The problem asks us to find the mass $m$ and the center of mass $(\bar{x}, \bar{y})$ of a lamina bounded by the given curves and with the indicated density. We will solve problem number 5. The curves are $y = e^{-x}$, $y = 0$, $x = 0$, $x = 1$, and the density is $\delta(x, y) = y^2$.

1. Problem Description

The problem asks us to find the mass $m$ and the center of mass $(\bar{x}, \bar{y})$ of a lamina bounded by the given curves and with the indicated density. We will solve problem number

5. The curves are $y = e^{-x}$, $y = 0$, $x = 0$, $x = 1$, and the density is $\delta(x, y) = y^2$.

2. Solution Steps

First, we calculate the mass $m$ using the formula:

$m = \iint_R \delta(x, y) dA$

In our case, $R$ is the region bounded by $y = e^{-x}$, $y = 0$, $x = 0$, $x = 1$, and $\delta(x, y) = y^2$. Thus,

$m = \int_{0}^{1} \int_{0}^{e^{-x}} y^2 dy dx$

$m = \int_{0}^{1} \left[ \frac{y^3}{3} \right]_{0}^{e^{-x}} dx$

$m = \int_{0}^{1} \frac{e^{-3x}}{3} dx$

$m = \frac{1}{3} \int_{0}^{1} e^{-3x} dx$

$m = \frac{1}{3} \left[ -\frac{1}{3} e^{-3x} \right]_{0}^{1}$

$m = \frac{1}{3} \left( -\frac{1}{3} e^{-3} + \frac{1}{3} e^{0} \right)$

$m = \frac{1}{9} (1 - e^{-3})$

Next, we calculate the moment about the y-axis, $M_y$:

$M_y = \iint_R x \delta(x, y) dA$

$M_y = \int_{0}^{1} \int_{0}^{e^{-x}} x y^2 dy dx$

$M_y = \int_{0}^{1} x \left[ \frac{y^3}{3} \right]_{0}^{e^{-x}} dx$

$M_y = \int_{0}^{1} x \frac{e^{-3x}}{3} dx$

$M_y = \frac{1}{3} \int_{0}^{1} x e^{-3x} dx$

Using integration by parts, let $u = x$, $dv = e^{-3x} dx$, so $du = dx$, $v = -\frac{1}{3} e^{-3x}$.

$\int u dv = uv - \int v du$

$M_y = \frac{1}{3} \left[ -\frac{x}{3} e^{-3x} \right]_{0}^{1} - \frac{1}{3} \int_{0}^{1} -\frac{1}{3} e^{-3x} dx$

$M_y = \frac{1}{3} \left( -\frac{1}{3} e^{-3} - 0 \right) + \frac{1}{9} \int_{0}^{1} e^{-3x} dx$

$M_y = -\frac{1}{9} e^{-3} + \frac{1}{9} \left[ -\frac{1}{3} e^{-3x} \right]_{0}^{1}$

$M_y = -\frac{1}{9} e^{-3} + \frac{1}{9} \left( -\frac{1}{3} e^{-3} + \frac{1}{3} \right)$

$M_y = -\frac{1}{9} e^{-3} - \frac{1}{27} e^{-3} + \frac{1}{27}$

$M_y = \frac{1}{27} (1 - 4e^{-3})$

Then, we calculate the moment about the x-axis, $M_x$:

$M_x = \iint_R y \delta(x, y) dA$

$M_x = \int_{0}^{1} \int_{0}^{e^{-x}} y y^2 dy dx$

$M_x = \int_{0}^{1} \int_{0}^{e^{-x}} y^3 dy dx$

$M_x = \int_{0}^{1} \left[ \frac{y^4}{4} \right]_{0}^{e^{-x}} dx$

$M_x = \int_{0}^{1} \frac{e^{-4x}}{4} dx$

$M_x = \frac{1}{4} \int_{0}^{1} e^{-4x} dx$

$M_x = \frac{1}{4} \left[ -\frac{1}{4} e^{-4x} \right]_{0}^{1}$

$M_x = \frac{1}{4} \left( -\frac{1}{4} e^{-4} + \frac{1}{4} \right)$

$M_x = \frac{1}{16} (1 - e^{-4})$

Finally, we calculate the coordinates of the center of mass:

$\bar{x} = \frac{M_y}{m} = \frac{\frac{1}{27}(1 - 4e^{-3})}{\frac{1}{9}(1 - e^{-3})} = \frac{1 - 4e^{-3}}{3(1 - e^{-3})}$

$\bar{y} = \frac{M_x}{m} = \frac{\frac{1}{16}(1 - e^{-4})}{\frac{1}{9}(1 - e^{-3})} = \frac{9(1 - e^{-4})}{16(1 - e^{-3})}$

3. Final Answer

$m = \frac{1}{9}(1 - e^{-3})$

$\bar{x} = \frac{1 - 4e^{-3}}{3(1 - e^{-3})}$

$\bar{y} = \frac{9(1 - e^{-4})}{16(1 - e^{-3})}$