Question

Using the divergence theorem, find the outward flux of f across the boundary of the region d, where d is the solid cube cut by the coordinate planes and the planes x2i y2j zk

Answer 1

I'll assume a generic cube of side length
`s`, so that

`D = \left\{(x,y,z) \in \Bbb R^3 ~:~ (x,y,z) \in [0,s]^3\right\}`

By the divergence theorem, the flux of

`\vec f(x,y,z) = x^2\,\vec\imath + y^2\,\vec\jmath + z\,\vec k`

across the boundary of
`D` is equivalent to the volume integral of
`\mathrm{div}(\vec f)` over
`D`.

We have

`\mathrm{div}(\vec f) = (\partial x^2)/(\partial x) + (\partial y^2)/(\partial y) + (\partial z)/(\partial z) = 2x + 2y + 1`

and so the flux is

`\displaystyle \iint_(\partial D) \vec f \cdot d\vec S = \iiint_D (2x+2y+1)\,\dV \\\\ ~~~~~~~~ = \int_0^s \int_0^s \int_0^s (2x+2y+1) \, dz \, dy \, dx \\\\ ~~~~~~~~ = s \int_0^s \int_0^s (2x + 2y + 1) \, dy \, dx \\\\ ~~~~~~~~ = s \int_0^s ((2x+1)s + s^2) \, dx \\\\ ~~~~~~~~ = s^2 \int_0^s (2x + 1 + s) \, dx \\\\ ~~~~~~~~ = s^2 (s^2 + s + s^2) = \boxed{2s^4 + s^3}`