Question

Compute the surface integral over the given oriented surface: F=y3i+8j−xk, portion of the plane x+y+z=1 in the octant x,y,z≥0, downward-pointing normal

Answer 1

Parameterize the surface (call it $S$) by

$\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(1-u-v)\,\vec k$

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

$\vec s_v\times\vec s_u=-\vec\imath-\vec\jmath-\vec k$

Then the integral of $\vec F$ over $S$ with the given orientation is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1(v^3\,\vec\imath+8\vec\jmath-u\,\vec k)\cdot(-\vec\imath-\vec\jmath-\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(-v^3-8+u)\,\mathrm du\,\mathrm dv=\boxed{-\frac{31}4}$