Question

Find the flux of F=(x^5+y^5+z^5-2x-3y-4z)i+sin(2y)j+4zsin^2(y)k across the surface of the tetrahedron bounded by the coordinate planes and the plane x+y+z=1

Answer 1

Use the divergence theorem.


`\mathbf F(x,y,z)=(x^5+y^5+z^5-2x-3y-4z)\,\mathbf i+\sin2y\,\mathbf j+4z\sin^2y\,\mathbf k`

`\implies(\\abla\cdot\mathbf F)(x,y,z)=(\partial(x^5+y^5+z^5-2x-3y-4z))/(\partial x)+(\partial(\sin2y))/(\partial y)+(\partial(4z\sin^2y))/(\partial z)=5x^4-2+2\cos2y+4\sin^2y`

The flux of
`\mathbf F` across the tetrahedron's surface
`S` is then given by the integral of
`\\abla\cdot\mathbf F` over the interior of the tetrahedron
`\mathbf R`.


`\displaystyle\iint_S\mathbf F\cdot\mathrm dS=\iiint_T\\abla\cdot\mathbf F\,\mathrm dV`

`=\displaystyle\int_(x=0)^(x=1)\int_(y=0)^(y=1-x)\int_(z=0)^(z=1-x-y)(5x^4-2+2\cos2y+4\sin^2y)\,\mathrm dz\,\mathrm dy\,\mathrm dx`

`=\frac1{42}`