# Integrate 2-Form over surface

Problem: Calculate $\int_S dx \wedge dy + dy \wedge dz$, where $S$ is the surface given by $S = \{(x,y,z) : x = z^2 +y^2 -1, x < 0\}$.

Wikipedia says:

Let $$\omega=f_{z}\, \mathrm dx \wedge \mathrm dy + f_{x}\, \mathrm dy \wedge \mathrm dz + f_{y}\, \mathrm dz \wedge \mathrm dx$$ be a 2-form on a surface with parametrization $$\mathbf{x} (s,t)=( x(s,t), y(s,t), z(s,t))\!$$

defined on some domain $D.$

Then, the surface integral of the two-form on the surface $S$ is given by

$$\int_{S} \omega = \int_D \left[ f_{z} ( \mathbf{x} (s,t)) \frac{\partial(x,y)}{\partial(s,t)} + f_{x} ( \mathbf{x} (s,t))\frac{\partial(y,z)}{\partial(s,t)} + f_{y} ( \mathbf{x} (s,t))\frac{\partial(z,x)}{\partial(s,t)} \right]\, \mathrm ds\, \mathrm dt$$.

I'm hoping someone can help me with the process of calculating these integrals. I think I need to parametrize S somehow, and the do that final computation accordingly? Any help would be greatly appreciated!

On ${\bf R}^3$, consider $\alpha =xdy + ydz$ so that $d\alpha =dxdy+dydz$
That is by Stokes, we have $$\int_S d\alpha =\int_{\partial S} \alpha$$
Boundary of $S$ is $x=0,\ z^2+y^2=1$ so that $$\alpha =ydz,\ \int_{\partial S} ydz =\int_0^{2\pi} (\cos\ t) d\sin\ t = \int_0^{2\pi}\frac{1}{2} dt= \pi$$