Question

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 = 7 − x2 − y2 that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation.

Answer 1

Parameterize $S$ by

$\vec r(u,v)=u\,\vec\imath+v\,\vec\jmath+(7-u^2-v^2)\,\vec k$

with $0\le u\le 1$ and $0\le v\le1$. Take the normal vector to be

$\vec r_u\times\vec r_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k$

Then the flux across $S$ is

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

$\displaystyle=\int_0^1\int_0^1(uv\,\vec\imath+(7-u^2-v^2)v\,\vec\jmath+(7-u^2-v^2)u\,\vec k)\cdot(\vec r_u\times\vec r_v)\,\mathrm du\,\mathrm dv$

$\displaystyle\int_0^1\int_0^1(2u^2v+(u+2v^2)(7-u^2-v^2))\,\mathrm du\,\mathrm dv=\frac{1343}{180}$