Question

Use the divergence theorem to calculate the surface integral z s ~f d~s ; that is, calculate the ux of ~f across s, where ~f = z~i + y~j + zx~k; and s is the surface of the tetrahedron enclosed by the coordinate planes and the plane x a + y b + z c = 1 with a; b; c > 0.

Answer 1

By the divergence theorem,

$\displaystyle\iint_S\mathbf f\cdot\mathrm d\mathbf S=\iiint_R(\nabla\cdot\mathbf f)\,\mathrm dV$

where $R$ is the solid whose boundary is $S$. We have

$\nabla\cdot\mathbf f=\dfrac{\partial z}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial zx}{\partial z}=1+x$

so we set up the volume integral as

$\displaystyle\iiint_R(\nabla\cdot\mathbf f)\,\mathrm dV=\int_{x=0}^{x=1/a}\int_{y=0}^{y=(1-ax)/b}\int_{z=0}^{z=(1-ax-by)/c}(1+x)\,\mathrm dz\,\mathrm dy\,\mathrm dx$
$=\dfrac{4a+1}{24a^2bc}$