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.
nice try bob
you're not that smart
Which is why I'm posting this. ._.
is this your question? or one you got from google?
pm me i have the answer
You can use the Divergence Theorem for this. Start with \(div ~ \mathbf F = 0 -1 + x \cos z = x \cos z -1 \) So you get: \(\Phi = \int_A \mathbf F \cdot d \mathbf A = \int_V div \mathbf F ~ dV\) \(\implies \int_V x cos z - 1 ~ dV = \int_V x cos z - 1 ~ dx ~dy ~ dz\) Then make some sensible subs and get the Jacobian right. Haven't done it myself, or had a look at the flux at the end caps, but it all tends to work out Ok on these type of questions. Quite nasty though.
Just realised that the flux through the end caps is zero so you can use a surface integral as well for the bottom end cap, \(\hat n = <0,0,-1> \) and z = 0 so \(\hat n \cdot \mathbf F = 0\) for the top cap \(\hat n = <0,0,1> \), z = 5 so \(\hat n \cdot \mathbf F = x sin 5\). With symmetry, positive and negative x-values cancel to give zero flux there too. still think you need to sort out that elliptical cylinder though with a substutition
If you're question is answered, please close it. Thank you.
I don't understand that language @Elsa213
not to mention your second word was a grammatical error
your*
better
Sorry cx What do you not understand?
yer layngwidge eez yewsholly theeez
Ef yewr question es answered, pl0x close et. Tenk yew.
That's better? e.e
yeee i wuz playeeng roblox dee udder day and i wuz infinite jumpeeeng
peeeple wer mad and repurting mee
Deed yew keel em?
yeees
Congratulationss, boi
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