My problem asks me to evaluate the integral (using direct integration) $$\iint_S (x)dy\wedge dz + (x+y)dz\wedge dx+(x^2+2z)dx\wedge dy$$ Being $S$ the surface of the solid limited by the following surfaces: $$S_1=\{(x,y,z)\in\mathbb{R}:2x^2+y^2=4z\},$$ $$S_2=\{(x,y,z)\in\mathbb{R}:x^2+2z=2\}.$$ I've first defined parameterizations for $S_1$ and $S_2$ (I'll call them $T$ for top and $B$ for bottom): $$T=(x,y,1-\frac{x^2}{2})$$ $$B=(x,y,\frac{2x^2+y^2}{4})$$ Being the region of integration $D:x^2+\frac{y^2}{4}\leq 1$ (ellipse).
I think (I'm not sure, correct me if I'm wrong) that: $$\iint_S (x)dy\wedge dz + (x+y)dz\wedge dx+(x^2+2z)dx\wedge dy=\iint_S (x)dydz + (x+y)dzdx+(x^2+2z)dxdy$$ And now integrate separately for $T$ and $B$ (being $\omega=(x)dydz + (x+y)dzdx+(x^2+2z)dxdy$): $$\iint_S \omega =\iint_T \omega - \iint_B \omega.$$ I'm stucked here, I don't know how to calculate this kind of integrals, and also I'm not sure if my work till here is correct. I think the result must gave $4\pi$. I will thank any help.
As a hint, you may pullback the form using a parametrization then integrate over a simpler domain.
$x=rsin(\theta)$, $y=2rcos(\theta)$, $D = \{ (r,\theta ): 0\leq r \leq 1$ and $-\pi \leq \theta \leq \pi$}.
$\phi (r, \theta)= (rsin\theta,2rcos(\theta), 1-r^2sin(\theta)/2)$
$\pm \iint_D \phi^{*}\omega = \iint_T \omega$
Sign depending on the orientation. A relevant definition is
Let $\omega$ be a $k$ -form on $\mathbf{R}^{n}$ and let $S=\phi(U)$ be an oriented $k$ -dimensional geometric set parameterized by a smooth, regular parameterization $\phi: U \rightarrow \mathbf{R}^{n},$ where $U \subset \mathbf{R}^{k}$ is a domain. Let $D \subset U$ be a region of integration. Then the integral of $\omega$ over $R=\phi(D)$ is defined to be $$ \int_{R} \omega=\pm \int_{D} \phi^{*} \omega $$ where the sign is positive or negative according to whether the given orientation on $S$ agrees with or does not agree with the orientation on $S$ induced by $\phi$.
You may refer to this document.