Let's solve the problem step by step.

(a) Calculating the Total Flux Through Each Face

We are calculating the flux of the vector field $\vec{v} = -3\hat{i} - 5\hat{j} - 2\hat{k}$ out of the cube $S$. The cube has side length 2, is centered at the origin, and its faces are parallel to the coordinate planes.

Flux Formula

The flux through a face of the cube is given by: $\text{Flux} = \int \int_S \vec{v} \cdot \vec{n} \, dA$ where $\vec{n}$ is the outward normal vector to the face, and $dA$ is the differential area element.

Since the vector field $\vec{v}$ is constant, the flux simplifies to: $\text{Flux} = (\vec{v} \cdot \vec{n}) \cdot A$ where $A$ is the area of the face. For a cube with side length 2, each face has area: $A = 2 \cdot 2 = 4$

Computing the Flux for Each Face

The faces are at $x = \pm 1$, $y = \pm 1$, and $z = \pm 1$. Let's compute the flux for each.

1. Face at $x = 1$:

   • Outward normal vector: $\vec{n} = \hat{i}$
   • $\vec{v} \cdot \vec{n} = (-3)(1) = -3$
   • Flux = $(-3) \cdot 4 = -12$

2. Face at $x = -1$:

   • Outward normal vector: $\vec{n} = -\hat{i}$
   • $\vec{v} \cdot \vec{n} = (-3)(-1) = 3$
   • Flux = $3 \cdot 4 = 12$

3. Face at $y = 1$:

   • Outward normal vector: $\vec{n} = \hat{j}$
   • $\vec{v} \cdot \vec{n} = (-5)(1) = -5$
   • Flux = $(-5) \cdot 4 = -20$

4. Face at $y = -1$:

   • Outward normal vector: $\vec{n} = -\hat{j}$
   • $\vec{v} \cdot \vec{n} = (-5)(-1) = 5$
   • Flux = $5 \cdot 4 = 20$

5. Face at $z = 1$:

   • Outward normal vector: $\vec{n} = \hat{k}$
   • $\vec{v} \cdot \vec{n} = (-2)(1) = -2$
   • Flux = $(-2) \cdot 4 = -8$

6. Face at $z = -1$:

   • Outward normal vector: $\vec{n} = -\hat{k}$
   • $\vec{v} \cdot \vec{n} = (-2)(-1) = 2$
   • Flux = $2 \cdot 4 = 8$

Total Flux

Now summing up the flux through all the faces: $\text{Total Flux} = -12 + 12 - 20 + 20 - 8 + 8 = 0$

(b) General Case for $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$

For a general constant vector field $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$, the flux through each face can be computed similarly. Due to symmetry:

• The flux through the face at $x = 1$ is $4a$, and at $x = -1$ is $-4a$.
• The flux through the face at $y = 1$ is $4b$, and at $y = -1$ is $-4b$.
• The flux through the face at $z = 1$ is $4c$, and at $z = -1$ is $-4c$.

The total flux is: $\text{Total Flux} = 4a - 4a + 4b - 4b + 4c - 4c = 0$

Explanation of Results

• The flux through the cube is zero because the vector field $\vec{v}$ is constant, and the outward flux through opposite faces cancels out. This symmetry ensures no net flux out of the cube.

Do you want further explanation or examples?

Here are 5 related questions:

1. How does the flux calculation change if the vector field is non-constant?
2. What is the flux through a sphere instead of a cube for a constant field?
3. Can divergence of a vector field explain the zero flux here?
4. How is the flux computed for faces not parallel to coordinate planes?
5. What happens to the flux if the cube's center is not at the origin?

Tip: Always check for symmetry in the geometry and vector field—it simplifies flux computations significantly.