \(\rho(x, y) = y\).

The mass \(M\) of the lamina is given by the double integral \( M = \int\int_D \rho(x, y) \, dA \). Since \(\rho(x, y) = y\), the mass is \( M = \int_0^1 \int_0^{e^x} y \, dy \, dx \). First, integrate with respect to \(y\):\[\int_0^{e^x} y \, dy = \left[ \frac{y^2}{2} \right]_0^{e^x} = \frac{(e^x)^2}{2} - 0 = \frac{e^{2x}}{2}\]Next, integrate with respect to \(x\):\[M = \int_0^1 \frac{e^{2x}}{2} \, dx = \frac{1}{2} \int_0^1 e^{2x} \, dx\]\[M = \frac{1}{2} \cdot \frac{1}{2} \left[ e^{2x} \right]_0^1 = \frac{1}{4} (e^2 - 1)\]

The center of mass \((\bar{x}, \bar{y})\) is given by the coordinates:\[\bar{x} = \frac{1}{M} \int\int_D x \rho(x, y) \, dA\]\[\bar{y} = \frac{1}{M} \int\int_D y \rho(x, y) \, dA\]Since \(\rho(x, y) = y\), we have \(x\rho(x,y) = xy\) and \(y\rho(x,y) = y^2\):1. **Calculate \(\bar{x}\):**\[\bar{x} = \frac{1}{M} \int_0^1 \int_0^{e^x} xy \, dy \, dx\]First, integrate with respect to \(y\):\[\int_0^{e^x} xy \, dy = x \left[ \frac{y^2}{2} \right]_0^{e^x} = x \cdot \frac{(e^x)^2}{2} = x \cdot \frac{e^{2x}}{2}\]Next, integrate with respect to \(x\):\[\int_0^1 \frac{x e^{2x}}{2} \, dx\ = \frac{1}{2} \left[ \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} \right]_0^1\ = \frac{1}{2} \left(\frac{1}{2} e^2 - \frac{1}{4} e^2 + \frac{1}{4} \right)\ = \frac{e^2/4}{2} - \frac{e^2/8} + \frac{1/8} \].2. **Calculate \(\bar{y}\):**\[\bar{y} = \frac{1}{M} \int_0^1 \int_0^{e^x} y^2 \, dy \, dx\]First, integrate with respect to \(y\):\[\int_0^{e^x} y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^{e^x} = \frac{(e^x)^3}{3} = \frac{e^{3x}}{3}\]Next, integrate with respect to \(x\):\[\int_0^1 \frac{e^{3x}}{3} \, dx = \frac{1}{3} \left[ \frac{1}{3} e^{3x} \right]_0^1 = \frac{1}{9} (e^3 - 1)\].

After performing the integrals for \(\bar{x}\) and \(\bar{y}\), substitute the mass \(M = \frac{1}{4} (e^2 - 1)\) to find the center of mass:\[\bar{x} = \frac{\text{integral result for } \bar{x}}{M}\]\[\bar{y} = \frac{1}{M} \cdot \frac{1}{9} (e^3 - 1)\]Calculate \(\bar{x}\) and \(\bar{y}\) to get the final values.