We are tasked with using the Divergence Theorem to evaluate the surface integral

$\iint_S \mathbf{F} \cdot d\mathbf{S},$

where $\mathbf{F} = xy^2 \mathbf{i} + x^2y \mathbf{j} + y \mathbf{k}$, and $S$ is the surface of the cylinder $x^2 + y^2 = 1$ bounded by the planes $z = 1$ and $z = -1$, and also including the circular portions $x^2 + y^2 \leq 1$ when $z = \pm 1$.

Step 1: Divergence Theorem

The Divergence Theorem states: $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F}) \, dV,$ where $V$ is the volume enclosed by the surface $S$ and $\nabla \cdot \mathbf{F}$ is the divergence of the vector field.

Step 2: Calculate $\nabla \cdot \mathbf{F}$

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

1. $\frac{\partial}{\partial x}(xy^2) = y^2$,
2. $\frac{\partial}{\partial y}(x^2y) = x^2$,
3. $\frac{\partial}{\partial z}(y) = 0$.

Thus: $\nabla \cdot \mathbf{F} = y^2 + x^2.$

Step 3: Set up the volume integral

The volume $V$ is the region bounded by the cylinder $x^2 + y^2 = 1$, and the planes $z = 1$ and $z = -1$. In cylindrical coordinates ($x = r\cos\theta$, $y = r\sin\theta$, $z = z$), this is described as: $0 \leq r \leq 1, \quad 0 \leq \theta \leq 2\pi, \quad -1 \leq z \leq 1.$

The volume element in cylindrical coordinates is: $dV = r \, dr \, d\theta \, dz.$

Substituting $\nabla \cdot \mathbf{F} = y^2 + x^2$ into the integral: $\iiint_V (y^2 + x^2) \, dV = \iiint_V (r^2 \sin^2\theta + r^2 \cos^2\theta) \, dV.$

Using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$: $y^2 + x^2 = r^2.$

Thus, the integral becomes: $\iiint_V r^2 \, dV = \int_{0}^{2\pi} \int_{0}^{1} \int_{-1}^{1} r^2 r \, dz \, dr \, d\theta.$

Step 4: Compute the integral

1. Integrate with respect to $z$: $\int_{-1}^{1} dz = 2.$

2. Integrate with respect to $r$: $\int_{0}^{1} r^3 \, dr = \left[\frac{r^4}{4}\right]_{0}^{1} = \frac{1}{4}.$

3. Integrate with respect to $\theta$: $\int_{0}^{2\pi} d\theta = 2\pi.$

Combining these results: $\iiint_V r^2 \, dV = 2 \cdot \frac{1}{4} \cdot 2\pi = \frac{\pi}{2}.$

Final Answer:

$\iint_S \mathbf{F} \cdot d\mathbf{S} = \frac{\pi}{2}.$

Let me know if you'd like further clarifications! Here are 5 related questions for practice:

1. How does the Divergence Theorem simplify surface integrals?
2. What is the role of cylindrical coordinates in this problem?
3. What are other applications of the Divergence Theorem in physics?
4. How do you compute divergence in spherical coordinates?
5. What happens if $\mathbf{F}$ has no divergence ($\nabla \cdot \mathbf{F} = 0$)?

Tip: Always ensure the surface completely encloses a volume before applying the Divergence Theorem!