Good morning, I have some questions about a surface integral with curl. The exercise is the following:
Be $(\Sigma, \omega)$ an oriented surface with boundary where $$\Sigma = \{(x, y, z): x^2 + y^2 = z^2+1 ,\ -1 \leq z \leq 3\}$$ Calculate $$ \int\int_{\Sigma}\langle \text{rot}F, \omega\rangle \text{d}\sigma$$ where $$ F(x, y, z) = -\dfrac{1}{3}(y, x, z)$$ and $$\omega(0, 1, 0) = (0, 1, 0)$$ Then he asks to verify the result also by applying Stokes (later).
Some details on the procedure
Well first of all it's not a big deal to find out that
$$\text{rot}F = \dfrac{1}{3}(1, 1, 1)$$
Then we have $$\int\int_{\Sigma} \langle \text{rot}F, \omega\rangle \text{d}\sigma = \dfrac{1}{3}\int\int_{\Sigma} \sum_{i = 1}^3 \omega_i \text{d}\sigma$$
Where $\omega = (\omega_1, \omega_2, \omega_3)$.
A parametrisation for $\Sigma$ is given by
$$\phi:[0, 2\pi) \times [-1, 3] \to \Sigma$$
where $$\phi(\theta, z) = (\sqrt{z^2+1}\cos\theta, \sqrt{z^2+1}\sin\theta, z)$$
In particular we find that the Jacobian is
$$ \begin{pmatrix} -\sqrt{z^2+1}\sin\theta & \dfrac{z}{\sqrt{z^2+1}}\cos\theta \\ \sqrt{z^2+1}\cos\theta & \dfrac{z}{\sqrt{z^2+1}}\sin\theta \\ 0 & 1 \end{pmatrix} $$
And its rank is two.
Thence:
$$\dfrac{\partial \phi}{\partial \theta} \wedge \dfrac{\partial \phi}{\partial z} = (\sqrt{z^2+1}\cos\theta, \sqrt{z^2+1}\sin\theta, -z)$$
Now:
$$\omega(0, 1, 0) = \omega(\phi(\pi/2, 0)) = \dfrac{\dfrac{\partial \phi}{\partial \theta} \wedge \dfrac{\partial \phi(\pi/2, 0)}{\partial z}}{||\dfrac{\partial \phi}{\partial \theta} \wedge \dfrac{\partial \phi(\pi/2, 0)}{\partial z}||} = (0, 1, 0)$$
He then now says that $\phi$ is compatible with $\omega$ hence the integral is
$$\dfrac{1}{3}\int \int_{[0, 2\pi]\times [-1, 3]}\left( \sqrt{z^2+1}\cos\theta + \sqrt{z^2+1}\sin\theta - z)\right) \text{d}\theta\text{d}z = -\dfrac{1}{3}\int\int_{\ldots}z \text{d}\theta\text{d}z = -\dfrac{8\pi}{3}$$
Now my questions
It's all clear until we need to calculate the norm of the cult of $\phi$, then its blackout.
1) I found NOWHERE that the compatibility between $\omega$ and $\omega(\phi)$ has to be verified through the ration between the cult of $\phi$ and its norm. So why do we have to do this?
2) Once we verified the compatibility... then what? I mean I do not need to know that $\omega(0, 1, 0) = \omega(\phi(\pi/2, 0)) = (0, 1, 0)$ do I? So why do I have to do this?
Thank you very much for your time, those are really critical points for me to understand...
Updates
I understood that the proof of the compatibility is irrelevant for the exercise.
So it remains the first question: why does that method tell me that they are compatible?
Using different notation, recall that the vector surface integral (or flux) of the vector field $\mathbf F$ over the surface $S$ is given by the Riemann double integral $$\iint_S (\mathbf F \cdot \mathbf n) \, dS = \iint_U \mathbf F(\mathbf G(u, v)) \cdot \mathbf N(u, v) \, dA,$$ where $\mathbf n$ is the (positively oriented) unit normal vector, $\mathbf G(u, v)$ is a parametrization of the surface $S$ over the region $U,$ and $\mathbf N(u, v) = \pm G_u(u, v) \times G_v(u, v)$ is the (positively oriented) normal vector.
Certainly, then, the orientation of the normal vector matters -- if $\mathbf N$ (or $\mathbf n$) were the opposite sign, then the integral would have the opposite sign -- so the question is how to detect what the "correct" orientation is. Of course, this depends on the orientation of $S$ (assuming that $S$ is orientable).
Our surface $S$ in question is the elliptic hyperboloid $x^2 + y^2 = z^2 + 1,$ so we are dealing with an orientable surface, and we must specify its orientation. Considering that $\mathbf n(0, 1, 0) = \langle 0, 1, 0 \rangle,$ the "correct" orientation is outward from the surface. Like you have observed, we have that $\mathbf G(u, v) = \langle \sqrt{v^2 + 1} \cos u, \sqrt{v^2 + 1} \sin u, v \rangle$ for $U = [0, 2 \pi] \times [-1, 3]$ so that $\mathbf N(u, v) = \langle \sqrt{v^2 + 1} \cos u, \sqrt{v^2 + 1} \sin u, -v \rangle.$ We can check that this is the "correct" normal vector by verifying that the sign of $\mathbf n(u, v) = \frac{\mathbf N(u, v)}{||\mathbf N(u, v)||}$ agrees with the "correct" orientation. $$\mathbf n(\pi/2, 0) = \frac{\mathbf N(\pi/2, 0)}{||\mathbf N(\pi/2, 0)||} = \frac{\langle 0, 1, 0 \rangle}{||\langle 0, 1, 0 \rangle||} = \langle 0, 1, 0 \rangle = \mathbf n(0, 1, 0)$$ We conclude that $\mathbf N(u, v)$ is the "correct" normal vector, and the first equation above holds.