PsychoBob
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| Joined: 08 Jul 2009 |
| Total Posts: 11047 |
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| 19 Aug 2012 02:23 AM |
Find the flux through a surface described as: z = x²y-2xy+x-5y 0 < x < 3 0 < y < 3 With a vector field described as: v(p, q, r) = pq[1, 0, 0] + (q²-p sin r)[0, 1, 0] + (s² + cos r cos p sin s)[0, 0, 1]
If anyone can even set up the proper flux integral I'll be more than impressed already |
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700gb
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| Joined: 08 Feb 2011 |
| Total Posts: 7503 |
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| 19 Aug 2012 02:24 AM |
| We'll get the anwser right after these messages! |
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PsychoBob
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| Joined: 08 Jul 2009 |
| Total Posts: 11047 |
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| 19 Aug 2012 02:33 AM |
Oh come on
None of you can even do the parametrization? |
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vienna456
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| Joined: 12 Sep 2010 |
| Total Posts: 15043 |
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| 19 Aug 2012 02:39 AM |
| Do your own homework.....lol |
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vienna456
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| Joined: 12 Sep 2010 |
| Total Posts: 15043 |
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| 19 Aug 2012 02:41 AM |
we could do it for you.....but then how would you learn...
lol |
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PsychoBob
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| Joined: 08 Jul 2009 |
| Total Posts: 11047 |
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| 19 Aug 2012 02:42 AM |
Alright, I'll at least get you started on the parametrization
The nice things about polynomial regions is that they parametrize incredibly easily
x = u y = v z = u²v - 2uv + u - 5v
So now we can make a vector valued function
F(x(u,v), y(u, v), z(u, v)) = u[1, 0, 0] + v[0, 1, 0] + (u²v - 2uv + u - 5v)[0, 0, 1] |
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| 19 Aug 2012 02:43 AM |
Sir, are you copy and pasting? it seems as if you are
Dr. Kuku, Scientist of OT |
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PsychoBob
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| Joined: 08 Jul 2009 |
| Total Posts: 11047 |
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| 19 Aug 2012 02:45 AM |
@Super
Believe it or not
I'm not
I do actually know all of this stuff and I am way too old for these forums |
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spot
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| Joined: 29 Dec 2006 |
| Total Posts: 37193 |
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| 19 Aug 2012 02:50 AM |
eh
AND I WAS LIKE, "YO DARTHNECRON, I'M COMIN FOR YOUR LOTION HOMIE" |
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