The problem requires evaluating the double integral $\int \int_R 6xy \, dA$, where the region $R$ is bounded by:

• $y = 6 - 3x$,
• $y = 0$ (the x-axis),
• $x = 4 - \frac{y^2}{9}$ (parabolic curve),
• In the first quadrant.

1. Region of Integration:

To find the correct graph:

• The line $y = 6 - 3x$ has a y-intercept of 6 and a slope of -3. This line meets the x-axis at $x = 2$.
• The parabola $x = 4 - \frac{y^2}{9}$ opens leftward with a vertex at $(4, 0)$. It crosses the x-axis at $x = 4$ and crosses the line $y = 6 - 3x$ somewhere below $y = 6$.
• The region is bounded by the x-axis and lies entirely in the first quadrant.

From the options, Graph D correctly represents this region.

2. Evaluate the Integral:

Now, we evaluate the double integral $\int \int_R 6xy \, dA$. The bounds of integration are:

• For $y$, it ranges from 0 to where the parabola intersects the line $y = 6 - 3x$. Setting $6 - 3x = 0$, we solve for $x = 2$.
• The parabola $x = 4 - \frac{y^2}{9}$ gives the limits for $x$, with $x$ ranging between $0$ and $4 - \frac{y^2}{9}$.

Therefore, the double integral becomes: $\int_0^6 \int_0^{4 - \frac{y^2}{9}} 6xy \, dx \, dy$

Step 1: Inner Integral over $x$: