To calculate the outward flux of $\mathbf{F} = \langle 3xy^2, 2xe^z, z^3 \rangle$ across the surface $S$, the Divergence Theorem states:

$\text{Flux of } \mathbf{F} \text{ across } S = \iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_V (\nabla \cdot \mathbf{F}) \, dV$

Step 1: Compute the divergence of $\mathbf{F}$

$\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(3xy^2) + \frac{\partial}{\partial y}(2xe^z) + \frac{\partial}{\partial z}(z^3)$

• $\frac{\partial}{\partial x}(3xy^2) = 3y^2$
• $\frac{\partial}{\partial y}(2xe^z) = 0$ (no $y$-dependence in $2xe^z$)
• $\frac{\partial}{\partial z}(z^3) = 3z^2$

Thus: $\nabla \cdot \mathbf{F} = 3y^2 + 3z^2$

Step 2: Set up the bounds of integration

The volume $V$ is bounded by:

• The cylinder $y^2 + z^2 = 4$, meaning $-2 \leq y \leq 2$ and $-\sqrt{4-y^2} \leq z \leq \sqrt{4-y^2}$
• The planes $x = 3$ and $x = 4$, meaning $3 \leq x \leq 4$

Step 3: Write the flux integral

Using the divergence theorem:

= \iiint_V (3y^2 + 3z^2) \, dx \, dy \, dz$$ In terms of $$dx \, dy \, dz$$, the integral becomes: $$\int_{x=3}^{x=4} \int_{y=-2}^{y=2} \int_{z=-\sqrt{4-y^2}}^{z=\sqrt{4-y^2}} (3y^2 + 3z^2) \, dz \, dy \, dx$$ ### Final Expression: $$\text{Flux of } \mathbf{F} = \int_{3}^{4} \int_{-2}^{2} \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} (3y^2 + 3z^2) \, dz \, dy \, dx$$ Would you like a detailed evaluation of this triple integral? --- **Questions to Expand:** 1. How does the Divergence Theorem simplify flux calculations over closed surfaces? 2. What are common scenarios where the Divergence Theorem is preferred over direct surface integrals? 3. What is the geometrical interpretation of divergence in this context? 4. How would the calculation change if the cylinder was centered around a different axis? 5. Can this integral be solved exactly, or is numerical approximation better? **Tip:** Always visualize the region of integration to confirm bounds before setting up the integral!