Question

Use the Divergence Theorem to compute the net outward flux of the vector field

F=(x2,- y2,z2) across the boundary of the region​ D, where D is the region in the first octant between the planes z= 9-x-y and

z = 3-x-y

Answer 1

`\vec F(x,y,z)=(x^2,-y^2,z^2)` has divergence

`\mathrm{div}\vec F(x,y,z)=2x-2y+2z`

so that by the divergence theorem, the flux of
`\vec F` across the boundary of
`D` is

`\displaystyle\iint_(\partial D)\vec F\cdot\mathrm d\vec S=2\iiint_D(x-y+z)\,\mathrm dV`

which you can compute by subtracting [the integral of
`\mathrm{div}\vec F` over the region bounded by
`9-x-y` and the coordinate axes] and [the integral of
`\mathrm{div}\vec F` over the region bounded by
`3-x-y` and the coordinate axes].

`=2\displaystyle\left\{\int_0^9\int_0^(9-x)\int_0^(9-x-y)-\int_0^3\int_0^(3-x)\int_0^(3-x-y)\right\}(x-y+z)\,\mathrm dz\,\mathrm dy\,\mathrm dx`

`=2\left(\frac{2187}8-\frac{27}8\right)=\boxed{540}`