To compute the flux of the vector field $\mathbf{F} = (xy, 4yz, 3zx)$ through the portion of the plane $3x + 2y + z = 6$ in the first octant, with a downward orientation, we'll follow these steps:

Step 1: Parameterize the Surface

The equation of the plane is: $3x + 2y + z = 6$ Solving for $z$ in terms of $x$ and $y$: $z = 6 - 3x - 2y$ This gives us a parameterization for $z$ on the surface in terms of $x$ and $y$.

Step 2: Find the Bounds in the First Octant

In the first octant, $x, y, z \geq 0$, so:

• When $x = 0$, $z = 6 - 2y$, meaning $y \leq 3$.
• When $y = 0$, $z = 6 - 3x$, meaning $x \leq 2$.

Thus, the region on the $xy$-plane is bounded by $0 \leq x \leq 2$ and $0 \leq y \leq 3 - \frac{3}{2}x$.

Step 3: Compute the Normal Vector

To find the flux, we need the normal vector. Since the surface is given by $z = 6 - 3x - 2y$, let: $\mathbf{r}(x, y) = (x, y, 6 - 3x - 2y)$ Then, the partial derivatives are: $\frac{\partial \mathbf{r}}{\partial x} = (1, 0, -3) \quad \text{and} \quad \frac{\partial \mathbf{r}}{\partial y} = (0, 1, -2)$ The cross product $\frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial y}$ gives the normal vector: $\mathbf{n} = \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -3 \\ 0 & 1 & -2 \end{matrix} \right| = (3, 2, 1)$ Since the orientation is downward, we take $\mathbf{n} = (-3, -2, -1)$.

Step 4: Set Up the Flux Integral

The flux $\iint_S \mathbf{F} \cdot d\mathbf{S}$ is given by: $\iint_D \mathbf{F} \cdot \mathbf{n} \, dA$ where $D$ is the projection of the surface onto the $xy$-plane, and $d\mathbf{S} = \mathbf{n} \, dA$.

Substitute $\mathbf{F} = (xy, 4yz, 3zx)$ and $\mathbf{n} = (-3, -2, -1)$: $\mathbf{F} \cdot \mathbf{n} = xy(-3) + 4y(6 - 3x - 2y)(-2) + 3x(6 - 3x - 2y)(-1)$

Step 5: Simplify the Integral and Evaluate

Given the setup,