Calculating $\iint_S \operatorname{curl} {\vec F} \cdot {\vec n}\, dS$, where $S$ is the union of three faces of a particular tetrahedron

Question

Let $$\vec F=xz \,\vec i-y \,\vec j +x^2y \,\vec k,$$ let $S$ be the surface consisting of the three faces not in the $xz$-plane of the tetrahedron bounded by the three co-ordinate planes and the plane $3x+y+3z=6$ . How to use Stokes' Theorem to evaluate $$\iint_S \operatorname{curl} {\vec F} \cdot {\vec n}\, dS$$ as a line integral? Should I evaluate on the three faces separately? what is the boundary curve? I am totally stuck. Please help. Thanks in advance.

Answer 1

Sketching a tetrahedron shows that the boundary $\partial S$ of the union $S$ of three of the four sides is precisely the (triangular) boundary of the omitted face $D$.

Like you say, with this fact in hand, one could invoke Stokes' Theorem to conclude that $$\iint_S \operatorname{curl} {\bf F} \cdot {\bf n} \,dS = \oint_{\partial S} {\bf F} \cdot d{\bf s},$$ and so we could just evaluate the line integral over $\partial S$. In this case it's even easier to use (again via Stokes' Theorem) that $$ \iint_S \operatorname{curl} {\bf F} \cdot {\bf n} \,dS = \iint_D \operatorname{curl} {\bf F} \cdot {\bf n} \, dS$$ (where $D$ is appropriately oriented), and evaluate the surface integral on the right because (1) this integral doesn't need to be decomposed into three separate path integrals to be evaluated, and (2) the normal vector field $\bf n$ is especially simple for $D$.