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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 :

LammettHash

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

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

[tex]\vec r(u,v)=\langle5(1-v),2(1-u)v,10uv\rangle[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Take the normal vector to [tex]\Sigma[/tex] to be

[tex]\vec r_v\times\vec r_u=\langle20v,50v,10v\rangle[/tex]

Then the flux of [tex]\vec F(x,y,z)=\langle x,y,z\rangle[/tex] across [tex]\Sigma[/tex] is

[tex]\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[/tex]

[tex]=\displaystyle\int_0^1\int_0^1100v\,\mathrm du\,\mathrm dv=\boxed{50}[/tex]

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