The problem asks to find the mass $m$ and the center of mass $(\bar{x}, \bar{y})$ of the lamina bounded by the curve $r = 2\sin\theta$ with density $\delta(r, \theta) = r$.

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 curve $r = 2\sin\theta$ with density $\delta(r, \theta) = r$.

2. Solution Steps

First, we need to find the mass $m$. The formula for mass in polar coordinates is:

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

The curve $r = 2\sin\theta$ traces a circle in the $xy$-plane centered at $(0, 1)$ with radius $1$. Therefore, $\theta$ ranges from $0$ to $\pi$. Also, $r$ ranges from $0$ to $2\sin\theta$. So,

$m = \int_0^\pi \int_0^{2\sin\theta} r \cdot r dr d\theta = \int_0^\pi \int_0^{2\sin\theta} r^2 dr d\theta$

$m = \int_0^\pi [\frac{1}{3}r^3]_0^{2\sin\theta} d\theta = \int_0^\pi \frac{1}{3} (2\sin\theta)^3 d\theta = \frac{8}{3} \int_0^\pi \sin^3\theta d\theta$

Now, we need to evaluate $\int_0^\pi \sin^3\theta d\theta$. We can use the identity $\sin^2\theta = 1 - \cos^2\theta$

$\int_0^\pi \sin^3\theta d\theta = \int_0^\pi \sin\theta (1 - \cos^2\theta) d\theta$

Let $u = \cos\theta$, then $du = -\sin\theta d\theta$. When $\theta = 0$, $u = 1$. When $\theta = \pi$, $u = -1$.

$\int_0^\pi \sin\theta (1 - \cos^2\theta) d\theta = -\int_1^{-1} (1 - u^2) du = \int_{-1}^1 (1 - u^2) du$

$= [u - \frac{1}{3}u^3]_{-1}^1 = (1 - \frac{1}{3}) - (-1 - \frac{1}{3}(-1)) = (1 - \frac{1}{3}) - (-1 + \frac{1}{3}) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$

So, $m = \frac{8}{3} \cdot \frac{4}{3} = \frac{32}{9}$

Next, we find the center of mass $(\bar{x}, \bar{y})$. The formulas are:

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

In polar coordinates, $x = r\cos\theta$ and $y = r\sin\theta$. Thus,

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

$\bar{x} = \frac{1}{m} \int_0^\pi [\frac{1}{4}r^4]_0^{2\sin\theta} \cos\theta d\theta = \frac{1}{m} \int_0^\pi \frac{1}{4} (2\sin\theta)^4 \cos\theta d\theta = \frac{1}{m} \int_0^\pi 4\sin^4\theta \cos\theta d\theta$

Let $u = \sin\theta$, then $du = \cos\theta d\theta$. When $\theta = 0$, $u = 0$. When $\theta = \pi$, $u = 0$.

So, $\int_0^\pi 4\sin^4\theta \cos\theta d\theta = 4 \int_0^0 u^4 du = 0$

Thus, $\bar{x} = \frac{1}{m} \cdot 0 = 0$

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

$\bar{y} = \frac{1}{m} \int_0^\pi [\frac{1}{4}r^4]_0^{2\sin\theta} \sin\theta d\theta = \frac{1}{m} \int_0^\pi \frac{1}{4} (2\sin\theta)^4 \sin\theta d\theta = \frac{1}{m} \int_0^\pi 4\sin^5\theta d\theta$

$\bar{y} = \frac{4}{m} \int_0^\pi \sin^5\theta d\theta$

Now, we need to evaluate $\int_0^\pi \sin^5\theta d\theta$.

$\int_0^\pi \sin^5\theta d\theta = \int_0^\pi \sin\theta (1 - \cos^2\theta)^2 d\theta = \int_0^\pi \sin\theta (1 - 2\cos^2\theta + \cos^4\theta) d\theta$

Let $u = \cos\theta$, then $du = -\sin\theta d\theta$. When $\theta = 0$, $u = 1$. When $\theta = \pi$, $u = -1$.

$\int_0^\pi \sin\theta (1 - 2\cos^2\theta + \cos^4\theta) d\theta = -\int_1^{-1} (1 - 2u^2 + u^4) du = \int_{-1}^1 (1 - 2u^2 + u^4) du$

$= [u - \frac{2}{3}u^3 + \frac{1}{5}u^5]_{-1}^1 = (1 - \frac{2}{3} + \frac{1}{5}) - (-1 + \frac{2}{3} - \frac{1}{5}) = 2(1 - \frac{2}{3} + \frac{1}{5}) = 2(\frac{15 - 10 + 3}{15}) = 2(\frac{8}{15}) = \frac{16}{15}$

So, $\bar{y} = \frac{4}{m} \cdot \frac{16}{15} = \frac{4}{\frac{32}{9}} \cdot \frac{16}{15} = \frac{36}{32} \cdot \frac{16}{15} = \frac{9}{8} \cdot \frac{16}{15} = \frac{9}{1} \cdot \frac{2}{15} = \frac{18}{15} = \frac{6}{5}$

3. Final Answer

$m = \frac{32}{9}$

$\bar{x} = 0$

$\bar{y} = \frac{6}{5}$

The mass is $\frac{32}{9}$ and the center of mass is $(0, \frac{6}{5})$.