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 9, the lamina is bounded by $r = 1$, $r = 2$, $\theta = 0$, and $\theta = \pi$, with density $\delta(r, \theta) = \frac{1}{r}$.

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 9, the lamina is bounded by $r = 1$, $r = 2$, $\theta = 0$, and $\theta = \pi$, with density $\delta(r, \theta) = \frac{1}{r}$.

2. Solution Steps

First, we find the mass $m$. The mass is given by the double integral of the density over the region. In polar coordinates, $dA = r dr d\theta$. So,

$m = \int_{0}^{\pi} \int_{1}^{2} \delta(r, \theta) r dr d\theta = \int_{0}^{\pi} \int_{1}^{2} \frac{1}{r} r dr d\theta = \int_{0}^{\pi} \int_{1}^{2} 1 dr d\theta$.

Evaluating the inner integral:

$\int_{1}^{2} 1 dr = [r]_{1}^{2} = 2 - 1 = 1$.

Evaluating the outer integral:

$m = \int_{0}^{\pi} 1 d\theta = [\theta]_{0}^{\pi} = \pi - 0 = \pi$.

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

$\bar{x} = \frac{M_y}{m}$, where $M_y = \iint x \delta(r, \theta) dA$. Since $x = r \cos\theta$,

$M_y = \int_{0}^{\pi} \int_{1}^{2} (r \cos\theta) (\frac{1}{r}) r dr d\theta = \int_{0}^{\pi} \int_{1}^{2} r \cos\theta dr d\theta = \int_{0}^{\pi} \cos\theta \left[ \frac{r^2}{2} \right]_{1}^{2} d\theta = \int_{0}^{\pi} \cos\theta (\frac{4}{2} - \frac{1}{2}) d\theta = \frac{3}{2} \int_{0}^{\pi} \cos\theta d\theta = \frac{3}{2} [\sin\theta]_{0}^{\pi} = \frac{3}{2} (\sin\pi - \sin0) = \frac{3}{2} (0 - 0) = 0$.

Therefore, $\bar{x} = \frac{M_y}{m} = \frac{0}{\pi} = 0$.

$\bar{y} = \frac{M_x}{m}$, where $M_x = \iint y \delta(r, \theta) dA$. Since $y = r \sin\theta$,

$M_x = \int_{0}^{\pi} \int_{1}^{2} (r \sin\theta) (\frac{1}{r}) r dr d\theta = \int_{0}^{\pi} \int_{1}^{2} r \sin\theta dr d\theta = \int_{0}^{\pi} \sin\theta \left[ \frac{r^2}{2} \right]_{1}^{2} d\theta = \int_{0}^{\pi} \sin\theta (\frac{4}{2} - \frac{1}{2}) d\theta = \frac{3}{2} \int_{0}^{\pi} \sin\theta d\theta = \frac{3}{2} [-\cos\theta]_{0}^{\pi} = \frac{3}{2} (-\cos\pi + \cos0) = \frac{3}{2} (-(-1) + 1) = \frac{3}{2} (1 + 1) = \frac{3}{2} (2) = 3$.

Therefore, $\bar{y} = \frac{M_x}{m} = \frac{3}{\pi}$.

3. Final Answer

$m = \pi$, $\bar{x} = 0$, $\bar{y} = \frac{3}{\pi}$.