Verification of Stokes Theorem I want to verify Stokes Theorem for the surface
$$
 \Phi = \{ (x,y,z) \in \mathbb R^3 : z = x^2 - y^2, x^2 + y^2 \le 1 \}
$$
and the vector field $F(x,y,z) := (y,z,x)$. For this I use the parametrisation
$$
 \varphi(r, \theta) = \begin{pmatrix} r\cos \theta \\ r \sin \theta \\ r^2(\cos^2 \theta - \sin^2\theta) \end{pmatrix} 
                             = \begin{pmatrix} r\cos \theta \\ r \sin \theta \\ r^2(2\cos^2\theta - 1) \end{pmatrix}  
$$
with $0 \le r \le 1, 0 \le \theta < 2\pi$.
Then
$$
 \varphi_r(r, \theta) = \begin{pmatrix} \cos\theta \\ \sin\theta \\ 2r(2\cos^2\theta - 1) \end{pmatrix} \qquad
 \varphi_{\theta}(r,\theta) = \begin{pmatrix} -r\sin\theta \\ r\cos\theta \\ -4r^2 \cos\theta \sin\theta \end{pmatrix}
$$
and
$$
 \varphi_r \times \varphi_{\theta} = \begin{pmatrix}
      -2r^2\cos\theta \\
       2r^2\sin\theta \\
       r 
      \end{pmatrix}. 
$$
and also
$$
 \mbox{rot}(F) = \begin{pmatrix} -1\\ -1\\ -1 \end{pmatrix}
$$
So now I have everything at hand to compute the two sides of 
$$
 \int_{\Phi} \mbox{rot}(F) \cdot \vec{n} ~ dS = \int_{\partial \Phi} F(r) \cdot dr.  
$$
For the RHS I have
$$
 \begin{align*}
 \int_0^{2\pi} F(\varphi(1, \theta)) \cdot \varphi_{\theta}(1, \theta) 
   & = r^3 \int_0^{2\pi} \cos\theta(2\cos^2\theta - 1) d\theta - r^2 \int_0^{2\pi} \sin\theta d\theta - 4r^3 \int_0^{2\pi} \cos^2\theta \sin\theta d\theta  \\
   & = 0
\end{align*}
$$
and for the LHS:
\begin{align*}
 \int_0^1 \int_0^{2\pi} \begin{pmatrix} -1\\-1\\-1\end{pmatrix} \cdot \begin{pmatrix}
      -2r^2\cos\theta \\
       2r^2\sin\theta \\
       r 
      \end{pmatrix}  d\theta dr
 & = - \int_0^1 \int_0^{2\pi} 2r^2(\sin(\theta)-\cos(\theta)) + r d\theta dr \\
 & = - \int_0^1 \int_0^{2\pi} r d\theta dr = \pi                         
\end{align*}
so these are not equal. But I looked at the calculations several times and I am sure they are right, so do you see what went wrong?

The main thing is that for your line integral you should have (remember that $r=1$)
$$\int_0^{2\pi} \big((\sin\theta)(-\sin\theta) + (2\cos^2\theta-1)(\cos\theta) + (\cos\theta)(-4\cos\theta\sin\theta)\big)\,d\theta = -\pi,$$
and you lost a minus sign in the surface integral at the end. I think you just made a copying error in part of it.
