The problem asks us 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 the density function $\delta(x,y) = y+1$.

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

2. Solution Steps

First, we calculate the mass $m$:

$$m = \int\int_R \delta(x,y) dA$$

where $R$ is the region bounded by the given curves.

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

Thus,

$$m = \int_0^4 \int_0^3 (y+1) dy dx = \int_0^4 [\frac{1}{2}y^2 + y]_0^3 dx = \int_0^4 (\frac{9}{2} + 3) dx = \int_0^4 \frac{15}{2} dx = \frac{15}{2} [x]_0^4 = \frac{15}{2}(4) = 30$$

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

$$M_x = \int\int_R y \delta(x,y) dA$$

$$M_y = \int\int_R x \delta(x,y) dA$$

So,

$$M_x = \int_0^4 \int_0^3 y(y+1) dy dx = \int_0^4 \int_0^3 (y^2+y) dy dx = \int_0^4 [\frac{1}{3}y^3 + \frac{1}{2}y^2]_0^3 dx = \int_0^4 (\frac{27}{3} + \frac{9}{2}) dx = \int_0^4 (9 + \frac{9}{2}) dx = \int_0^4 \frac{27}{2} dx = \frac{27}{2} [x]_0^4 = \frac{27}{2}(4) = 54$$

$$M_y = \int_0^4 \int_0^3 x(y+1) dy dx = \int_0^4 x \int_0^3 (y+1) dy dx = \int_0^4 x [\frac{1}{2}y^2 + y]_0^3 dx = \int_0^4 x (\frac{9}{2} + 3) dx = \int_0^4 x (\frac{15}{2}) dx = \frac{15}{2} \int_0^4 x dx = \frac{15}{2} [\frac{1}{2}x^2]_0^4 = \frac{15}{2} (\frac{16}{2}) = \frac{15}{2}(8) = 60$$

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

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

$$\bar{y} = \frac{M_x}{m} = \frac{54}{30} = \frac{9}{5} = 1.8$$

3. Final Answer

$m = 30$

$\bar{x} = 2$

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

The mass is $30$ and the center of mass is $(2, \frac{9}{5})$.