Question

Find the flux of the vector field Bold Upper F equals left angle 0 comma 0 comma negative 2 right angle across the slanted face of the tetrahedron z equals 2 minus x minus y in the first octant with the normal vectors pointing in the positive​ z-direction. You may use either an explicit or parametric description of the surface.The flux is_______.

Answer 1

The plane $z=2-x-y$ has vertices at the intercepts (2, 0, 0), (0, 2, 0), and (0, 0, 2). Parameterize the tetrahedral face (call it $T$) by

$\vec s(u,v)=(1-v)((1-u)\langle2,0,0\rangle+u\langle0,2,0\rangle)+v\langle0,0,2\rangle$

$\vec s(u,v)=\langle2(1-u)(1-v),2u(1-v),2v\rangle$

for $0\le u\le1$ and $0\le v\le1$. Take the normal vector to $T$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=4(1-v)\langle1,1,1\rangle$

Then the flux of $\vec F$ across $T$ is

$\displaystyle\iint_T\vec F\cdot\mathrm d\vec S=4\int_0^1\int_0^1(1-v)\langle0,0,-2\rangle\cdot\langle1,1,1\rangle\,\mathrm du\,\mathrm dv$

$=\displaystyle-8\int_0^1(1-v)\,\mathrm dv=\boxed{-4}$