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anonymous
 3 years ago
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
anonymous
 3 years ago
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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the curl is 3i7j4k

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you want to go from line integral to surface integral, shouldn't you need the anticurl?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What is the anticurl?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Nevermind, they gave the anticurl

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I mean A such that \[ B = \nabla \times A \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But apparently they gave it to you already.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do 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
 3 years ago
Best ResponseYou've already chosen the best response.0So to use stokes theorem, you need to parametrize the surface.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0They don't want you to set up the line integral, but I can help you do that if you want.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The surface is the triangle?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I want to be able to do it using stoke's theorem/how they want it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here is stokes theorem: \[ \iint_S\mathbf{F}\cdot d\mathbf{S}=\int_{\partial S}\nabla \times \mathbf{F}\cdot d\mathbf{r} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 3 years ago
Best ResponseYou've already chosen the best response.0No. I guess I was making a mistake. So how would I parametrize the surface? After that would the surface integral require a normal?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm getting line integrals and surface and green's theorem andstoke's theorem confused since we are barely covering them.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You can easily find two parallel vectors for the triangle right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, you do it by finding a vector between two points.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Between two points. So would I have to use points not contained in the triangle?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh wow, you're right.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah so we want the bounds on the yz plane.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay so since I know the yz plane is parallel to the trianle would I be able to choose any 2 points there?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh so it is a projection onto the yz plane

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0of the triangle, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362708042969:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay 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
 3 years ago
Best ResponseYou've already chosen the best response.0You could parametrize it as \[ \Phi(u,v) = (6,u,v),\quad 0\le u \le 5,\quad u\le v \le 5 \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh 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
 3 years ago
Best ResponseYou've already chosen the best response.0In this case \(\mathbf{F}\) doesn't need to be messed with, you only take the curl to do the line integral part.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.0How many times do I have to say you don't take the curl?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry I keep looking up at the equation. Okay. What would be the next step in my process?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \iint_S \mathbf{F}\cdot d\mathbf{S}= \iint_D\mathbf{F}\circ\Phi(u,v)\cdot \frac{\partial \Phi}{\partial u}\times \frac{\partial \Phi}{\partial v}dudv \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \mathbf{F}\circ \Phi(u,v) = \mathbf{F}(x(u,v),y(u,v), z(u,v)) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(D\) corresponds to the bounds of \(u,v\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay so it is the vector field f in terms of u and v dotted with the normal of the parametrization?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay I'm going to work it out really quick.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay 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 \[\int\limits_{0}^{5}20u4u^2du\] from there I got \[10u^24/3u^3\] and my answer was 250/3. Did I mess up somewhere because my answer seems incorrect? Was my set up correct?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You have to consider the orientation.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362709722289:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes it should be negative.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay 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
 3 years ago
Best ResponseYou've already chosen the best response.0Do you know what the answer is?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay 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 74 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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