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

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explanation: