To evaluate $\iint\limits_S (\nabla \times \mathbf{F}) \cdot d \mathbf{S}$, we will use Stokes' theorem, which states:

$\iint\limits_S (\nabla \times \mathbf{F}) \cdot d \mathbf{S} = \oint\limits_C \mathbf{F} \cdot d\mathbf{r},$

where $C$ is the boundary curve of the surface $S$, oriented such that the normal vector to $S$ follows the right-hand rule with respect to $C$. The surface $S$ is given as $z = 5 - x^2 - y^2$ for $z \geq 1$, which is a portion of a paraboloid.

Steps:

1. Boundary Curve $C$: The boundary $C$ occurs where $z = 1$, which corresponds to $1 = 5 - x^2 - y^2$, or $x^2 + y^2 = 4$. This is a circle in the $z = 1$ plane of radius 2.

2. Parameterize $C$: Let the curve be parameterized as: $\mathbf{r}(t) = 2\cos t\, \mathbf{i} + 2\sin t\, \mathbf{j} + \mathbf{k}, \quad t \in [0, 2\pi].$ Then, $d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = (-2\sin t\, \mathbf{i} + 2\cos t\, \mathbf{j}) dt.$

3. Evaluate $\mathbf{F}$ along $C$: Substitute $x = 2\cos t$, $y = 2\sin t$, and $z = 1$ into $\mathbf{F}$: $\mathbf{F} = (z^2 + yz) \mathbf{i} - (3xy + xz) \mathbf{j} + (x^3y^3 + xy) \mathbf{k}.$ Substituting: $\mathbf{F} = \left(1^2 + (2\sin t)(1)\right) \mathbf{i} - \left(3(2\cos t)(2\sin t) + (2\cos t)(1)\right) \mathbf{j} + \left((2\cos t)^3(2\sin t)^3 + (2\cos t)(2\sin t)\right) \mathbf{k}.$ Simplify each term: $\mathbf{F} = (1 + 2\sin t) \mathbf{i} - (12\cos t \sin t + 2\cos t) \mathbf{j} + (8\cos^3 t \sin^3 t + 4\cos t \sin t) \mathbf{k}.$

4. Dot product $\mathbf{F} \cdot d\mathbf{r}$: Substitute $\mathbf{F}$ and $d\mathbf{r}$ into $\mathbf{F} \cdot d\mathbf{r}$: $\mathbf{F} \cdot d\mathbf{r} = (1 + 2\sin t)(-2\sin t) + (-12\cos t \sin t - 2\cos t)(2\cos t).$ Simplify: $\mathbf{F} \cdot d\mathbf{r} = -2\sin t - 4\sin^2 t - 24\cos t \sin t \cos t - 4\cos^2 t.$ Further simplifying: $\mathbf{F} \cdot d\mathbf{r} = -2\sin t - 4\sin^2 t - 12\sin(2t) - 4\cos^2 t.$

5. Integrate over $t \in [0, 2\pi]$: Integrals of odd functions ($\sin t$ or $\sin(2t)$) over $[0, 2\pi]$ are 0. Using $\sin^2 t + \cos^2 t = 1$: $\int_0^{2\pi} \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} (-4) dt = -4 \times 2\pi = -8\pi.$

Final Answer:

$\iint\limits_S (\nabla \times \mathbf{F}) \cdot d \mathbf{S} = -8\pi.$

Would you like further clarifications or a breakdown of a specific step?

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5 Related Questions:

1. What is Stokes' theorem and when can it be applied?
2. How do you parameterize a circle in 3D space?
3. What is the significance of divergence and curl in vector calculus?
4. How do you simplify dot products involving trigonometric functions?
5. Why do integrals of odd functions over symmetric intervals vanish?

Tip:

To verify your parameterizations, always check that they satisfy the original equations defining the curve or surface.