Let S be cylinder given by $x^2+y^2=1$ between $z=1$ and $z=3.$ For $\varphi=e^xdx\wedge dy+ ydz\wedge dx+xdy\wedge dz$ find $\int_S\varphi$.
I managed to finish the problem, but I'm getting strange results. It would be great if someone could verify what I did.
First, parameterize the surface: $k_1(u,v)=(u,\sqrt{1-u^2},v)$ and $k_2=\{u,-\sqrt{1-u^2},v\}$.
Then $$k_{1u}=\left(1,\frac{-u}{\sqrt{1-u^2}},0\right),k_{1v}=\left(0,0,1\right)\\k_{2u}=\left(1,\frac{u}{\sqrt{1-u^2}},0\right),k_{2v}=\left(0,0,1\right)$$
And is $\displaystyle\int_S\varphi=\displaystyle\int \varphi (k_{1u},k_{1v})+\displaystyle\int \varphi(k_{2u},k_{2v})$.
Calculating each one of the differential forms I got:
$$\varphi (k_{1u},k_{1v})=-\sqrt{1-u^2}+u\left(\frac{-u}{\sqrt{1-u^2}}\right)=\frac{-1}{\sqrt{1-u^2}}\\ \varphi(k_{2u},k_{2v})=-\sqrt{1-u^2}+\frac{u^2}{\sqrt{1-u^2}}$$. And calculating the integrals I got
$$\int \varphi (k_{1u},k_{1v})=\int_1^3\int_0^1 \frac{-1}{\sqrt{1-u^2}}dudv = - \int_1^3 (\arcsin u |_0^1) dv=-\frac{\pi}{2}\int_1^3dv=-\pi \\ \int \varphi (k_{2u},k_{2v}) = \int_1^3 \left(-\int_0^1 \sqrt{1-u^2}du+\int_0^1 \frac{u^2}{\sqrt{1-u^2}}du\right)dv\\=\int_1^3\left(-\frac{\pi}{4}+\int_0^{\frac{\pi}{2}}\frac{\sin^2\theta}{\sqrt{1-\sin^2\theta}}\cos\theta\right)dv\\=\int_1^3\left( -\frac{\pi}{4}+\int_0^{\pi/2} \sin^2 u du\right)dv=0$$
Then $\displaystyle\int_S \varphi = - \pi $?
Why do you refuse to use natural parametrizations, like cylindrical coordinates $(\theta,z)$? And the problem has to specify an orientation on $S$ if you are to get a single answer. (Otherwise, we can only say $\pm$.)
In cylindrical coordinates, this integral becomes, quite simply, $$\int_0^{2\pi}\int_1^3 dz\,d\theta = 4\pi,$$ assuming we orient $S$ with the normal pointing radially outward.
There are two problems with your parametrization. One is that you need $-1\le u\le 1$. The other is that you have not thought about orientation. The two parametrizations of the circle both go left-to-right and do not give a coherent orientation on the circle.