To evaluate the double integral $\iint_D xy \, dA$, where $D$ is the triangular region with vertices $(0, 0)$, $(4, 0)$, and $(8, 6)$, we'll proceed as follows:

Step 1: Identify the Boundaries of the Region $D$

The triangular region $D$ has vertices at:

• $(0, 0)$
• $(4, 0)$
• $(8, 6)$

To set up the bounds, we need the equation of the line that forms the hypotenuse of the triangle between $(0,0)$ and $(8,6)$.

Equation of the Hypotenuse Line

The line passing through $(0, 0)$ and $(8, 6)$ has a slope: $\text{slope} = \frac{6 - 0}{8 - 0} = \frac{3}{4}$ Thus, the equation of the line is: $y = \frac{3}{4}x$

Step 2: Set Up the Integral

Now we can set up the bounds for the integral. For each fixed $x$ between 0 and 8, $y$ ranges from 0 up to $\frac{3}{4}x$.

So the double integral becomes: $\iint_D xy \, dA = \int_{x=0}^{8} \int_{y=0}^{\frac{3}{4}x} xy \, dy \, dx$

Step 3: Evaluate the Inner Integral with Respect to $y$

$\int_{x=0}^{8} \int_{y=0}^{\frac{3}{4}x} xy \, dy \, dx$ First, integrate $xy$ with respect to $y$: $= \int_{x=0}^{8} x \left[ \frac{y^2}{2} \right]_{y=0}^{y=\frac{3}{4}x} \, dx$ $= \int_{x=0}^{8} x \cdot \frac{\left( \frac{3}{4}x \right)^2}{2} \, dx$ $= \int_{x=0}^{8} x \cdot \frac{\frac{9}{16}x^2}{2} \, dx$ $= \int_{x=0}^{8} x \cdot \frac{9}{32}x^2 \, dx$ $= \int_{x=0}^{8} \frac{9}{32} x^3 \, dx$

Step 4: Evaluate the Outer Integral with Respect to $x$

Now we integrate $\frac{9}{32} x^3$ with respect to $x$: $= \frac{9}{32} \int_{x=0}^{8} x^3 \, dx$ The antiderivative of $x^3$ is $\frac{x^4}{4}$, so: $= \frac{9}{32} \left[ \frac{x^4}{4} \right]_{0}^{8}$ $= \frac{9}{32} \cdot \frac{8^4}{4}$ Since $8^4 = 4096$: $= \frac{9}{32} \cdot \frac{4096}{4}$ $= \frac{9}{32} \cdot 1024$ $= 288$

Final Answer

$\iint_D xy \, dA = 288$

Would you like more details on any step or have any questions?

Here are five related questions:

1. How would we change the bounds if the triangular region had a different set of vertices?
2. What if the integrand was $x^2y$ instead of $xy$?
3. How does the slope of the line affect the integration bounds?
4. What is the geometric interpretation of the value obtained from this integral?
5. How would polar coordinates change the setup for a similar integral?

Tip: When setting up double integrals, start by carefully defining the region $D$ and its boundaries.