Question

What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the portion of the elliptic cylinder r(u,v) = (2cos v) i + (sin v) j + (u) k with 0 ≤ u ≤ 5, 0 ≤ v ≤ 2pi.

Answer 1

`\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS`

`\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS`

`\displaystyle\iint_\sigma\mathbf F\cdot\left((\mathbf r_u*\mathbf r_v)/(\|\mathbf r_u*\mathbf r_v\|)\right)\|\mathbf r_u*\mathbf r_v\|\,\mathrm dA`

`\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u*\mathbf r_v)\,\mathrm dA`

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have


`\mathbf r_u=\frac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k`

`\mathbf r_v=\frac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j`

The cross product is


`\mathbf r_u*\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j`

So, the flux is given by


`\displaystyle\iint_\sigma(e^(-\sin v)\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA`

`\displaystyle\int_0^5\int_0^(2\pi)(-e^(-\sin v)\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du`

`\displaystyle-5\int_0^(2\pi)e^(-\sin v)\cos v\,\mathrm dv+10\int_0^(2\pi)\sin^2v\,\mathrm dv`

`\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^(2\pi)(1-\cos2v)\,\mathrm dv`

where
`t=-\sin v` in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with


`\displaystyle5\int_0^(2\pi)(1-\cos2v)\,\mathrm dv=5\left(v-\frac12\sin2v\right)\bigg|_(v=0)^(v=2\pi)=10\pi`