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Mathematics 38 Online
Bob:

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.

AMQ:

nice try bob

AMQ:

you're not that smart

Bob:

Which is why I'm posting this. ._.

AMQ:

is this your question? or one you got from google?

SkyVoltage43:

pm me i have the answer

sillybilly123:

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.

sillybilly123:

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

Elsa213:

If you're question is answered, please close it. Thank you.

Bob:

I don't understand that language @Elsa213

Bob:

not to mention your second word was a grammatical error

Elsa213:

your*

Bob:

better

Elsa213:

Sorry cx What do you not understand?

Bob:

yer layngwidge eez yewsholly theeez

Elsa213:

Ef yewr question es answered, pl0x close et. Tenk yew.

Elsa213:

That's better? e.e

Bob:

yeee i wuz playeeng roblox dee udder day and i wuz infinite jumpeeeng

Bob:

peeeple wer mad and repurting mee

Elsa213:

Deed yew keel em?

Bob:

yeees

Elsa213:

Congratulationss, boi

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