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 the cube, we apply the Divergence Theorem:

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

Step 1: Compute the divergence of $\mathbf{H}$

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

Step 2: Define the volume of integration

The cube $V$ has corners at $(0,0,0)$ and $(1,1,1)$, with side length 1. The volume is given by: $V = \{(x, y, z) \, | \, 0 \leq x \leq 1, \, 0 \leq y \leq 1, \, 0 \leq z \leq 1\}$

Step 3: Set up the integral

Using the Divergence Theorem: $\int_S \mathbf{H} \cdot d\mathbf{A} = \int_V (\nabla \cdot \mathbf{H}) \, dV$ Substitute $\nabla \cdot \mathbf{H} = 2x + 2y + 2z$: $\int_S \mathbf{H} \cdot d\mathbf{A} = \int_0^1 \int_0^1 \int_0^1 (2x + 2y + 2z) \, dx \, dy \, dz$

Step 4: Compute the integral

Break the integral into three parts: $\int_0^1 \int_0^1 \int_0^1 (2x + 2y + 2z) \, dx \, dy \, dz = \int_0^1 \int_0^1 \int_0^1 2x \, dx \, dy \, dz + \int_0^1 \int_0^1 \int_0^1 2y \, dx \, dy \, dz + \int_0^1 \int_0^1 \int_0^1 2z \, dx \, dy \, dz$

First term:

$\int_0^1 \int_0^1 \int_0^1 2x \, dx \, dy \, dz = \int_0^1 \int_0^1 \left[ x^2 \right]_0^1 \, dy \, dz = \int_0^1 \int_0^1 1 \, dy \, dz = \int_0^1 \int_0^1 1 \, dy \, dz = 1$

Second term:

#### Second term: \[ \int_0^1 \int_0^1 \int_0^1 2y \, dx \, dy \, dz$$ Since $$2y$$ does not depend on $$x$$, the integration over $$x$$ contributes only a factor of 1: $$\int_0^1 \int_0^1 \int_0^1 2y \, dx \, dy \, dz = \int_0^1 \int_0^1 2y \cdot 1 \, dy \, dz = \int_0^1 \int_0^1 2y \, dy \, dz$$ Now integrate with respect to $$y$$: $$\int_0^1 2y \, dy = \left[y^2\right]_0^1 = 1^2 - 0^2 = 1$$ Thus: $$\int_0^1 \int_0^1 \int_0^1 2y \, dx \, dy \, dz = \int_0^1 1 \, dz = 1$$ #### Third term: $$\int_0^1 \int_0^1 \int_0^1 2z \, dx \, dy \, dz$$ Similarly, $$2z$$ does not depend on $$x$$ or $$y$$, so integration over these variables contributes factors of 1: $$\int_0^1 \int_0^1 \int_0^1 2z \, dx \, dy \, dz = \int_0^1 \int_0^1 2z \cdot 1 \, dy \, dz = \int_0^1 \int_0^1 2z \, dz = \int_0^1 2z \, dz$$ Now integrate with respect to $$z$$: $$\int_0^1 2z \, dz = \left[z^2\right]_0^1 = 1^2 - 0^2 = 1$$ #### Total flux: Adding all three contributions: $$\int_S \mathbf{H} \cdot d\mathbf{A} = 1 + 1 + 1 = 3$$ ### Final Answer: The flux of the vector field $$\mathbf{H}$$ through the surface $$S$$ of the cube is: $$\boxed{3}$$ Would you like further details or explanations? Here are some related questions to deepen your understanding: 1. What is the physical significance of the flux of a vector field? 2. How does the Divergence Theorem simplify surface integrals in general? 3. How would the calculation change if the cube's dimensions or orientation were different? 4. Can this result be verified by directly computing the surface integral? 5. What happens if the vector field $$\mathbf{H}$$ is non-linear in a more complex way? **Tip:** When working with symmetry or simple geometric boundaries, consider using the Divergence Theorem to reduce computation!