To compute the flux of the vector field $\mathbf{F} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ through the curved surface of the cylinder $x^2 + y^2 = 4$, we proceed as follows:

Problem Breakdown

1. Surface Geometry:

   • The cylinder is described by $x^2 + y^2 = 4$, extending vertically in the $z$-direction.
   • The surface is bounded:
     • Below by the plane $x + y + z = 2$,
     • Above by the plane $x + y + z = 8$.

2. Orientation:

   • The flux is to be computed through the curved surface of the cylinder, oriented away from the $z$-axis.

3. Flux Formula: The flux of $\mathbf{F}$ through a surface $S$ is given by: $\text{Flux} = \iint_S \mathbf{F} \cdot \mathbf{n} \, dS$ where $\mathbf{n}$ is the outward-pointing unit normal vector, and $dS$ is the area element.

Approach Using Divergence Theorem

Since the curved surface forms part of a closed surface with the planes $x + y + z = 2$ and $x + y + z = 8$, we can apply the Divergence Theorem: $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_V (\nabla \cdot \mathbf{F}) \, dV$ Here, $V$ is the volume enclosed by the cylinder and the planes.

Step 1: Compute $\nabla \cdot \mathbf{F}$

$\mathbf{F} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}, \quad \nabla \cdot \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3$

Step 2: Volume of the Region

The volume is the intersection of:

• The cylinder $x^2 + y^2 = 4$, which has a radius of 2,
• The planes $x + y + z = 2$ and $x + y + z = 8$.

Rewriting $x + y + z = k$ in terms of $z$: $z = k - x - y$ The limits for $x^2 + y^2 \leq 4$ describe a disk in the $xy$-plane of radius 2.

The volume can be expressed as: $V = \int_{k=2}^{k=8} \int_{x^2+y^2 \leq 4} (z_\text{top} - z_\text{bottom}) \, dA$ where $z_\text{top} = 8 - x - y$ and $z_\text{bottom} = 2 - x - y$.

Step 3: Solve for $V$

The height of the cylinder between the planes is constant: $\Delta z = (8 - x - y) - (2 - x - y) = 6$

The base area is the disk $x^2 + y^2 \leq 4$: $\text{Area} = \int_{x^2+y^2 \leq 4} dA = \pi (2)^2 = 4\pi$

Thus, the volume is: $V = \Delta z \cdot \text{Area} = 6 \cdot 4\pi = 24\pi$

Step 4: Flux Calculation

The flux is: $\text{Flux} = \iiint_V (\nabla \cdot \mathbf{F}) \, dV = \iiint_V 3 \, dV = 3 \cdot V = 3 \cdot 24\pi = 72\pi$

Final Answer:

$\text{Flux} = 72\pi$

Would you like detailed explanations of any step? Here are 5 related questions to expand your understanding:

1. How does the Divergence Theorem simplify flux calculations for closed surfaces?
2. What is the physical significance of $\nabla \cdot \mathbf{F}$ (the divergence) in this context?
3. How can we directly parametrize the curved surface of the cylinder for verification?
4. Why does the orientation of the surface affect the direction of the normal vector?
5. How is the volume of a region bounded by curved and planar surfaces calculated?

Tip: Always check if the region is closed when considering the Divergence Theorem—it saves effort by converting a surface integral into a volume integral!