# Compute $\displaystyle\int_{S^1}i^*\omega$ where $\omega\in\Omega^1(\mathbb{R^2})$

Let $\omega\in\Omega^1(\mathbb{R^2})$ be a 1-form such that $d\omega=dx\wedge dy$. Let $i:S^1\to \mathbb{R^2}$ be the inclusion and $S^1$ be the unit circle. I want to compute $\displaystyle\int_{S^1}i^*\omega$. By Stoke's we have that $$\displaystyle\int_{S^1}i^*\omega=\int_{B}d\omega\,\,\,\,\,,B\text{ is the unit disc}.$$

We can parametrize $B$ by $Y(r,\theta)=(r\cos\theta,r\sin\theta)$ where $0\leq r\leq1,$ $0\leq\theta\leq 2\pi.$ Now, I need to check if this parametrization is coherent with the orientation on $B$. I am not sure how to do this and would like some help on that. Assuming this is coherent, $$\int_B d\omega=\int_0^1 \int_0^{2\pi}\begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix}drd\theta=2\pi$$. Is this correct?

• The area of the unit disk in $\Bbb R^2$ is $\pi$. – Ted Shifrin Jul 22 '17 at 0:38

## 1 Answer

To check that the orientation induced by this parameterization is consistent with the orientation on $B$ it suffices to show that the determinant of the Jacobian of the transition matrix between these two charts is positive, which you already did!

I think there is a small mistake in your computation though since the $d\omega$ is the usual volume form on $\mathbb{R}^2$, and hence on $B$. So the integral should be the area of the circle, which is $\pi$. You have the limits of your integrals switched ($r$ doesn't go from $0$ to $2\pi$, $\theta$ does!), what you should have is $$\int_{B}d\omega=\int_{0}^{2\pi}\int_{0}^{1}\begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix}dr~d\theta=\int_{0}^{2\pi}\frac{1}{2}~d\theta=\pi.$$

• Determinant of the Jacobian being the one inside the integral? – Heisenberg Jul 19 '17 at 18:54
• Yes. Your map $Y:[0,1]\times[0,2\pi)\to B$ is the transition function between the polar coordinates chart and the usual Euclidean coordinates chart on $B$. – yousuf soliman Jul 19 '17 at 18:56
• There is a slight issue in that $Y$ is not smooth at $r=0$, but this doesn't matter since integrating over $B\setminus\{0\}$ is the same as integrating over $B$ since $\{0\}$ has Lebesgue measure zero. In some sense, it would have been better to write $\int_{B}d\omega = \int_{B}dxdy$, and then use the usual change of variables for the Lebesgue integral to argue that this is the correct result. – yousuf soliman Jul 19 '17 at 19:01
• Why don't you need to consider a outward pointing normal vector/ – Heisenberg Jul 19 '17 at 19:20
• I'm not sure why I would need to? What is your definition of orientation? The definition I'm using can be found here: map.mpim-bonn.mpg.de/… . Then we can consider $B$ as a smooth manifold with two charts: (1) the Euclidean coordinates chart $\mathrm{id}:B\to B$ and (2) the polar coordinates chart $Y:(0,1]\times[0,2\pi)\to B\setminus\{0\}$. Then we see that these charts form an orientation on $B$ since the determinant of the Jacobian of the transition function is positive. – yousuf soliman Jul 19 '17 at 19:27