The problem asks to find the mass $m$ and the center of mass $(\bar{x}, \bar{y})$ of the lamina bounded by the curves $x=0$, $x=4$, $y=0$, $y=3$, with density function $\delta(x, y) = y+1$.

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 curves $x=0$, $x=4$, $y=0$, $y=3$, with density function $\delta(x, y) = y+1$.

2. Solution Steps

First, we calculate the mass $m$ using the formula:

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

In this case, the region $R$ is a rectangle defined by $0 \le x \le 4$ and $0 \le y \le 3$. Thus,

$m = \int_0^4 \int_0^3 (y+1) \, dy \, dx$

$m = \int_0^4 \left[ \frac{y^2}{2} + y \right]_0^3 \, dx$

$m = \int_0^4 \left( \frac{9}{2} + 3 \right) \, dx$

$m = \int_0^4 \frac{15}{2} \, dx$

$m = \frac{15}{2} \left[ x \right]_0^4$

$m = \frac{15}{2} (4)$

$m = 30$

Next, we calculate the moments $M_x$ and $M_y$:

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

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

$M_x = \int_0^4 \int_0^3 y(y+1) \, dy \, dx$

$M_x = \int_0^4 \int_0^3 (y^2 + y) \, dy \, dx$

$M_x = \int_0^4 \left[ \frac{y^3}{3} + \frac{y^2}{2} \right]_0^3 \, dx$

$M_x = \int_0^4 \left( \frac{27}{3} + \frac{9}{2} \right) \, dx$

$M_x = \int_0^4 \left( 9 + \frac{9}{2} \right) \, dx$

$M_x = \int_0^4 \frac{27}{2} \, dx$

$M_x = \frac{27}{2} \left[ x \right]_0^4$

$M_x = \frac{27}{2} (4)$

$M_x = 54$

$M_y = \int_0^4 \int_0^3 x(y+1) \, dy \, dx$

$M_y = \int_0^4 x \int_0^3 (y+1) \, dy \, dx$

$M_y = \int_0^4 x \left[ \frac{y^2}{2} + y \right]_0^3 \, dx$

$M_y = \int_0^4 x \left( \frac{9}{2} + 3 \right) \, dx$

$M_y = \int_0^4 x \left( \frac{15}{2} \right) \, dx$

$M_y = \frac{15}{2} \int_0^4 x \, dx$

$M_y = \frac{15}{2} \left[ \frac{x^2}{2} \right]_0^4$

$M_y = \frac{15}{2} \left( \frac{16}{2} \right)$

$M_y = \frac{15}{2} (8)$

$M_y = 60$

Finally, we can calculate the center of mass $(\bar{x}, \bar{y})$:

$\bar{x} = \frac{M_y}{m}$

$\bar{y} = \frac{M_x}{m}$

$\bar{x} = \frac{60}{30} = 2$

$\bar{y} = \frac{54}{30} = \frac{9}{5}$

3. Final Answer

The mass $m$ is $30$. The center of mass $(\bar{x}, \bar{y})$ is $(2, \frac{9}{5})$.