Integral of a differential form along a non-defined path

Question

Let $R>0$ and $\Omega=\{(x,y)\in \mathbb{R^2}:x^2+y^2<R^2,y>0\}$. Consider also $\omega (x,y)=x^2dx+2xydy$. My goal is to prove that $\int_{\partial_{+}\Omega}\omega=\frac{4}{3}R^3$, where $\partial _+ \Omega$ denotes the "edge" of $\Omega$ positively oriented. I know that using the Green-Riemann I can show that, but could I use the definition? That is: $$\int_{\partial_{+}\Omega}\omega=\int_0^{\pi}(\omega \space\circ\space\gamma)(\theta)\gamma'(\theta)d\theta$$ Where $\gamma(\theta)=(R\cos(\theta),R\sin(\theta))$. On impulse I'd say it's not possible since it's not defined, but then again, the edge has $0$ volume, hence any semi circle suficiently together would be a pretty good approximation, thus it should work, what is your opinion?

Answer 1

Yes, we can do the computation that way. The boundary actually has two components:

• The straight component $C_1$, parameterised by $\gamma_1 : [-1, 1] \to C_1$ sending $t \mapsto (Rt, 0)$.

• The curved component $C_2$, parameterised by $\gamma_2 : [0, \pi] \to C_2$ sending $t \mapsto (R\cos t, R\sin t)$.

We can now evaluate the integral by pulling back the $1$-form $\omega$ to $[-1, 1]$ and $[0, \pi]$ via $\gamma_1$ and $\gamma_2$. \begin{align} \oint_{\partial_+ \Omega} \omega &= \int_{[-1, 1]} \gamma_1^\star (\omega)+\int_{[0,\pi]}\gamma_2^\star (\omega) \\ &= \int_{-1}^1 (Rt)^2(Rdt) + \int_0^\pi \left( (R^2\cos^2 t).(-R\sin t\ dt) + (2R^2 \sin t \cos t).(R\cos t \ dt) \right) \\ &= \int_{-1}^1 R^3 t^2 dt + \int_{0}^{\pi} R^3 \sin t \cos^2 t \ dt \\ &= \frac{4}{3} R^3\end{align}

[To spell it out: $$ \int_{[-1,1]} \gamma_1^\star (x^2 dx) = \int_{[-1, 1]} ((x \circ \gamma_1)(t))^2 \left( \frac{d(x \circ \gamma_1)}{dt}(t). dt \right) = \int_{-1}^1 (Rt)^2 \left( \frac{d(Rt)}{dt} dt\right)$$ and so on...]

A final remark: these integrals are not ill-defined at all. $\omega$ is a $1$-form, so you should expect to be able to integrate it along one-dimensional things like $\partial_+ \Omega$. The fact that $\partial_+ \Omega $ has zero measure as a subset of $\mathbb R^2$ should not bother you.