We are asked to find the mass $m$ and center of mass $(\bar{x}, \bar{y})$ of the lamina bounded by the curve $r = 1 + \cos\theta$ with density $\delta(r, \theta) = r$.

1. Problem Description

We are asked to find the mass $m$ and center of mass $(\bar{x}, \bar{y})$ of the lamina bounded by the curve $r = 1 + \cos\theta$ with density $\delta(r, \theta) = r$.

2. Solution Steps

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

$m = \iint_R \delta(r, \theta) \, dA$

Since the region is described in polar coordinates, we have $dA = r \, dr \, d\theta$. The curve $r = 1 + \cos\theta$ traces out a cardioid, and to traverse the entire region, $\theta$ varies from $0$ to $2\pi$ and $r$ varies from $0$ to $1 + \cos\theta$. Thus,

$m = \int_0^{2\pi} \int_0^{1+\cos\theta} r \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^{1+\cos\theta} r^2 \, dr \, d\theta$

First, we integrate with respect to $r$:

$\int_0^{1+\cos\theta} r^2 \, dr = \frac{1}{3} r^3 \Big|_0^{1+\cos\theta} = \frac{1}{3} (1+\cos\theta)^3$

Now, we integrate with respect to $\theta$:

$m = \frac{1}{3} \int_0^{2\pi} (1+\cos\theta)^3 \, d\theta = \frac{1}{3} \int_0^{2\pi} (1 + 3\cos\theta + 3\cos^2\theta + \cos^3\theta) \, d\theta$

We know that $\int_0^{2\pi} \cos\theta \, d\theta = 0$.

Also, $\int_0^{2\pi} \cos^2\theta \, d\theta = \int_0^{2\pi} \frac{1+\cos(2\theta)}{2} \, d\theta = \frac{1}{2} \left[ \theta + \frac{1}{2}\sin(2\theta) \right]_0^{2\pi} = \frac{1}{2} (2\pi) = \pi$.

Furthermore, $\int_0^{2\pi} \cos^3\theta \, d\theta = \int_0^{2\pi} \cos\theta (1-\sin^2\theta) \, d\theta = \int_0^{2\pi} (\cos\theta - \cos\theta \sin^2\theta) \, d\theta = \left[\sin\theta - \frac{1}{3}\sin^3\theta \right]_0^{2\pi} = 0$.

Therefore,

$m = \frac{1}{3} \int_0^{2\pi} (1 + 3\cos^2\theta) \, d\theta = \frac{1}{3} (2\pi + 3\pi) = \frac{5\pi}{3}$

Now we calculate $\bar{x}$ and $\bar{y}$:

$\bar{x} = \frac{1}{m} \iint_R x \delta(r, \theta) \, dA$ and $\bar{y} = \frac{1}{m} \iint_R y \delta(r, \theta) \, dA$

We have $x = r \cos\theta$ and $y = r \sin\theta$, so

$\bar{x} = \frac{1}{m} \int_0^{2\pi} \int_0^{1+\cos\theta} (r\cos\theta) r \cdot r \, dr \, d\theta = \frac{1}{m} \int_0^{2\pi} \int_0^{1+\cos\theta} r^3 \cos\theta \, dr \, d\theta$

$\bar{y} = \frac{1}{m} \int_0^{2\pi} \int_0^{1+\cos\theta} (r\sin\theta) r \cdot r \, dr \, d\theta = \frac{1}{m} \int_0^{2\pi} \int_0^{1+\cos\theta} r^3 \sin\theta \, dr \, d\theta$

$\bar{x} = \frac{3}{5\pi} \int_0^{2\pi} \int_0^{1+\cos\theta} r^3 \cos\theta \, dr \, d\theta = \frac{3}{5\pi} \int_0^{2\pi} \frac{1}{4} (1+\cos\theta)^4 \cos\theta \, d\theta$

$= \frac{3}{20\pi} \int_0^{2\pi} (1+4\cos\theta + 6\cos^2\theta + 4\cos^3\theta + \cos^4\theta)\cos\theta \, d\theta$

$= \frac{3}{20\pi} \int_0^{2\pi} (\cos\theta + 4\cos^2\theta + 6\cos^3\theta + 4\cos^4\theta + \cos^5\theta) \, d\theta$

We know $\int_0^{2\pi} \cos\theta d\theta = 0$, $\int_0^{2\pi} \cos^3\theta d\theta = 0$, $\int_0^{2\pi} \cos^5\theta d\theta = 0$, and $\int_0^{2\pi} \cos^2\theta d\theta = \pi$. Also $\int_0^{2\pi} \cos^4\theta \, d\theta = \frac{3\pi}{4}$.

So, $\bar{x} = \frac{3}{20\pi} (0 + 4\pi + 0 + 4 \cdot \frac{3\pi}{4} + 0) = \frac{3}{20\pi} (4\pi + 3\pi) = \frac{3}{20\pi} (7\pi) = \frac{21}{20}$.

$\bar{y} = \frac{3}{5\pi} \int_0^{2\pi} \int_0^{1+\cos\theta} r^3 \sin\theta \, dr \, d\theta = \frac{3}{5\pi} \int_0^{2\pi} \frac{1}{4} (1+\cos\theta)^4 \sin\theta \, d\theta$

$= \frac{3}{20\pi} \int_0^{2\pi} (1+\cos\theta)^4 \sin\theta \, d\theta$

Let $u = 1 + \cos\theta$. Then $du = -\sin\theta \, d\theta$.

$\bar{y} = \frac{3}{20\pi} \int_{2}^{2} -u^4 \, du = 0$.

3. Final Answer

$m = \frac{5\pi}{3}$

$(\bar{x}, \bar{y}) = (\frac{21}{20}, 0)$