Stoke theorem and exterior derivative

Question

$w=x \, dy\wedge dz - 2z f(y) \, dx \wedge dy + y f(y) \, dz \wedge dx$ where $f: \mathbb{R} \to \mathbb{R}$ belong $C^1$ (differentiable and derivative is continuous) with $f(1)=1$. Find $f$ so that $dw=0$. With that $f$, what is $\int_S w$, where $S$ is the surface of $\left\{x^2+y^2+z^2=1,z \geq \frac{\sqrt{2}}{2} \right\}$?

I know the solution of the first part is $ f=c y +1 $ with $c$ is constant so that $dw=0$. I know The second one applying stoke theorem. But i can not solve it. Can someone help me?

Answer 1

You can check that $w = d\gamma$ with $$\gamma = (yz+3cy^2z/4)\,dx - (xz+cxyz/2)\,dy - (cxy^2/4)\,dz$$ (use the explicit formula that appears in the proof of Poincaré Lemma or another usual trick). Then $$\int_S w = \int_Sd\gamma = \int_{\partial S}\gamma$$ with $\partial S =$ the circle $x^2+y^2 = 1/2$, $z=\sqrt2/2$. This circle can be parametrized as $$ x = (\sqrt2/2)\cos t,\qquad y = (\sqrt2/2)\sin t,\qquad z = \sqrt2/2,\qquad t\in[0,2\pi]. $$ Can you continue?

Answer 2

Even easier, if $\partial S = \partial S'$ (as oriented curves) and $\omega$ is a $2$-form with $d\omega=0$, then $$\int_S \omega = \int_{S'} \omega.$$ It is easy to integrate $\omega$ over the disk $x^2+y^2\le 1/2$, $z=1/\sqrt 2$ (oriented, presumably, upwards, just like $S$). There is no need to find a $1$-form $\gamma$ with $d\gamma=\omega$.

This method can be used, more generally, even if $d\omega\ne 0$, provided one can integrate $d\omega$ over the $3$-dimensional region whose boundary is $S-S'$.