- §1. Vectors and Forms
- §2. Line & Surface Integrals
- §3. Integrands Revisited
- §4. Stokes' Theorem
- §5. Calculus Theorems
- §6. Integration by Parts
- §7. Further Corollaries

### Corollaries of Stokes' Theorem

If $M$ is compact (and open), it has no boundary, so $\partial{M}=\emptyset$. The boundary of a disk is a circle, but a circle has no boundary. Similarly, the boundary of a ball is a sphere but a sphere has no boundary. Circles and spheres are therefore examples of compact surfaces with no boundary.

If $M$ has no boundary, then \begin{equation} \int_M d\alpha = \int_{\partial M} \alpha = 0 \end{equation} for *any* ($n-1$)-form $\alpha$. The contrapositive of this statement says that if an $n$-form $\beta$ satisfies \begin{equation} \int_M\beta \ne 0 \end{equation} then $\beta\ne d\alpha$ for any ($n-1$)-form $\alpha$; as discussed in §9.3, we say that $\beta$ is not *exact*. For example, the orientation itself satisfies \begin{equation} \int_M \omega > 0 \end{equation} which shows that $\omega$ is not exact.

Similarly, suppose that an ($n-1$)-form $\alpha$ is such that \begin{equation} d\alpha = h\,\omega \end{equation} for some function $h$. Since the integral of the left-hand-side is zero, and the integral of $\omega$ is nonzero, either $h$ is identically zero, or it must change sign; in either case, $h=0$ somewhere.

As an example, *all* smooth functions $f$ on a circle must satisfy $df=0$ somewhere. And any smooth vector field $\FF$ on a sphere must have vanishing curl somewhere, since $F=\FF\cdot d\rr$ implies that $dF=\grad\times\FF\cdot d\AA$.