We are asked to find the mass $m$ and the center of mass $(\bar{x}, \bar{y})$ of a lamina bounded by the curves $x=0$, $x=4$, $y=0$, and $y=3$. The density function is given by $\delta(x, y) = y + 1$.

Okay, I will solve problem #1 from the image.

1. Problem Description

We are asked to find the mass $m$ and the center of mass $(\bar{x}, \bar{y})$ of a lamina bounded by the curves $x=0$, $x=4$, $y=0$, and $y=3$. The density function is given by $\delta(x, y) = y + 1$.

2. Solution Steps

First, we find the mass $m$ using the double integral:

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

In this case,

$m = \int_{0}^{4} \int_{0}^{3} (y+1) \, dy \, dx$

We first evaluate the inner integral:

$\int_{0}^{3} (y+1) \, dy = \left[ \frac{1}{2}y^2 + y \right]_{0}^{3} = \frac{1}{2}(3^2) + 3 - 0 = \frac{9}{2} + 3 = \frac{9}{2} + \frac{6}{2} = \frac{15}{2}$

Now we evaluate the outer integral:

$m = \int_{0}^{4} \frac{15}{2} \, dx = \frac{15}{2} \int_{0}^{4} dx = \frac{15}{2} [x]_{0}^{4} = \frac{15}{2} (4 - 0) = \frac{15}{2} \cdot 4 = 15 \cdot 2 = 30$

Next, we find $M_y$ using the formula:

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

$M_y = \int_{0}^{4} \int_{0}^{3} x(y+1) \, dy \, dx$

First, we evaluate the inner integral:

$\int_{0}^{3} x(y+1) \, dy = x \int_{0}^{3} (y+1) \, dy = x \left[ \frac{1}{2}y^2 + y \right]_{0}^{3} = x \cdot \frac{15}{2}$

Now we evaluate the outer integral:

$M_y = \int_{0}^{4} x \cdot \frac{15}{2} \, dx = \frac{15}{2} \int_{0}^{4} x \, dx = \frac{15}{2} \left[ \frac{1}{2}x^2 \right]_{0}^{4} = \frac{15}{2} \cdot \frac{1}{2} (4^2 - 0^2) = \frac{15}{4} \cdot 16 = 15 \cdot 4 = 60$

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

Next, we find $M_x$ using the formula:

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

$M_x = \int_{0}^{4} \int_{0}^{3} y(y+1) \, dy \, dx$

First, we evaluate the inner integral:

$\int_{0}^{3} (y^2+y) \, dy = \left[ \frac{1}{3}y^3 + \frac{1}{2}y^2 \right]_{0}^{3} = \frac{1}{3}(3^3) + \frac{1}{2}(3^2) = \frac{27}{3} + \frac{9}{2} = 9 + \frac{9}{2} = \frac{18}{2} + \frac{9}{2} = \frac{27}{2}$

Now we evaluate the outer integral:

$M_x = \int_{0}^{4} \frac{27}{2} \, dx = \frac{27}{2} \int_{0}^{4} dx = \frac{27}{2} [x]_{0}^{4} = \frac{27}{2} (4-0) = \frac{27}{2} \cdot 4 = 27 \cdot 2 = 54$

$\bar{y} = \frac{M_x}{m} = \frac{54}{30} = \frac{27}{15} = \frac{9}{5}$

3. Final Answer

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