The problem involves evaluating a double integral of the function $4xy$ over the triangular region $D$, with vertices at $(0, 0)$, $(1, 2)$, and $(0, 3)$.

Step 1: Identify the boundaries of the region $D$

We are given a triangular region with vertices at:

• $(0, 0)$
• $(1, 2)$
• $(0, 3)$

To describe this region in terms of inequalities, we first find the equation of the line between $(0, 0)$ and $(1, 2)$. The slope of this line is: $m = \frac{2 - 0}{1 - 0} = 2$ Thus, the equation of the line is: $y = 2x$

Now, the limits for $x$ will range from $0$ to $1$ (the $x$-coordinates of the triangular region), and for a given $x$, $y$ will range between $y = 2x$ and $y = 3$ (the vertical height of the triangle).

Step 2: Set up the double integral

The double integral is set up as: $\int_0^1 \int_{2x}^3 4xy \, dy \, dx$

Step 3: Evaluate the inner integral (with respect to $y$)

The inner integral is: $\int_{2x}^3 4xy \, dy$ To solve this, treat $x$ as a constant. The antiderivative of $4xy$ with respect to $y$ is: $\frac{4xy^2}{2} = 2xy^2$ Now, evaluate it at the bounds $y = 3$ and $y = 2x$: [ \left[ 2x(3)^2 \right] - \left[ 2x(2x)^2 \right] = 18x - 16x^3 ]

Step 4: Evaluate the outer integral (with respect to $x$)

Now, we integrate $18x - 16x^3$ with respect to $x$: $\int_0^1 (18x - 16x^3) \, dx$ The antiderivative of $18x$ is $9x^2$, and the antiderivative of $16x^3$ is $4x^4$. So we have: $9x^2 - 4x^4$ Evaluating this at the bounds $x = 1$ and $x = 0$: [ \left[ 9(1)^2 - 4(1)^4 \right] - \left[ 9(0)^2 - 4(0)^4 \right] = 9 - 4 = 5 ]

Thus, the value of the double integral is $\boxed{5}$.

Would you like more details on any of these steps? Here are 5 related questions that expand on this topic:

1. How do we determine the boundaries of a triangular region for a double integral?
2. What are the general steps for evaluating a double integral over a region?
3. How do changes in the function being integrated affect the calculation of double integrals?
4. What is the significance of the order of integration (i.e., integrating with respect to $y$ first, then $x$)?
5. How does the shape of the region (e.g., triangular, rectangular) influence the setup of the limits for the integral?

Tip: Always visualize the region of integration before setting up the limits for the double integral. This can help avoid mistakes with the boundaries.