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

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Divergence Theorem Statement:

The flux of a vector field $\mathbf{H}$ across a closed surface $S$ is given by: $\iint_S \mathbf{H} \cdot \mathbf{n} \, dS = \iiint_V (\nabla \cdot \mathbf{H}) \, dV$ where $V$ is the volume enclosed by $S$, and $\nabla \cdot \mathbf{H}$ is the divergence of $\mathbf{H}$.

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

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

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Step 2: Volume of the Cube

The cube has corners at the origin $(0,0,0)$ and $(1,1,1)$, so its volume $V$ is the region: $V = \{(x, y, z) \,|\, 0 \leq x, y, z \leq 1 \}$

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Step 3: Compute $\iiint_V (\nabla \cdot \mathbf{H}) \, dV$

Substitute $\nabla \cdot \mathbf{H} = 2x + 2y + 2z$ into the triple integral: $\iiint_V (\nabla \cdot \mathbf{H}) \, dV = \iiint_V (2x + 2y + 2z) \, dV$ Expand the integral: $\iiint_V (2x + 2y + 2z) \, dV = 2 \iiint_V x \, dV + 2 \iiint_V y \, dV + 2 \iiint_V z \, dV$

Each integral is computed separately over the cube $0 \leq x, y, z \leq 1$:

1. Integral of $x$: $\iiint_V x \, dV = \int_0^1 \int_0^1 \int_0^1 x \, dz \, dy \, dx = \int_0^1 \int_0^1 \left[ xz \right]_0^1 \, dy \, dx = \int_0^1 \int_0^1 x \, dy \, dx$ [ = \int_0^1 \left[ xy \right]_0^1 , dx = \int_0^1 x , dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} ]

2. Integral of $y$: By symmetry, the integral of $y$ is identical: $\iiint_V y \, dV = \frac{1}{2}$

3. Integral of $z$: Similarly, the integral of $z$ is: $\iiint_V z \, dV = \frac{1}{2}$

Substitute these results back: $\iiint_V (2x + 2y + 2z) \, dV = 2 \left(\frac{1}{2}\right) + 2 \left(\frac{1}{2}\right) + 2 \left(\frac{1}{2}\right) = 1 + 1 + 1 = 3$

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Final Answer:

The flux of the vector field $\mathbf{H}$ over the surface $S$ of the cube is: $\boxed{3}$

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Would you like more details about any of the steps? Here are some related questions:

1. How does the Divergence Theorem simplify flux calculations?
2. What happens if the cube's dimensions or position change?
3. Can the flux be computed directly without the Divergence Theorem?
4. How is symmetry used to simplify the integral calculations here?
5. What are the conditions for applying the Divergence Theorem?

Tip: Always check for symmetry in a problem before setting up integrals—it can significantly reduce computation.