Question: (ii) Hence or otherwise calculate the outward flux of $F$
$$\iint_S F \cdot dS$$
across the boundary of the region $S$, where $F(x,y,z) = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k}$ and $S$ is the region cut from the solid cylinder $x^2 + y^2 \leq a^2$ by the planes $z=0$ and $z=1$.
1. **State the problem:** Calculate the outward flux of the vector field $F(x,y,z) = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k}$ across the closed surface $S$ which is the boundary of the solid cylinder defined by $x^2 + y^2 \leq a^2$, bounded between $z=0$ and $z=1$.
2. **Use the Divergence Theorem:** The outward flux across $S$ can be computed using the Divergence Theorem:
$$\iint_S F \cdot dS = \iiint_V \nabla \cdot F \, dV$$
where $V$ is the volume enclosed by $S$.
3. **Calculate the divergence $\nabla \cdot F$:**
$$\nabla \cdot F = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z^2) = 2x + 2y + 2z$$
4. **Set up the volume integral:** The volume $V$ is the cylinder $x^2 + y^2 \leq a^2$ with $0 \leq z \leq 1$.
It is easier to use cylindrical coordinates:
$$x = r \cos \theta, \quad y = r \sin \theta, \quad z = z$$
with $0 \leq r \leq a$, $0 \leq \theta \leq 2\pi$, $0 \leq z \leq 1$.
The divergence in cylindrical coordinates becomes:
$$\nabla \cdot F = 2r \cos \theta + 2r \sin \theta + 2z$$
The volume element is:
$$dV = r \, dr \, d\theta \, dz$$
5. **Write the integral:**
$$\iiint_V \nabla \cdot F \, dV = \int_0^1 \int_0^{2\pi} \int_0^a (2r \cos \theta + 2r \sin \theta + 2z) r \, dr \, d\theta \, dz$$
6. **Separate the integral:**
$$= \int_0^1 \int_0^{2\pi} \int_0^a 2r^2 \cos \theta \, dr \, d\theta \, dz + \int_0^1 \int_0^{2\pi} \int_0^a 2r^2 \sin \theta \, dr \, d\theta \, dz + \int_0^1 \int_0^{2\pi} \int_0^a 2zr \, dr \, d\theta \, dz$$
7. **Evaluate the $r$ integrals:**
$$\int_0^a r^2 \, dr = \frac{a^3}{3}, \quad \int_0^a r \, dr = \frac{a^2}{2}$$
So the integrals become:
$$2 \frac{a^3}{3} \int_0^1 \int_0^{2\pi} \cos \theta \, d\theta \, dz + 2 \frac{a^3}{3} \int_0^1 \int_0^{2\pi} \sin \theta \, d\theta \, dz + 2 \int_0^1 z \int_0^{2\pi} \frac{a^2}{2} \, d\theta \, dz$$
8. **Evaluate the $\theta$ integrals:**
$$\int_0^{2\pi} \cos \theta \, d\theta = 0, \quad \int_0^{2\pi} \sin \theta \, d\theta = 0, \quad \int_0^{2\pi} d\theta = 2\pi$$
9. **Simplify:**
The first two integrals vanish because their $\theta$ integrals are zero.
The third integral becomes:
$$2 \int_0^1 z \cdot \frac{a^2}{2} \cdot 2\pi \, dz = 2 \pi a^2 \int_0^1 z \, dz$$
10. **Evaluate the $z$ integral:**
$$\int_0^1 z \, dz = \frac{1}{2}$$
11. **Final flux value:**
$$\text{Flux} = 2 \pi a^2 \cdot \frac{1}{2} = \pi a^2$$
**Answer:** The outward flux of $F$ across the surface $S$ is
$$\boxed{\pi a^2}$$