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 = 2

Answer 1

`\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+10\,\mathbf k`

`\implies\\abla\cdot\mathbf f(x,y,z)=1+1+0=2`

By the divergence theorem, the surface integral along
`S` is equivalent to the triple integral over the region
`R` bounded by
`S`:


`\displaystyle\iint_S\mathbf f(x,y,z)\,\mathrm dS=\iiint_R\\abla\cdot\mathbf f(x,y,z)\,\mathrm dV=2\iiint_R\mathrm dV`

Convert to cylindrical coordinates, setting


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

The triple integral is then equivalent to


`=\displaystyle2\int_(\theta=0)^(\theta=2\pi)\int_(r=0)^(r=1)\int_(Y=0)^(Y=2-r\cos\theta)r\,\mathrm dY\,\mathrm dr\,\mathrm\theta`

`=\displaystyle2\int_(\theta=0)^(\theta=2\pi)\int_(r=0)^(r=1)r(2-r\cos\theta)\,\mathrm dr\,\mathrm\theta`

`=\displaystyle\frac23\int_(\theta=0)^(\theta=2\pi)(3-\cos\theta)\,\mathrm dr\,\mathrm\theta`

`=4\pi`