The problem asks to find the mass $m$ and the center of mass $(\bar{x}, \bar{y})$ of the lamina bounded by the given curves and with the indicated density for problem 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 to find the mass $m$ and the center of mass $(\bar{x}, \bar{y})$ of the lamina bounded by the given curves and with the indicated density for problem

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 find the mass $m$ using the double integral:

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

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

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

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

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

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

Next, we find $M_x$ using the double integral:

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

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

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

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

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

$$M_x = \frac{1}{4} \left( -\frac{1}{4} e^{-4} - \left( -\frac{1}{4} e^{0} \right) \right)$$

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

Then, we find $M_y$ using the double integral:

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

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

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

We use integration by parts: $u = x$, $dv = e^{-3x} dx$. Then $du = dx$ and $v = -\frac{1}{3} e^{-3x}$.

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

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

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

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

$$M_y = -\frac{1}{9} e^{-3} + \frac{1}{9} \left( -\frac{1}{3} e^{-3} + \frac{1}{3} \right) = -\frac{1}{9} e^{-3} - \frac{1}{27} e^{-3} + \frac{1}{27} = \frac{1}{27} (1 - 4e^{-3})$$

Now, we can find the center of mass $(\bar{x}, \bar{y})$:

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

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

3. Final Answer

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

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

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