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) = xy i + yz j + zx k S is the part of the paraboloid z = 2 − x2 − y2 that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation

Respuesta :

Parameterize [tex]S[/tex] by

[tex]\vec r(u,v)=u\,\vec\imath+v\,\vec\jmath+(2-u^2-v^2)\,\vec k[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Take the normal vector to [tex]S[/tex] to be

[tex]\vec r_u\times\vec r_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k[/tex]

Then the flux of [tex]\vec F[/tex] across [tex]S[/tex] is

[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1(uv\,\vec\imath+v(2-u^2-v^2)\,\vec\jmath+u(2-u^2-v^2)\,\vec k)\cdot(\vec r_u\times\vec r_v)\,\mathrm du\,\mathrm dv[/tex]

[tex]=\displaystyle\int_0^1\int_0^1(2u^2v+(u+2v^2)(2-u^2-v^2))\,\mathrm du\,\mathrm dv=\boxed{\frac{293}{180}}[/tex]

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