I'm getting two different results by using two methods that should be equivalent...
"Find the line integral $\int x.dx+y.dy+(x^2+y^2).dz$ along the curve C. C is the intersection between paraboloid $z=2-x^2-y^2$ and cone $z=\sqrt(x^2+y^2)$. Verify using Stokes theorem."
So I have the vector field $\overrightarrow F (x,y,z) = <x;y;x^2+y^2>$ and from an equation I got the intersection between the two surfaces: $$ \begin{align} z &= 2-x^2-y^2\\ x^2+y^2 &= 2-z\\ \sqrt{x^2+y^2} &= \sqrt{2-z}\\ z &= \sqrt{2-z}\\ z^2 &= 2-z\\ z^2+z-1 &= 0 \end{align} $$ So either $z=1$ or $z=-2$, but $z=-2$ is not in the positive half of the cone, so I discard it. From there I get that: $$ \begin{align} x^2+y^2 &= 2-1\\ x^2+y^2 &= 1 \end{align} $$
If I parameterize the surfaces using cylindrical coordinates: $$ x^2+y^2=r^2\\ x= r. cos(\theta)\\ y= r. sin(\theta)\\ z= 2-r^2 $$ with $0 \leqslant \theta \leqslant 2 \pi$ and $0\leqslant r \leqslant 1$.
Then my parameterization is $\alpha(\theta, r)= <r.cos (\theta); r. sin (\theta); 2-r^2>$
To get the normal vector I have to find $\alpha_r \times \alpha_\theta$:
$$\alpha_r = <cos(\theta); sin(\theta); -2r)>\\ \alpha_\theta = <-r.sin(\theta); r.cos(\theta); 0>$$ and this is done by: $$ \begin{vmatrix} cos(\theta) & sin(\theta) & -2r \\ -r.sin(\theta) & r.cos(\theta) & 0 \\ \end{vmatrix} $$ So from this operation, my normal vector is $\overrightarrow N = <2.r^2 . cos(\theta); 2.r^2.sin(\theta); r>$
Then, I proceed to obtain the line integral by the "regular" method and by Stokes theorem. First, by the regular method I got:
$$\int_C \overrightarrow F . \overrightarrow{ds} = \iint_S \overrightarrow F(\alpha(\theta, r)) \cdot (\alpha_r \times \alpha_\theta) \, dA$$ $$ \int_0^{2\pi} \int_0^1 <r.cos(\theta); r.sin(\theta)>; r^2> \cdot <2.r^2.cos(\theta); 2.r^2.sin(\theta); r> dr \, d\theta =\\ = \int_0^{2\pi} \int_0^1 (2.r^3.cos^2(\theta)+2.r^3.sin^2(\theta)+r^3) dr \, d\theta =\\ = \int_0^{2\pi} \int_0^1 2.r^3(cos(\theta)+sin(\theta)+r^3)dr \, d\theta = \\ = \int_0^{2\pi} \int_0^1 (2.r^3+r^3)dr \, d\theta= \\ = \int_0^{2\pi} \int_0^1 3.r^3 dr \, d\theta = \\ = \int_0^{2\pi} ( \frac 34 r^4) |_0^1 . d\theta =\\ = \int_0^{2\pi} \frac 34 d\theta =\\ = ( \frac 34 \theta ) |_0^{2\pi}= \frac 32 \pi $$
Then, by Stokes theorem I have to find the curl of $\overrightarrow F$ to calculate $\iint_S curl \overrightarrow F . \overrightarrow{ds}$
$$ curl \overrightarrow F = 2y-2x $$
And by plugging the parameterization into this curl I get: $curl \overrightarrow F (r,\theta)=<2.r.sin(\theta); -2.r.cos(\theta); 0>$
Now I did $\int_C \overrightarrow F \cdot \overrightarrow {ds} = \iint_S curl \overrightarrow F \cdot \overrightarrow N \cdot ds$ and this is what I got: $$ \int_0^{2\pi} \int_0^1 <2.r^2.cos(\theta); 2.r^2.sin(\theta)> \cdot <2.r.sin(\theta); -2.r.cos(\theta);0> dr \, d\theta= \\ =\int_0^{2\pi} \int0_1 (4.r^3.cos(\theta).sin(\theta)-4.r^3.cos(\theta).sin(\theta)+0) dr \, d\theta=\\ \int_0^{2\pi} \int_0^1 0 \, dr \, d\theta = 0 $$
So I got two different results and I'm not sure why...
You're mixing the procedures of Stokes' theorem when evaluating line integrals. First, we have
$$\underline{F}(x,y)=\left<x,y,x^2+y^2\right>$$
Now let $\underline{\gamma}(\theta)$ trace $C$, a circle of radius $1$ sitting at $z=1$, so
$$\underline{\gamma}(\theta)=\left<\cos\theta,\sin\theta,1\right>$$
The line integral is then
$$\oint_C\underline{F}(x,y)\cdot d\underline{\gamma}=\int_0^{2\pi}\underline{F}(\theta)\cdot\underline{\gamma}'(\theta)d\theta$$
Substituting and writing on the dot product gives
$$\int_0^{2\pi}\left(-\sin\theta\cos\theta+\sin\theta\cos\theta\right)d\theta=0$$
I think you got it, but remember curl of a vector field is another vector field, not a scalar field
$$\nabla\times\underline{F}(x,y)=2y\underline{i}-2x\underline{j}+0\underline{k}$$
Also, the dot product requires that both vectors are of the same dimension. $\left<a,b\right>\cdot\left<c,d,e\right>$ makes no sense