Question

Find the flux of F = x^3 i  + y^3 j  + z^3k through the closed surface bounding the solid region x^2 + y^2 ≤ 4, 0 ≤ z ≤ 4

Answer 1

Use the divergence theorem. Let
`R` be the cylindrical region, then


`\displaystyle\iint_(\partial R)\mathbf F\cdot\mathbf n\,\mathrm dS=\iiint_R\\abla\cdot\mathbf F\,\mathrm dV`

(where
`\mathbf n` denotes the unit normal vector to
`\partial R`, but we don't need to worry about it now)

We have


`\mathrm{div }\mathbf F=(\\abla\cdot\mathbf F)(x,y,z)=(\partial\mathbf F)/(\partial x)+(\partial\mathbf F)/(\partial y)+(\partial\mathbf F)/(\partial z)`

`\\abla\cdot\mathbf F=3x^2+3y^2+3z^2`

For the solid
`R` with boundary
`\partial R`, we can set up the following volume integral in cylindrical coordinates for ease:


`\displaystyle3\iiint_R(x^2+y^2+z^2)\,\mathrm dV=3\int_(z=0)^(z=4)\int_(\theta=0)^(\theta=2\pi)\int_(r=0)^(r=2)(r^2+z^2)r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz`

`=\displaystyle6\pi\int_(z=0)^(z=4)\int_(r=0)^(r=2)(r^3+rz^2)\,\mathrm dr\,\mathrm dz`

`=\displaystyle12\pi\int_(z=0)^(z=4)(2+z^2)\,\mathrm dz`

`=352\pi`