Calculate the flux of the vector field f(x, y, z) = 3hx + y, x − y, x2 + y 2 − 2zi through the surface s parametrized by φ(u, v) = u + 2v, u − 2v, u2 + 2v 2 with 0 ≤ u, v ≤ 1, and oriented by φu × φv

Respuesta :

I'm not sure what to make of the "h" and "i" in your question, so I'll just ignore them (and the 3). Looks like we have

[tex]\vec f(x,y,z)=(x+y,x-y,x^2+y^2-2z)[/tex]

and a surface parameterzied by

[tex]\varphi(u,v)=(u+2v,u-2v,u^2+2v^2)[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Then

[tex]\varphi_u\times\varphi_v=(4u+4v,4u-4v,-4)[/tex]

so that the flux is given by the integral

[tex]\displaystyle\iint_S\vec f\cdot\mathrm d\vec S=\int_0^1\int_0^1(2u,4v,4v^2)\cdot(4u+4v,4u-4v,-4)\,\mathrm du\,\mathrm dv[/tex]

[tex]=\displaystyle8\int_0^1\int_0^1(u^2+3uv-4v^2)\,\mathrm du\,\mathrm dv=\boxed{-2}[/tex]