# Evaluating a surface integral with differential forms

Let $\alpha=x dy-\frac{1}{2}(x^2+y^2)dz$ be a differential form in $\mathbb{R}^3=\{(x,y,z)\;|\;x,y,z\in\mathbb{R}\}$and let $Z=\{(\cos\theta,\sin\theta,s)\;|\;0\leq\theta \leq2\pi, 0\leq s\leq 1\}$ be a cylinder. I'd like to compute $\int_Z d\alpha$.

First, I need to find $d\alpha$. This is fairly straightforward: $$d\alpha=dx\wedge dy-\frac{1}{2}(2x dx+2ydy)\wedge dz = dx\wedge dy - x dx\wedge dz - y dy\wedge dz$$ Next, I need to rewrite the differential using the parameterization of $Z$. I think this means that I need to rewrite each $dx, dy,$ and $dz$ as $$dx = \frac{dx}{d\theta}d\theta + \frac{dx}{ds} ds$$ (replacing $x$ with $y$ and $z$ to get the corresponding expressions). Is this correct?

Assuming it is, then I get the straightforwards calculations that $$dx = -\sin\theta d\theta \quad dy=\cos\theta d\theta \quad dz=ds$$ So I can rewrite $d\alpha$ in terms of $d\theta\wedge ds$: $$d\alpha = (-\sin\theta d\theta\wedge \cos\theta d\theta)-(\cos\theta(-\sin\theta) d\theta\wedge ds)-(\sin\theta\cos\theta d\theta\wedge ds) = 0$$ This looks strange. I thought that I was supposed to be left with a differential form $d\theta\wedge ds$ that I could then integrate over. But what I'm left with now is the following double integral: $$\int_0^1\int_0^{2\pi} 0 d\theta\wedge ds =0$$ This just doesn't seem right. I'm not very familiar with differential forms, so maybe I'm missing something obvious. Is this analogous to the case where the integration of an exact form over a closed curve is $0$? Clearly $d\alpha$ is exact because we're given $\alpha$. We're also integrating over a cylinder, which I guess is sort of like a closed curve, only it's a closed surface instead.

In short, I have two questions: is the procedure I outlined above a correct way to go about evaluating surface integrals of differential forms? Second, was there a short cut I could have recognized that would have made the final answer obvious (assuming that $0$ is in fact the correct answer)?

• Using Stoke's theorem it is easier to integrate $\alpha$ over the boundary of the cylinder which consists of two circles. Since the orientation on the two circles is oposite you should expect things to cancel out. – PAD Oct 29 '12 at 6:18