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

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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

Physics

1 answer:

inna [3.1K] · Answer 1 · 2026-03-28T18:03:45+03:00

Answer:

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

Explanation:

To calculate the flux of F (vector field) across 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 will evaluate the following integral:

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

Substituting for the surface

x = u cos v

y = u sin v

z = v

Then

F(S(u,v)) = u sin v i - u cos v j + $v^{2}$ k

The normal vector N is computed as

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

Where:

$S_{u} = =$

$S_{v} = =$

N = < cos v, sin v, 0 > X <- u sin v, u cos v, 2v

N = < 2v sin v, -2v cos v, u >

F(S(u,v)).N = < u sin v, -u cos v, $v^{2}$ >. < 2v sin v, -2v cos v, 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