anonymous
  • anonymous
Use Stokes' Theorem to find the circulation of F =4yi +3zj +7xk around the triangle obtained by tracing out the path (6,0,0) to (6,0,5), to (6,5,5) back to (6,0,0). Circulation = ∫CF ⋅dr
Calculus1
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
so the curl is -3i-7j-4k
anonymous
  • anonymous
If you want to go from line integral to surface integral, shouldn't you need the anticurl?
anonymous
  • anonymous
What is the anticurl?

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anonymous
  • anonymous
Nevermind, they gave the anticurl
anonymous
  • anonymous
I see
anonymous
  • anonymous
I mean A such that \[ B = \nabla \times A \]
anonymous
  • anonymous
But apparently they gave it to you already.
anonymous
  • anonymous
do you know how I would set up the line integral. I know that I have to use the points to make an equation of the triangle but after that I'm not sure
anonymous
  • anonymous
So to use stokes theorem, you need to parametrize the surface.
anonymous
  • anonymous
They don't want you to set up the line integral, but I can help you do that if you want.
anonymous
  • anonymous
The surface is the triangle?
anonymous
  • anonymous
I want to be able to do it using stoke's theorem/how they want it
anonymous
  • anonymous
Here is stokes theorem: \[ \iint_S\mathbf{F}\cdot d\mathbf{S}=\int_{\partial S}\nabla \times \mathbf{F}\cdot d\mathbf{r} \]
anonymous
  • anonymous
Now, the only way to use stokes theorem here is to do the left (surface integral) side. If you did the right (line integral side) would you really be using stokes?
anonymous
  • anonymous
No. I guess I was making a mistake. So how would I parametrize the surface? After that would the surface integral require a normal?
anonymous
  • anonymous
I'm getting line integrals and surface and green's theorem andstoke's theorem confused since we are barely covering them.
anonymous
  • anonymous
You can easily find two parallel vectors for the triangle right?
anonymous
  • anonymous
the yz-plane?
anonymous
  • anonymous
No, you do it by finding a vector between two points.
anonymous
  • anonymous
Between two points. So would I have to use points not contained in the triangle?
anonymous
  • anonymous
?
anonymous
  • anonymous
oh wow, you're right.
anonymous
  • anonymous
yeah so we want the bounds on the yz plane.
anonymous
  • anonymous
okay so since I know the yz plane is parallel to the trianle would I be able to choose any 2 points there?
anonymous
  • anonymous
oh so it is a projection onto the yz plane
anonymous
  • anonymous
of the triangle, right?
anonymous
  • anonymous
|dw:1362708042969:dw|
anonymous
  • anonymous
okay so does that tell me what the limits of integration for z are going to be or am I supposed to use it to parametrize the triangle?
anonymous
  • anonymous
You could parametrize it as \[ \Phi(u,v) = (6,u,v),\quad 0\le u \le 5,\quad u\le v \le 5 \]
anonymous
  • anonymous
Oh I see. Because the x value is constant, right? Then from there will I have to calculte the normal to dot it with the curl of F?
anonymous
  • anonymous
In this case \(\mathbf{F}\) doesn't need to be messed with, you only take the curl to do the line integral part.
anonymous
  • anonymous
So I just dot the parametrization of the triangle with the curl and then I just use the limits of integration for u and v?
anonymous
  • anonymous
How many times do I have to say you don't take the curl?
anonymous
  • anonymous
Sorry I keep looking up at the equation. Okay. What would be the next step in my process?
anonymous
  • anonymous
\[ \iint_S \mathbf{F}\cdot d\mathbf{S}= \iint_D\mathbf{F}\circ\Phi(u,v)\cdot \frac{\partial \Phi}{\partial u}\times \frac{\partial \Phi}{\partial v}dudv \]
anonymous
  • anonymous
\[ \mathbf{F}\circ \Phi(u,v) = \mathbf{F}(x(u,v),y(u,v), z(u,v)) \]
anonymous
  • anonymous
\(D\) corresponds to the bounds of \(u,v\).
anonymous
  • anonymous
okay so it is the vector field f in terms of u and v dotted with the normal of the parametrization?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Okay I'm going to work it out really quick.
anonymous
  • anonymous
Okay so when I did the normal I got <1,0,0> which, then dotted with <4u,3v,42> So I got 4u then I took the double integral \[\int\limits_{0}^{5}\int\limits_{u}^{5}4udv du\] that became \[\int\limits_{0}^{5}20u-4u^2du\] from there I got \[10u^2-4/3u^3\] and my answer was 250/3. Did I mess up somewhere because my answer seems incorrect? Was my set up correct?
anonymous
  • anonymous
You have to consider the orientation.
anonymous
  • anonymous
the orientation is counter is wingspanwise so that means the normal is supposed to be supposed to be negative, right, or the parametrization of the triangle?
anonymous
  • anonymous
|dw:1362709722289:dw|
anonymous
  • anonymous
Yes it should be negative.
anonymous
  • anonymous
the normal?
anonymous
  • anonymous
yeah.
anonymous
  • anonymous
Okay I change the normal to negative and then my answer just ended up being -250/3 but it is still wrong. Does it look like I did a mistake in setting it up?
anonymous
  • anonymous
Do you know what the answer is?
anonymous
  • anonymous
Okay I got it I decided to use the parametrization that you gave me. Then I took the gradient of it and dotted it with the curl I got. so it was <-3,-7,-4> dotted with the gradient of <6,-u,v> which gave me 7-4 which is 3. Then I used all the limits of integration you gave me and got the correct answer which is 75/2 . I'm not sure if this is another part of the formula but it worked. Thank you very much! Would you gladly help me with one more wuestion if I post it up?

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