Since the field vectors $\displaystyle F(x,y,z) = xi +0j +zk$ and the cylinder obtained by rotation the straight: $\displaystyle x=1; y=t; z=0$ with $\displaystyle -1 \leq t \leq 1$ around the axis OY. Determine the flow of the field F in the direction normal outside of the cylinder

My solution:

Rotating

$\displaystyle F(r(t,u)) = (cosu,t,sinu)$

normal: $\displaystyle N = (\frac{ \partial (r)}{ \partial (t)} X \frac{ \partial (r)}{ \partial (u)}) = (cosu,0,sinu)$

$\displaystyle \int_0^{2 \pi} \int_{-1}^1 (cosu,t,sinu).(cosu,0,sinu) = 4 \pi$

It is correct ?