Question

Find the flux of the following vector fields across the given surface with the specified orientation. You may use either an explicit or parametric description of the surface.

44. F=〈x,y,z〉across the slanted face of the tetrahedron z=10−2x−5y in the first octant; normal vectors point upward.

Answer 1

The tetrahedron passes through the intercepts (5, 0, 0), (0, 2, 0), and (0, 0, 10). Parameterize the surface (call it
`\Sigma`) by

`\vec r(u,v)=(1-v)\langle5,0,0\rangle+v\left((1-u)\langle0,2,0\rangle+u\langle0,0,10\rangle\right)`

`\vec r(u,v)=\langle5(1-v),2(1-u)v,10uv\rangle`

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

`\vec r_v*\vec r_u=\langle20v,50v,10v\rangle`

Then the flux of
`\vec F(x,y,z)=\langle x,y,z\rangle` across
`\Sigma` is

`\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\langle5(1-v),2(1-u)v,10uv\rangle\cdot\langle20v,50v,10v\rangle\,\mathrm du\,\mathrm dv`

`=\displaystyle\int_0^1\int_0^1100v\,\mathrm du\,\mathrm dv=\boxed{50}`