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.

$\mathbf F(x,y,z)=(x^5+y^5+z^5-2x-3y-4z)\,\mathbf i+\sin2y\,\mathbf j+4z\sin^2y\,\mathbf k$
$\implies(\nabla\cdot\mathbf F)(x,y,z)=\dfrac{\partial(x^5+y^5+z^5-2x-3y-4z)}{\partial x}+\dfrac{\partial(\sin2y)}{\partial y}+\dfrac{\partial(4z\sin^2y)}{\partial z}=5x^4-2+2\cos2y+4\sin^2y$

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

$\displaystyle\iint_S\mathbf F\cdot\mathrm dS=\iiint_T\nabla\cdot\mathbf F\,\mathrm dV$
$=\displaystyle\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}\int_{z=0}^{z=1-x-y}(5x^4-2+2\cos2y+4\sin^2y)\,\mathrm dz\,\mathrm dy\,\mathrm dx$
$=\dfrac1{42}$