Question

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

Answer 1

Answer:

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu$ ≈ 3077.34

Explanation:

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

F(x,y,z) =  y i − x j + $z^{2}$ 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:  

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu$

How in the surface

x = ucosv

y = usinv

z = v

Then

F(S(u,v)) = usinv i - ucosvj +$v^{2}$k

The normal vector N is equal to

$N = S_{u}XS_{v}$

Where:

$S_{u} = =$

$S_{v} = =$

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

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

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

F(S(u,v)) .N = 2uv + u$v^{2}$

Thus

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {2uv}+u v^{2}\,dvdu$ ≈ 3077.34