Vector field and the cylinder obtained by rotation the straight with around the OY axis. Determine the flow of the field F in the direction normal outside of the cylinder. (the normal vector points to out)
If you know the divergence theorem it is simply 2*V = 4pi
Even though this is not a closed surface the flux through the right and left faces is 0 (j component of F is 0) you can close up the cylinder and use the divergence theorem
If Not
The equation of the cylinder is then x^2 +z^2 = 1 -1< y <1
You'll have to break this up into the upper half and lower half over the rectangle -1 < x < 1 and -1 < y <1
For the upper half N = x/z i + k For the lower half N = -x/z i - k
You'll get 2pi in both cases
Divergence Theorem : Flux is triple integral of divergence
Here div = 2 volume is pi (1)^2 *2 = 4pi
Not sure where the integral comes from but even with this integral is 2*2pi is still 4pi
If you're not going to use the divergence theorem
then you're just going to have to look up calculating N for a surface
where z = f(x,y)---do the work