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

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

Then the flux of $\vec F$ across $S$ is

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

$\displaystyle=\int_0^1\int_0^1(uv\,\vec\imath+v(2-u^2-v^2)\,\vec\jmath+u(2-u^2-v^2)\,\vec k)\cdot(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv$

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