Going with the Flow

The vector field $\textbf{F}$ is tangent to the surface $S$ , a part of a sphere. To verify this, take the dot product of $\textbf{F}$ with the radial vector $x\textbf{i}+y\textbf{j}+z\textbf{k}$ . Thus the flux is $\boxed{0}$ .

(Recall that the flux is defined as $\int_{s}\textbf{F}\cdot\textbf{n}dS$ , where $\textbf{n}$ is the upward unit normal; the flux depends only on the component of $\textbf{F}$ perpendicular to $S$ ) . In a physical sense, the flux represents the rate at which things are flowing through the surface.

Note that $\mathbf{F} \,=\, \nabla\times \mathbf{G}$ , where $\mathbf{G}(\mathbf{r}) \; = \; \tfrac12(x^2 + y^2)\mathbf{k} \;.$ Using Stokes' Theorem, it follows that $I \; = \; \iint_S \mathbf{F} \cdot d\mathbf{S} \; = \; \oint_C \mathbf{G} \cdot d\mathbf{r} \;,$ where $C$ is the curve defining the intersection of the sphere $x^2 + y^2 + z^2 = 4$ and the surface $z = x^2 + 4y^2$ , oriented in such a manner that $S$ is a positively oriented capping surface for $C$ .

But then $I \; = \; \tfrac12\oint_C(x^2 + y^2)\,dz \; = \; \tfrac12\oint_C (4 - z^2)\,dz \;.$ If the curve $C$ is parametrized as $\mathbf{r} \,=\, \big(x(t),y(t),z(t)\big)$ for $0 \le t \le 1$ , then $I \; = \; \tfrac12\int_0^1 \big(4 - z(t)^2\big)\,\dot{z}(t)\,dt \; = \; \tfrac12\Big[4z(t) - \tfrac13z(t)^3\Big]_0^1 \;.$ Since the curve $C$ is closed, we have $z(0) = z(1)$ , and hence the integral $I \,=\, \boxed{0}$ .