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Lerok [7]
2 years ago
9

Evaluate the surface integraliintegral.gifSF � dSfor the given vector field F and the oriented surface S. In other words, find t

he flux of F across S. For closed surfaces, use the positive (outward) orientation.F(x, y, z) = xy i + yz j + zx kS is the part of the paraboloidz = 2 ? x2 ? y2 that lies above the square 0 ? x ? 1, 0 ? y ? 1,and has upward orientation
Mathematics
1 answer:
ehidna [41]2 years ago
8 0

Looks like the paraboloid has equation

z=2-x^2-y^2

and S is the part of this surface with 0\le x\le1 and 0\le y\le1. Parameterize S by

\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(2-u^2-v^2)\,\vec k

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

\vec s_u\times\vec s_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k

Then the flux of \vec F across S is

\displaystyle\iint_S\vec F\cdot\mathrm d\vec S

\displaystyle=\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(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv

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

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