Question

Evaluate the surface integraliintegral.gifSF � dSfor 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 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

Answer 1

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*\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{(293)/(180)}`