Question

Evaluate the surface integral ∫∫s F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = x i + y j + 10 k. S is the boundary of the region enclosed by the cylinder x2 + z2 = 1 and the planes y = 0 and x + y = 8.

Answer 1

Use the divergence theorem.

`\vec F(x,y,z)=x\,\vec\i\math+y\,\vec\jmath+10\,\vec k\implies\\abla\cdot\vec F(x,y,z)=(\partial(x))/(\partial x)+(\partial(y))/(\partial y)+(\partial(10))/(\partial z)=2`

The div theorem says the integral of
`\vec F` across
`S` is equal to the integral of
`\\abla\cdot\vec F` over the region with boundary
`S` (call it
`R`):

`\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R(\\abla\cdot\vec F(x,y,z))\,\mathrm dx\,\mathrm dy\,\mathrm dz`

`=\displaystyle2\iiint_R\mathrm dx\,\mathrm dy\,\mathrm dz`

Convert to cylindrical coordinates:

`\begin{cases}x=r\cos\theta\\y=y\\z=r\sin\theta\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dy`

Then the integral is

`=\displaystyle2\int_0^(2\pi)\int_0^1\int_0^(8-r\cos\theta)r\,\mathrm dy\,\mathrm dr\,\mathrm d\theta=\boxed{16\pi}`