To calculate the flux of the vector field $\mathbf{H} = x^2\mathbf{i} + y^2\mathbf{j} + z^2\mathbf{k}$ through the surface of the cube $S$, we use Gauss's Divergence Theorem:

$\int_S \mathbf{H} \cdot d\mathbf{A} = \int_V (\nabla \cdot \mathbf{H}) \, dV$

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

The divergence of $\mathbf{H} = x^2\mathbf{i} + y^2\mathbf{j} + z^2\mathbf{k}$ is: $\nabla \cdot \mathbf{H} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z^2)$ $\nabla \cdot \mathbf{H} = 2x + 2y + 2z$

Step 2: Set up the volume integral

The cube $S$ has one corner at the origin $(0, 0, 0)$ and extends to $(1, 1, 1)$, so the volume $V$ is defined by: $0 \leq x \leq 1, \, 0 \leq y \leq 1, \, 0 \leq z \leq 1$ The volume integral is: $\int_V (\nabla \cdot \mathbf{H}) \, dV = \int_0^1 \int_0^1 \int_0^1 (2x + 2y + 2z) \, dx \, dy \, dz$

Step 3: Break up the integral

We can separate the terms: $\int_0^1 \int_0^1 \int_0^1 (2x + 2y + 2z) \, dx \, dy \, dz = 2 \int_0^1 \int_0^1 \int_0^1 x \, dx \, dy \, dz + 2 \int_0^1 \int_0^1 \int_0^1 y \, dx \, dy \, dz + 2 \int_0^1 \int_0^1 \int_0^1 z \, dx \, dy \, dz$

Step 4: Evaluate each term

1. For the $x$-term: $\int_0^1 \int_0^1 \int_0^1 x \, dx \, dy \, dz = \int_0^1 \int_0^1 \left[\frac{x^2}{2}\right]_0^1 \, dy \, dz = \int_0^1 \int_0^1 \frac{1}{2} \, dy \, dz = \int_0^1 \frac{1}{2} \, dz = \frac{1}{2}$

2. For the $y$-term: $\int_0^1 \int_0^1 \int_0^1 y \, dx \, dy \, dz = \int_0^1 \int_0^1 y \, dy \, dz = \int_0^1 \left[\frac{y^2}{2}\right]_0^1 dz = \int_0^1 \frac{1}{2} dz = \frac{1}{2}$

3. For the $z$-term: $\int_0^1 \int_0^1 \int_0^1 z \, dx \, dy \, dz = \int_0^1 \int_0^1 z \, dz = \int_0^1 \left[\frac{z^2}{2}\right]_0^1 dy = \int_0^1 \frac{1}{2} dy = \frac{1}{2}$

Step 5: Sum the contributions

Now, sum the contributions from all terms: $2 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} = 1 + 1 + 1 = 3$

Final Answer:

The flux of $\mathbf{H}$ through the surface of the cube is: $\boxed{3}$

Let me know if you want further clarification or have more questions! 😊

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Expansion Questions:

1. How would the flux change if the vector field were $\mathbf{H} = xyz \mathbf{i} + x^2y \mathbf{j} + yz^2 \mathbf{k}$?
2. What is the physical interpretation of divergence in the context of this problem?
3. Can you compute the flux directly by evaluating the surface integral without using the divergence theorem?
4. How does the result change if the cube is shifted to have a corner at $(a, a, a)$?
5. What would the flux be if the vector field had an additional constant term $\mathbf{H} = x^2\mathbf{i} + y^2\mathbf{j} + z^2\mathbf{k} + \mathbf{c}$?

Tip: Always check the vector field's divergence before applying the divergence theorem—it simplifies the calculation dramatically!