Consider the vector field $\vec{F}(x,y,z)=(x,y,z)$, and the surface parametrized by $\Phi(u,v)=(uv,\frac{u^{2}+v^{2}}{2},\frac{u^{2}-v^{2}}{2})$ where $0\leq u\leq 2 $ and $0\leq v \leq 4$. Evaluate the surface integral $\iint_{S}\vec{F}\cdot d\vec{S}$.
This question doesn't seem too bad. $\iint_{S}\vec{F}\cdot d\vec{S}=\iint_D{F(\vec\Phi(u,v))\cdot(\vec{T_{u}}\times \vec{T_v})}$ where $\vec{T_{u}}$ and $\vec{T_{v}}$ are tangent vectors to the surface, and D is the domain of the surface in u v space.
$\vec{T_{u}}\times \vec{T_v}=(-2uv,u^{2}+v^{2},v^2-u^2)$ and $F(\vec\Phi(u,v))=(uv,\frac{u^{2}+v^{2}}{2},\frac{u^{2}-v^{2}}{2})$. Taking the dot product gives $F(\vec\Phi(u,v))\cdot(\vec{T_{u}}\times \vec{T_v})=0$ so, we should have: $\iint_{S}\vec{F}\cdot d\vec{S}=\iint_D{F(\vec\Phi(u,v))\cdot(\vec{T_{u}}\times \vec{T_v})}=\iint_D(0)=0$
But apparently the answer is $\frac{-1024}{9}$. I've been looking at this for ages and I have no idea where I went wrong. Any help would be much appreciated.