The problem asks us 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 number 4. The bounding curves are $y = 1/x$, $y = x$, $y = 0$, and $x = 2$. The density function is given by $\delta(x, y) = x$.

1. Problem Description

The problem asks us 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 number

4. The bounding curves are $y = 1/x$, $y = x$, $y = 0$, and $x = 2$. The density function is given by $\delta(x, y) = x$.

2. Solution Steps

First, we need to find the intersection points of the curves. The intersection of $y=x$ and $y=1/x$ is found by setting $x = 1/x$, which gives $x^2 = 1$, so $x = 1$ (since $x>0$). Thus, the intersection point is $(1, 1)$. The intersection of $y=1/x$ and $x=2$ is at $(2, 1/2)$.

The mass $m$ is given by the double integral of the density function over the region:

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

In our case, $m = \int_1^2 \int_0^{1/x} x \, dy \, dx + \int_0^1 \int_0^x x \, dy \, dx$.

Evaluating the inner integrals:

$\int_0^{1/x} x \, dy = x [y]_0^{1/x} = x(1/x) = 1$

$\int_0^x x \, dy = x [y]_0^x = x(x) = x^2$

Now, evaluating the outer integrals:

$\int_1^2 1 \, dx = [x]_1^2 = 2 - 1 = 1$

$\int_0^1 x^2 \, dx = [\frac{1}{3}x^3]_0^1 = \frac{1}{3}$

So, $m = 1 + \frac{1}{3} = \frac{4}{3}$.

Next, we find $\bar{x}$ and $\bar{y}$.

$\bar{x} = \frac{1}{m} \iint_R x \delta(x, y) \, dA = \frac{1}{m} \iint_R x^2 \, dA$.

$\bar{y} = \frac{1}{m} \iint_R y \delta(x, y) \, dA = \frac{1}{m} \iint_R xy \, dA$.

$\iint_R x^2 \, dA = \int_1^2 \int_0^{1/x} x^2 \, dy \, dx + \int_0^1 \int_0^x x^2 \, dy \, dx$

$\int_0^{1/x} x^2 \, dy = x^2 [y]_0^{1/x} = x^2 (1/x) = x$

$\int_0^x x^2 \, dy = x^2 [y]_0^x = x^3$

$\int_1^2 x \, dx = [\frac{1}{2} x^2]_1^2 = \frac{1}{2}(4 - 1) = \frac{3}{2}$

$\int_0^1 x^3 \, dx = [\frac{1}{4} x^4]_0^1 = \frac{1}{4}$

$\iint_R x^2 \, dA = \frac{3}{2} + \frac{1}{4} = \frac{6}{4} + \frac{1}{4} = \frac{7}{4}$

$\bar{x} = \frac{1}{m} (\frac{7}{4}) = \frac{3}{4} \cdot \frac{7}{4} = \frac{21}{16}$.

$\iint_R xy \, dA = \int_1^2 \int_0^{1/x} xy \, dy \, dx + \int_0^1 \int_0^x xy \, dy \, dx$

$\int_0^{1/x} xy \, dy = x [\frac{1}{2} y^2]_0^{1/x} = \frac{x}{2x^2} = \frac{1}{2x}$

$\int_0^x xy \, dy = x [\frac{1}{2} y^2]_0^x = \frac{x^3}{2}$

$\int_1^2 \frac{1}{2x} \, dx = \frac{1}{2} [\ln x]_1^2 = \frac{1}{2} (\ln 2 - \ln 1) = \frac{1}{2} \ln 2$

$\int_0^1 \frac{x^3}{2} \, dx = \frac{1}{2} [\frac{1}{4} x^4]_0^1 = \frac{1}{8}$

$\iint_R xy \, dA = \frac{1}{2} \ln 2 + \frac{1}{8}$

$\bar{y} = \frac{1}{m} (\frac{1}{2} \ln 2 + \frac{1}{8}) = \frac{3}{4} (\frac{1}{2} \ln 2 + \frac{1}{8}) = \frac{3}{8} \ln 2 + \frac{3}{32}$.

3. Final Answer

$m = \frac{4}{3}$

$\bar{x} = \frac{21}{16}$

$\bar{y} = \frac{3}{8} \ln 2 + \frac{3}{32}$