Evaluate the surface integral s f · ds for the given vector field f and the oriented surface s. in other words, find the flux of f across s. for closed surfaces, use the positive (outward) orientation. f(x, y, z) = y i − x j + z2 k s is the helicoid (with upward orientation) with vector equation r(u, v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 5π

≈ 3077.34

Explanation:

For calculating the flux of F (vector field) across the surface S, where

F(x,y,z) =  y i − x j + k

and S(u,v) =  u cos v i + u sin v j + v k,  0 ≤ u ≤ 2, 0 ≤ v ≤ 5π

We evaluate the following integral:  

How in the surface

x = ucosv

y = usinv

z = v

Then

F(S(u,v)) = usinv i - ucosvj +k

The normal vector N is equal to

Where:

N = <cosv, sinv, 0> X <-usinv, ucosv, 2v

N = <2vsinv, -2vcosv, u>

F(S(u,v)) .N = <usinv, -ucosv,>.<2vsinv, -2vcosv, u>

F(S(u,v)) .N = 2uv + u

Thus

≈ 3077.34

explanation:

newton's third law, for every action there is an equal and opposite reaction.

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it