A smooth surface S is parametrized by
$r(u,v) = \cos(2u)(2 + v\cos(u)) \vec{i} + \sin(2u)(2 + v\cos(u)) \vec{j} + v\sin(u) \vec{k}$
where $0 \le u \le 2 \pi, -1 \le v \le 1$. Prove that for every smooth vector field $\vec{F}$ on S, $\iint_{S} \vec{F} \cdot d \vec{S} = 0$ (actually the original question is $\int_{S} \vec{F} \cdot d \vec{S} = 0$, but I think it's wrong since S has two dimension?)
My attempt is to calculate $r_u$, $r_v$, and $r_u \times r_v$
$r_u \times r_v = \{4\cos(2u)\sin(u) + 2v\cos(u)\cos(2u)\sin(u) - v\sin(2u) \} \vec{i} + \{v\cos(2u) + 4\sin(u)\sin(2u) + v\sin^2(2u) \} \vec{j} + \{ -4\cos(u)-2v\cos^2(u) \} \vec{k}$
and let $\vec{F} = P \vec{i} + Q \vec{j} + R \vec{k}$
then $$\iint_{S} \vec{F} \cdot d \vec{S} = $$ $$\int_{-1}^{1} \int_0^{2\pi} P\{4\cos(2u)\sin(u) + 2v\cos(u)\cos(2u)\sin(u) - v\sin(2u) \} + Q\{v\cos(2u) + 4\sin(u)\sin(2u) + v\sin^2(2u) \} + R\{ -4\cos(u)-2v\cos^2(u) \} \,du\,dv = $$ $$ \int_{-1}^{1} 0 \,dv = 0$$
However I don't know whether it is right way to do this question. Can anyone tell me? Thanks !
btw if there's a better way to do this, please tell me, thanks!
When I read this question I started thinking: This cannot be. That's why I made a picture of half your surface. Looking at this picture you should be able to find out why all these flux integrals have the value $0$.
The essential point is that $${\bf r}(\pi+u,v)={\bf r}(u,-v)\qquad(0\leq u\leq\pi, \ -1\leq v\leq 1)\ .$$ This means that any any spacial surface element ${\rm d}S$ occurring in the total computation is covered twice by the parametrization, but with unit normals of opposite direction. The total flux will then be $=0$ for any velocity field ${\bf v}$ defined in a neighborhood of $S$.