2

The Generalized Stokes Theorem. If $\omega$ is any smooth $(k-1)$ form on $X$, then $$\int_{\partial X} \omega = \int_X d\omega.$$

Let $C \subset \mathbb{R}^2$ be a (smooth) simple closed curve in the plane and $\Omega$ the closed subset that it bounds. Thus $\Omega$ is a $2$-manifold with boundary $C$. Assume the origin, $O \in \mathbb{R}^2$, lies in the interior of $\Omega$.

By Stokes Theorem, $$\iint_{\Omega}r dr \wedge d\theta = \int_C \frac{1}{2}r^2 d\theta.$$

But how can I justify the integrand is well-defined, and what is the orientation?

Thank you!

Tumbleweed
  • 1,001
  • 1
    You have a singularity at $0$. Apply Stokes' theorem to $\Omega_\varepsilon = {x \in \Omega : \lvert x\rvert > \varepsilon}$ and let $\varepsilon \to 0$. Show that the area integral converges, and that the integral over ${\lvert x\rvert = \varepsilon}$ tends to $0$. – Daniel Fischer Aug 03 '13 at 09:05
  • Oh, so you mean by "Justify the integrand is well-defnied", the question is asking for the integral converges, and the integral over ${|x|=\epsilon}$ tends to 0? Thank you, @DanielFischer. – Tumbleweed Aug 03 '13 at 18:10
  • And also, do you mind demystify orientation as well? I am totally at lost.. – Tumbleweed Aug 03 '13 at 18:12
  • Well, at the origin, you have the problem that $dr$ and $d\theta$ aren't defined (no problem if you write it in terms of $dx$ and $dy$). So there you can't apply Stokes as formulated without justification. But Stokes holds more generally, $d\omega$ not being defined at some points doesn't matter if it behaves nicely near these points. – Daniel Fischer Aug 03 '13 at 18:33

1 Answers1

1

The curve $c$ is oriented by the direction it is parametrized in. Depending on this orientation, there is a unique orientation on $\Omega$ such that the induced orientation on the boundary $c$ coincides with this parametrization.

Stoke's theorem holds whenever the form $\omega$ is $C^1$. That said, the form $r^2d\theta$ is in fact smooth, as $$ r^2d\theta = x dy - y dx.$$ Its differential $rdr \wedge d\theta$ is in fact the standard volume form on $\mathbb{R}^2$. So no differentiability problems arise in the first place.

\Edit: I will be more elaborate on both points.

1.) A regular curve in the plane is something different than an immersed $1$-fold. A $1$-fold defined by a curve $c$ comes with a natural orientation defined by saying that the basis $\dot{c}(t)$ of the $1$-dimensional tangent space is positively oriented. On the other hand, for the Stokes formula, this is quite irrelevant: You just have to make sure that the orientations of $\Omega$ and the bounding curve fit together in the sense you learned it when talking about the orientation that a manifold induces on its boundary.

2.) If you ask if the integrand is well-defined, shouldn't you know what well-defined means in the first place? The only thing that I can say now is somewhat "philosophical": What exactly is meant by the phrase "We define the one-form $r d \theta$ on $\Omega$", when clearly $d\theta$ is not defined on all of $\Omega$. There could be different interpretations: For example, we defined a $1$-form on almost all on $\Omega$ (everywhere except on a zero set) which suffices for integration. Or, it should be understood that we mean the smooth $1$-form that is defined on all of $\Omega$ and coincides with $r^2d\theta$ everywhere where the latter is defined. This $1$-form exists as I showed you and this seems to me the natural interpretation.

Kofi
  • 1,818
  • Thanks Kofi. May I ask for more details about your answer on orientation? What do you mean by "the direction it is parametrized in"? For example, if we parametrize a circle as $(r \cos \theta, r\sin \theta)$, doea that mean it has CCW orientation? And also, I don't quite understand your answer on well-definedness. Actually, would you mind explaining for me what does the question mean by justify the integrand is well-defined? Thank you! – Tumbleweed Aug 03 '13 at 17:55