# The Unapologetic Mathematician

## The Classical Stokes Theorem

At last we come to the version of Stokes’ theorem that people learn with that name in calculus courses. Ironically, unlike the fundamental theorem and divergence theorem special cases, Stokes’ theorem only works in dimension $n=3$, where the differential can take us straight from a line integral over a $1$-dimensional region to a surface integral over an $n-1$-dimensional region.

So, let’s say that $S$ is some two-dimensional oriented surface inside a three-dimensional manifold $M$, and let $c=\partial S$ be its boundary. On the other side, let $\alpha$ be a $1$-form corresponding to a vector field $F$. We can easily define the line integral $\displaystyle\int\limits_c\alpha$

and Stokes’ theorem tells us that this is equal to $\displaystyle\int\limits_{\partial S}\alpha=\int\limits_Sd\alpha$

Now if we define $\beta=*d\alpha$ as another $1$-form then we know it corresponds to the curl $\nabla\times F$. But on the other hand we know that in dimension $3$ we have $*^2=1$, and so we find $*\beta=**d\alpha=d\alpha$ as well. Thus we have $\displaystyle\int\limits_c\alpha=\int\limits_S*\beta$

which means that the line integral of $F$ around the (oriented) boundary $c$ of $S$ is the same as the surface integral of the curl $\nabla\times F$ through $S$ itself. And this is exactly the old Stokes theorem from multivariable calculus.

November 23, 2011 - Posted by | Differential Geometry, Geometry

## 1 Comment »

1. […] the left we can use Stokes’ theorem, while on the right we can pull the derivative outside the […]

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