We are tasked to compute the flux of the vector field $\mathbf{H} = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k}$ through the surface $S$ of a cube with corners at $(0,0,0)$ and $(1,1,1)$, meaning it has side lengths of 1 and its faces are aligned with the coordinate axes.

The flux through the surface $S$ is given by the surface integral:

$\iint_S \mathbf{H} \cdot \mathbf{dA}$

Solution Steps

1. Gauss's Divergence Theorem

The divergence theorem allows us to convert a surface integral to a volume integral over the cube. Mathematically,

$\iint_S \mathbf{H} \cdot \mathbf{dA} = \iiint_V \nabla \cdot \mathbf{H} \, dV$

Here:

• $V$ is the volume enclosed by the surface $S$,
• $\nabla \cdot \mathbf{H}$ is the divergence of $\mathbf{H}$.

2. Calculate the Divergence $\nabla \cdot \mathbf{H}$

The vector field $\mathbf{H}$ is given as:

$\mathbf{H} = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k}$

The divergence of $\mathbf{H}$ is:

$\nabla \cdot \mathbf{H} = \frac{\partial}{\partial x} (x^2) + \frac{\partial}{\partial y} (y^2) + \frac{\partial}{\partial z} (z^2)$

Computing each term:

• $\frac{\partial}{\partial x} (x^2) = 2x$,
• $\frac{\partial}{\partial y} (y^2) = 2y$,
• $\frac{\partial}{\partial z} (z^2) = 2z$.

Thus,

$\nabla \cdot \mathbf{H} = 2x + 2y + 2z.$

3. Set Up the Volume Integral

The volume integral over the cube $0 \leq x, y, z \leq 1$ is:

$\iiint_V \nabla \cdot \mathbf{H} \, dV = \iiint_V (2x + 2y + 2z) \, dV.$

Split the integral into three parts:

$\iiint_V (2x + 2y + 2z) \, dV = 2 \iiint_V x \, dV + 2 \iiint_V y \, dV + 2 \iiint_V z \, dV.$

The region $V$ is the cube $0 \leq x, y, z \leq 1$. The integral of $x$, $y$, or $z$ over this cube is:

$\int_0^1 \int_0^1 \int_0^1 x \, dx \, dy \, dz = \int_0^1 x \, dx \cdot \int_0^1 dy \cdot \int_0^1 dz.$

Each component evaluates as follows:

• $\int_0^1 x \, dx = \frac{x^2}{2} \Big|_0^1 = \frac{1}{2}$,
• $\int_0^1 dy = 1$,
• $\int_0^1 dz = 1$.

Thus,

$\iiint_V x \, dV = \frac{1}{2}, \quad \iiint_V y \, dV = \frac{1}{2}, \quad \iiint_V z \, dV = \frac{1}{2}.$

4. Combine the Results

Now summing up the contributions:

$\iiint_V (2x + 2y + 2z) \, dV = 2 \left( \frac{1}{2} \right) + 2 \left( \frac{1}{2} \right) + 2 \left( \frac{1}{2} \right).$

Simplify:

$\iiint_V (2x + 2y + 2z) \, dV = 2 \cdot \frac{3}{2} = 3.$

Final Answer:

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

$\boxed{3}.$

Do you need further clarification or details? Let me know!

Here are 5 related questions to explore further:

1. How is Gauss's divergence theorem applied to other vector fields?
2. What is the physical interpretation of flux in vector fields?
3. How would the result change if the vector field $\mathbf{H}$ had an additional component?
4. Can the flux be computed directly via surface integrals for each face of the cube?
5. What happens if the domain $V$ is a sphere instead of a cube?

Tip: Gauss's theorem simplifies surface integrals greatly when symmetry and enclosed volumes are involved!