# Finding integral of exact form on punctured 3-space

I'm trying to solve the following quals problem, and I think I have the general idea, but there's one step I'm not sure about, and I highlight it below.

The problem:

Let $$M=\mathbb{R}^3\setminus0$$. Let $$\Sigma_1\subset M$$ be the unit sphere, and $$\sigma_1$$ its volume form such that $$$$\int_{\Sigma_1}\sigma_1 = 1$$$$ Suppose $$\rho:M\rightarrow\Sigma_1$$ is the radial retraction, and $$\omega=\rho^\ast(\sigma_1)$$. If $$f:\mathbb{R}^3\rightarrow\mathbb{R}$$ is a smooth compactly-supported function, then $$$$\int_M df\wedge\omega$$$$ depends only on $$f(0)$$.

My attempt is as follows: since $$f$$ is compactly supported, there's an $$R$$ such that $$f\equiv0$$ on $$\Sigma_R$$. Define \begin{align} A_r &= \{x\in M: r\le|x|\le R\}\\ B_r &= \{x\in M: |x|\le r\} \end{align} By Stokes' theorem then $$$$\int_{A_r}df\wedge\omega = \int_{\Sigma_r}f\omega$$$$ Because $$\rho:\Sigma_r\rightarrow\Sigma_1$$ is a diffeomorphism, it must be that $$\rho^\ast(\sigma_1)=\sigma_r$$ on $$\Sigma_r$$. So the integral on the right (and hence the left) satisfies $$$$m_r \le \int_{\Sigma_r}f\omega \le M_r$$$$ with $$m_r=\min_{B_r}\{f(x)\}$$ and $$M_r=\max_{B_r}\{f(x)\}$$.

Here's the part I'm unsure of. I want to claim $$$$\int_M df\wedge\omega = \lim_{r\rightarrow0^+}\int_{A_r} df\wedge\omega = f(0)$$$$ I believe I can prove the second equality via some squeeze-type argument, but how can I rigorously shows the first equality holds? It seems intuitive to me, but I'm not even sure how to prove the limit exists. I'm also concerned because this seems a much stronger conclusion than the problem asked for: the original integral was supposed to depend on $$f(0)$$, and I've shown it is $$f(0)$$!

• Is $\int_M \phi = \lim\limits_{r\to 0^+}\int_{\Sigma_r}\phi$? Commented Sep 5 at 0:52
• If $f$ has spherical symmetry then this is easy, so maybe use $f$ to construct a spherically symmetric function I.e. by setting $g(r)$ to be equal to integral of $f$ over the sphere of radius $r$? Commented Sep 5 at 6:06
• Although now I have confused myself…why is this not just zero? We have that $f\omega$ is compactly supported and equals $df\wedge\omega$. $M$ is a manifold without boundary, so stokes theorem should imply this goes to zero right? Commented Sep 5 at 6:15
• @TedShifrin: thank you for the comment. Do you mean the integral over $A_r$ or $\Sigma_r$? I'm not sure if what you said is true, and maybe that's the crux of the issue? More help (or hints) would be most appreciated. Commented Sep 5 at 14:40
• Not quite. But you should use the fact that $\int_{A_r} \phi + \int_{B_r} \phi = \int_{A_r\cup B_r} \phi$ and then use continuity of $\phi$ at $0$ to deduce what you need. Commented Sep 5 at 22:13

Write everything in spherical coordinates, then $$\omega=\frac{1}{4\pi^2}\sin\theta d\theta d\phi$$, and $$df=\partial_rfdr+\partial_\theta fd\theta+\partial_\phi fd\phi$$. It follows that: \begin{align} \int_Mdf \wedge \omega=&\frac{1}{4\pi}\int_0^{2\pi}\int_0^{\pi}\int_0^\infty (\partial_rf(r,\theta,\phi))\sin\theta drd\theta d\phi \end{align} Now we can integrate with respect to $$r$$ holding all other variables constant, so regardless of whether or not $$f$$ depends on $$\theta$$ and $$\phi$$, we know that $$\int_0^\infty\partial_rf dr=f(0)$$. However, $$f(0)$$ doesn't depend on $$\theta$$ or $$\phi$$ (as for all values of $$\theta$$ and $$\phi$$, $$f$$ must be the same at the orign, so this is a constant and we have that: $$\int_Mdf \wedge \omega=f(0)$$ Moreover, regarding my previous comment, for $$f$$ to have compact support on $$M$$, we must have that $$f(0)=0$$, as other wise the support of $$f$$ in $$M$$ can't be compact as it will be a compact set minus a point. It follows that this is not a counter example to stokes theorem which is nice.
Edit: A priori I guess there’s some finicky Lebesque integration stuff one “should” know and be careful about when integrating non compactly supported forms, but since $$f$$ has compact support on $$\mathbb R^3$$ everything is “probably” fine.
• Hi Chris, thank you for the answer. I haven't had a chance to work through your solution, but I wanted to clear up a misunderstanding: $f$ is defined on all of $\mathbb{R}^3$, not just $M$. So it's not true that $f(0)=0$, and also not necessarily true that $f$ has compact support on $M$ (so Stokes' theorem might not apply). Commented Sep 5 at 14:38