Problem calculating an integral over a surface I've been trying to solve this for awhile and can't find a way.
Given $ S={(x,y,z) \in R^3 : z = x^2 - y^2 , x^2 + y^2 \leq 1 } $ and $\phi :R^3 \to R $ defined as $\phi (x,y,z)= (4z +8y^2 + 1)^{3/2}$, find $\iint_{S} \phi dS $.
I think I have to use $ \int_{0}^{2\pi} \int_{0}^{1} \phi (T(r,\theta)) \left \| T_{\theta} \times T_{r} \right \| dr d\theta $
So I first parametrized S with $x= r \cos \theta $, $y= r \sin \theta $ and $ z = r [2\cos^2 \theta - 1] $. So I had $ T(r,\theta)= (r\cos\theta , r\sin\theta, r[\cos^2\theta - 1]) $.     
If I'm not mistaken I have to use $ \int_{0}^{2\pi} \int_{0}^{1} \phi (T(r,\theta) \left \| T_{\theta} \times T_{r} \right \| dr d\theta $.  
Now I need to find $ || T_{r} \times T_{\theta} || = || (r\cos\theta[2\cos^2 \theta-1+2\sin\theta], r\sin\theta, -r) ||$   
Now, as you can imagine that ends up being one messed up thing once you square the first term, and I haven't found anyway to simplify anything, which leaves me with a pretty tough integral to solve.    
This is the integral I got, but Wolfram Alpha couldn't solve it and neither could I.    
$ \int_{0}^{2\pi} \int_{0}^{1} (8 r^2 \sin^2(\theta)+4 r(2 \cos^2(\theta)-1)+1)^{3/2} \sqrt((r \sin(\theta))^2+(2 r \cos^3(\theta)+2 r \sin^2(\theta) \cos(\theta)-r \cos(\theta))^2+(-r \cos^2(\theta)-r \sin^2(\theta)|^2) dr d\theta $   
I'm guessing I either used a wrong parametrization, or there's another way to solve this not using $ \iint_{S} \phi(T(r,\theta)) \left || T_{r} \times T_{\theta} \right || dS $.   
Any help would be greatly appreciated.   
EDIT: As Didier pointed out, $z = r^2 [2 \cos^2 \theta +1] $. But that doesn't make matters a whole lot simpler since now $ || T_{r} \times T_{\theta} || = || (-4r^2 \cos \theta \sin^2 \theta - 2 r^2 \cos\theta [2\cos^2 \theta +1], -4 r^{2} \cos^{2} \theta \sin \theta + 2r^2 \sin \theta [2\cos^2\theta +1], r) || $.   
EDIT2: Another retarded mistake, fixed by Didier.
$ || T_{r} \times T_{\theta} || = (-4r^2 \cos\theta\sin^2\theta-2r^2 \cos\theta[2\cos^2 \theta -1], 4r^2 \cos^2 \theta \sin\theta-2r^2\sin\theta[2\cos^2\theta-1], r) $, still not totally right though.
I must have another error somewhere but just can't seem to find it.
 A: I believe you're still looking for something to find your mistake with. Now your things aren't looking that bad from here so I guess just giving you the solution will let you see where you're wrong in your numbers crunching.
Let $S = \{ T(r,\theta) = (r \cos \theta, r \sin \theta, r^2 (\cos^2 \theta - \sin^2 \theta)) \, | \, \theta \in [0,2\pi], r \in [0,1] \}$ be the chosen parametrization of $S$. Therefore
$\phi(T(r,\theta)) = (4(r^2(\cos^2 \theta - \sin^2 \theta)) + 8 (r \sin \theta)^2 + 1)^{3/2} $
$ = (4r^2 \cos^2 \theta - 4 r^2 \sin^2 \theta + 8 r^2 \sin^2 \theta + 1)^{3/2} = (4r^2 + 1)^{3/2}$
and
$
T_r \times T_{\theta} = 
\begin{vmatrix}
\vec i & \vec j & \vec k \\
\cos \theta & \sin \theta & 2r(\cos^2 \theta - \sin^2 \theta) \\
- r \sin \theta & r \cos \theta  & r^2(-4 \sin \theta \cos \theta)
\end{vmatrix}
$
$
= (-4r^2\sin^2 \theta \cos \theta - 2r^2 \cos^3 \theta  + 2r^2 \sin^2 \theta \cos \theta, 4r^2 \sin \theta \cos^2 \theta - 2r^2 \sin \theta \cos^2 \theta + 2r^2 \sin^3 \theta, r)
$
$
=(-2r^2 \cos \theta, 2 r^2 \sin \theta,r) \quad
$
so that $\quad ||T_r \times T_{\theta}|| = \sqrt{4r^4 + r^2} = r \sqrt{4r^2 + 1}$.
Now it's easy because
$$
\int_0^{2\pi} \int_0^1 (4r^2+1)^{3/2} (r\sqrt{4r^2 +1}) \, dr d\theta = 2\pi \int_0^1 r(4r^2 + 1)^2 \, dr
$$
which gives a number I am not in the mood for computing right now, but I think you'll be fine. =)
