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 3. The given curves are $y = 0$, $y = \sin x$, and $0 \le x \le \pi$. The density function is $\delta(x, y) = y$.

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

3. The given curves are $y = 0$, $y = \sin x$, and $0 \le x \le \pi$. The density function is $\delta(x, y) = y$.

2. Solution Steps

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

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

In this case, $R$ is defined by $0 \le x \le \pi$ and $0 \le y \le \sin x$, and $\delta(x, y) = y$. Therefore,

$m = \int_{0}^{\pi} \int_{0}^{\sin x} y \, dy \, dx$

$m = \int_{0}^{\pi} \left[ \frac{1}{2} y^2 \right]_{0}^{\sin x} dx$

$m = \int_{0}^{\pi} \frac{1}{2} \sin^2 x \, dx$

Using the identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$,

$m = \frac{1}{2} \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} \, dx$

$m = \frac{1}{4} \int_{0}^{\pi} (1 - \cos(2x)) \, dx$

$m = \frac{1}{4} \left[ x - \frac{1}{2} \sin(2x) \right]_{0}^{\pi}$

$m = \frac{1}{4} \left[ (\pi - \frac{1}{2} \sin(2\pi)) - (0 - \frac{1}{2} \sin(0)) \right]$

$m = \frac{1}{4} [\pi - 0 - 0 + 0] = \frac{\pi}{4}$

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

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

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

First, let's find $\bar{x}$:

$\bar{x} = \frac{1}{m} \int_{0}^{\pi} \int_{0}^{\sin x} x y \, dy \, dx$

$\bar{x} = \frac{1}{\pi/4} \int_{0}^{\pi} x \left[ \frac{1}{2} y^2 \right]_{0}^{\sin x} dx$

$\bar{x} = \frac{4}{\pi} \int_{0}^{\pi} \frac{1}{2} x \sin^2 x \, dx$

$\bar{x} = \frac{2}{\pi} \int_{0}^{\pi} x \sin^2 x \, dx$

Using the identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$,

$\bar{x} = \frac{2}{\pi} \int_{0}^{\pi} x \frac{1 - \cos(2x)}{2} \, dx$

$\bar{x} = \frac{1}{\pi} \int_{0}^{\pi} (x - x \cos(2x)) \, dx$

$\bar{x} = \frac{1}{\pi} \left[ \int_{0}^{\pi} x \, dx - \int_{0}^{\pi} x \cos(2x) \, dx \right]$

$\int_{0}^{\pi} x \, dx = \left[ \frac{1}{2} x^2 \right]_{0}^{\pi} = \frac{\pi^2}{2}$

$\int_{0}^{\pi} x \cos(2x) \, dx$: Using integration by parts, let $u=x$, $dv = \cos(2x) dx$, then $du = dx$ and $v = \frac{1}{2} \sin(2x)$.

$\int_{0}^{\pi} x \cos(2x) \, dx = \left[ \frac{1}{2} x \sin(2x) \right]_{0}^{\pi} - \int_{0}^{\pi} \frac{1}{2} \sin(2x) \, dx$

$= \left[ \frac{1}{2} x \sin(2x) \right]_{0}^{\pi} - \frac{1}{2} \left[ -\frac{1}{2} \cos(2x) \right]_{0}^{\pi}$

$= \left[ \frac{1}{2} \pi \sin(2\pi) - 0 \right] + \frac{1}{4} [\cos(2\pi) - \cos(0)] = 0 + \frac{1}{4} [1 - 1] = 0$

So, $\bar{x} = \frac{1}{\pi} \left[ \frac{\pi^2}{2} - 0 \right] = \frac{\pi}{2}$

Next, let's find $\bar{y}$:

$\bar{y} = \frac{1}{m} \int_{0}^{\pi} \int_{0}^{\sin x} y^2 \, dy \, dx$

$\bar{y} = \frac{1}{\pi/4} \int_{0}^{\pi} \left[ \frac{1}{3} y^3 \right]_{0}^{\sin x} dx$

$\bar{y} = \frac{4}{\pi} \int_{0}^{\pi} \frac{1}{3} \sin^3 x \, dx$

$\bar{y} = \frac{4}{3\pi} \int_{0}^{\pi} \sin^3 x \, dx$

$\sin^3 x = \sin x (1 - \cos^2 x) = \sin x - \sin x \cos^2 x$

$\int_{0}^{\pi} \sin^3 x \, dx = \int_{0}^{\pi} (\sin x - \sin x \cos^2 x) \, dx$

$= \left[ -\cos x \right]_{0}^{\pi} - \int_{0}^{\pi} \sin x \cos^2 x \, dx$

$= [-\cos(\pi) - (-\cos(0))] - \int_{0}^{\pi} \sin x \cos^2 x \, dx$

$= [-(-1) - (-1)] - \int_{0}^{\pi} \sin x \cos^2 x \, dx = 2 - \int_{0}^{\pi} \sin x \cos^2 x \, dx$

Let $u = \cos x$, $du = -\sin x \, dx$. When $x=0$, $u=1$; when $x=\pi$, $u=-1$.

$\int_{0}^{\pi} \sin x \cos^2 x \, dx = \int_{1}^{-1} -u^2 \, du = \int_{-1}^{1} u^2 \, du = \left[ \frac{1}{3} u^3 \right]_{-1}^{1} = \frac{1}{3} (1 - (-1)) = \frac{2}{3}$

$\int_{0}^{\pi} \sin^3 x \, dx = 2 - \frac{2}{3} = \frac{4}{3}$

$\bar{y} = \frac{4}{3\pi} \left( \frac{4}{3} \right) = \frac{16}{9\pi}$

3. Final Answer

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

$\bar{x} = \frac{\pi}{2}$

$\bar{y} = \frac{16}{9\pi}$