Respuesta :
Answer:
[tex]\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu[/tex] ≈ 3077.34
Explanation:
For calculating the flux of F (vector field) across the surface S, where
F(x,y,z) = y i − x j + [tex]z^{2}[/tex] 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:
[tex]\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu[/tex]
How in the surface
x = ucosv
y = usinv
z = v
Then
F(S(u,v)) = usinv i - ucosvj +[tex]v^{2}[/tex]k
The normal vector N is equal to
[tex]N = S_{u}XS_{v}[/tex]
Where:
[tex]S_{u} = <\frac{dx}{du} , \frac{dy}{du}, \frac{dz}{du}> = <cosv, sinv, 0>[/tex]
[tex]S_{v} = <\frac{dx}{dv} , \frac{dy}{dv}, \frac{dz}{dv}> = <-usinv, ucosv, 2v>[/tex]
N = <cosv, sinv, 0> X <-usinv, ucosv, 2v
N = <2vsinv, -2vcosv, u>
F(S(u,v)) .N = <usinv, -ucosv,[tex]v^{2}[/tex]>.<2vsinv, -2vcosv, u>
F(S(u,v)) .N = 2uv + u[tex]v^{2}[/tex]
Thus
[tex]\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {2uv}+u v^{2}\,dvdu[/tex] ≈ 3077.34
