To find the mass $M$ and the center of mass $(\bar{x}, \bar{y})$ of the lamina occupying the region $D$, we proceed by setting up integrals based on the density function $\rho(x, y) = 5(x + y)$.

1. Mass of the lamina $M$:

   The mass $M$ is given by: $M = \iint_D \rho(x, y) \, dA$ where $\rho(x, y) = 5(x + y)$.

2. Determining the region $D$:

   The region $D$ is a triangle with vertices at $(0, 0)$, $(2, 1)$, and $(0, 3)$. Let's determine the bounds for $x$ and $y$ within this region.

   The line connecting $(0, 3)$ and $(2, 1)$ has the equation: $y = -\frac{1}{2}x + 3$ So, for a fixed $x$-value, $y$ ranges from $y = 0$ to $y = -\frac{1}{2}x + 3$, and $x$ ranges from $0$ to $2$.

3. Setting up the mass integral $M$:

   Now, we can express $M$ as: $M = \int_0^2 \int_0^{-\frac{1}{2}x + 3} 5(x + y) \, dy \, dx$

4. Coordinates for the center of mass $(\bar{x}, \bar{y})$:

   The coordinates $(\bar{x}, \bar{y})$ are calculated as follows: $\bar{x} = \frac{1}{M} \iint_D x \rho(x, y) \, dA = \frac{1}{M} \int_0^2 \int_0^{-\frac{1}{2}x + 3} 5x(x + y) \, dy \, dx$ $\bar{y} = \frac{1}{M} \iint_D y \rho(x, y) \, dA = \frac{1}{M} \int_0^2 \int_0^{-\frac{1}{2}x + 3} 5y(x + y) \, dy \, dx$

Let's compute these integrals step-by-step.The mass $M$ of the lamina is: $M = 55$

The coordinates for the center of mass $(\bar{x}, \bar{y})$ are: $\bar{x} \approx 1.045$ $\bar{y} \approx 1.485$

If you need further details or clarifications, please let me know!

Here are 5 related questions that expand on this topic:

1. How would the center of mass change if the density function were constant?
2. What impact would a different shape for $D$ (e.g., a rectangle) have on the solution approach?
3. How can polar coordinates be used in finding the center of mass of circular regions?
4. How is the density function related to physical properties in real-life applications?
5. What methods can be applied to find the moment of inertia of this lamina about the origin?

Tip: When integrating over triangular regions, setting up the integration bounds carefully based on vertices simplifies the calculation process.