Given the vector field $F$ compute the flux of the curl of $F$ through the surface $\Sigma$

Question

I was having some problems understanding how he found $\gamma(t)$ from the given $\Sigma$ and i was hoping someone could explain to me how if that is ok

So the problem goes like this:

Given the vector field $F(x, y, z) = (z, x, y)$, compute the flux of the curl of F through the surface $ Σ = (x, y, z) ∈ R^ 3 : z = xy, x^2 + y^ 2 ≤ 1 $

oriented so that the normal versor points upward

So what the professor did was first he computed $\gamma(t)$ using parametrization and he immediately writes

$\gamma(t)=(cos(t),sin (t), cos(t)sin(t) )$ with $t\in[0,2\pi]$ and from here he finds $\gamma$' and from there he computes

$\int _\Sigma rot F *nd\sigma$=$\int_0^{2\pi} F(\gamma(t))\gamma'(t)dt$

and from there is history i can do it myself

But what i couldn't understand is how did he get the $\gamma$

Answer 1

The boundary of the surface is given by $$\partial \Sigma=\{(x, y, z) ∈ R^ 3 : z = xy,\ x^2 + y^ 2 =1\}.$$ Now since the $z$ coordinate'only depends' on $x$ and $y$, you first parametrize $x^2 + y^ 2 =1$ as usual by letting $x=\cos t$ and $y=\sin t$ with $t\in [0, 2\pi]$ (since you consider the whole circle) and then just replace $z=\cos t\sin t$, which is precisely the parametrization your professor found.

Answer 2

Your professor is applying Stokes' Theorem which says that the line integral of a vector field over the boundary of a surface $S$ is equal to the surface integral of the curl of the vector field over surface $S$.

In this case, surface is $z = xy, x^2 + y^2 \leq 1$

So the boundary curve is $x^2 + y^2 = 1, z = xy$

$x^2 + y^2 = 1 \implies x = \cos t, y = \sin t, 0 \leq t \leq 2\pi$

$z = xy = \cos t \cdot \sin t$

Hence the parametrization,

$\lambda(t) = (\cos t, \sin t, \cos t \cdot \sin t), 0 \leq t \leq 2 \pi$