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I'm asked to evaluate the flux $$int_{S} rot K domega,$$ where $S$ is the region $-2 leq z leq 1, sqrt{x^2+y^2}=1+z^2$ and $K$ is the vector field $(-zy,zx,z^2)$.

So when I evaluate directly the integral, I use as parametrization $$x=rcos(theta), y=rsin(theta), z=sqrt{r-1}.$$ As rotation, I get $(-x,-y,2z)$.

As normal vector, I get $$(cos(theta),sin(theta),frac{1}{2sqrt{r-1}})X(-rsin(theta),rcos(theta),0)=(frac{-rcos(theta)}{2sqrt{r-1}},frac{-rsin(theta)}{2sqrt{r-1}},r)$$ Now, if I take the dot product of that with $(-rcos(theta), -rsin(theta),sqrt{r-1})$, I get $$frac{r^2}{2sqrt{r-1}}+rsqrt{r-1}$$ which seems really ugly, so I wanted to know if until now, everything I did is correct, and if yes, if I'm supposed to integrate this from $r=5$ to $r=2$ (given that $z=sqrt{r-1}$) times $2pi$.

Using the relation with Stoke's theorem, I get the integral over the border. As parametrization, I use $((1+z^2)cos(theta),(1+z^2)sin(theta),z)$. So the normal vector is $(2zcos(theta), 2zsin(theta),1)X(-(1+z^2)sin(theta), (1+z^2)cos(theta),0)=(-(1+z^2)cos(theta),-(1+z^2)sin(theta),2z(1+z^2))$.

The dot product of that with $(-z(1+z^2)sin(theta),z(1+z^2)cos(theta),z^2)$ is $z(1+z^2)^2sin(theta)cos(theta)-z(1+z^2)cos(theta)sin(theta)+2z^3(1+z^2)$. If we integrate that, we're just left with $2z^3(1+z^2)$. If we integrate that from $-2$ to $1$, we get $frac{1}{4}+frac{1}{6}-4-frac{64}{6}$ times $4pi$ which again seems a bit weird.

Direct approach:

Can I suggest you use this parametrisation:
$$ x = (1 + u^2) cos phi, y = (1 + u^2) sin phi, z = u (u in [-2, 1], phi in [0, 2pi])$$

[Notice how my parametrisation avoids square roots! The parametrisation that you chose, with $z = pm sqrt{1 - r}$, is problematic because you have to treat the separate cases where $z$ is positive or negative, giving you two integrals to evaluate instead of one. And besides, avoiding square roots makes the algebra nicer.]

As you say, the curl is

$$ nabla times vec K=begin{bmatrix} -x \ -y \2z end{bmatrix}=begin{bmatrix} -(1+u^2)cos phi \ -(1+u^2)sinphi\2uend{bmatrix}$$
while is area element is
$$ d vec A= frac{partialvec r}{partial u} times frac{partial vec r}{partial phi} dudphi = begin{bmatrix} 2ucos phi \ 2usinphi\ 1end{bmatrix} times begin{bmatrix}-(1+u^2)sinphi \ (1+u^2)cosphi \ 0end{bmatrix} dudphi= begin{bmatrix} -(1+u^2)cosphi \ -(1+u^2)sinphi \ 2u(1+u^2)end{bmatrix}dudphi$$
so the integral is
$$ iint (nabla times vec K).dvec A=int_{u=-2}^{u=1}int_{phi=0}^{phi=2pi}left( 1 + 6u^2 + 5u^4right) = 108pi $$

Using Stoke's theorem:

The region has two boundary components:
$$ C_1: x = 2cos phi, y = 2sinphi, z = +1 (phi in [0, 2pi])$$
$$ C_2: x = 5cos phi, y = 5sinphi, z = -2 (phi in [0, 2pi])$$

[Notice how these boundary curves are parameterised by a single parameter $phi$, and notice how the parameter $u$ that we used in the direct approach is now treated as a constant. (It is equal to $1$ on $C_1$ and $-2$ on $C_2$).]

For the orientations to be consistent, $C_1$ must be traversed anticlockwise, and $C_2$ clockwise. So
begin{align} iint (nabla times vec K) . dvec A &= oint_{C_1} vec K . dvec l - oint_{C_2} vec K . dvec l \ &= oint_{phi = 0}^{phi = 2pi} begin{bmatrix}-2sinphi \ 2cos phi \ 1 end{bmatrix} . begin{bmatrix}-2sinphi \ 2 cos phi \ 0 end{bmatrix}dphi - oint_{phi = 0}^{phi = 2pi} begin{bmatrix}10sinphi \ -10cos phi \ 4 end{bmatrix} . begin{bmatrix}-5sinphi \ 5 cos phi \ 0 end{bmatrix}dphi \ &= 108piend{align}

[Notice that the line elements $dvec l = frac{partial vec r}{partial phi}$ are tangential to the boundary curves. They are not normals!]

The dot product shouldn't be so ugly.

With the parameterization
$K = (-yz, xz, z^2) = ((rsin theta)sqrt{r-1},(-rcos theta)sqrt{r-1}, |r-1|)$

When you dot it with the Jacobian.

$((rsin theta)sqrt{r-1},(-rcos theta)sqrt{r-1}, |r-1|)cdot(frac {-rcostheta}{2sqrt{r-1}},frac{-rsin(theta)}{2sqrt{r-1}},r)\
r^2-r$

However, I see a different problem

$-2le z le 1$ is not consistent with $z = sqrt {r-1}$