## ?12 Group Title 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 one year ago one year ago

1. ?12 Group Title

so the curl is -3i-7j-4k

2. wio Group Title

If you want to go from line integral to surface integral, shouldn't you need the anticurl?

3. ?12 Group Title

What is the anticurl?

4. wio Group Title

Nevermind, they gave the anticurl

5. ?12 Group Title

I see

6. wio Group Title

I mean A such that $B = \nabla \times A$

7. wio Group Title

But apparently they gave it to you already.

8. ?12 Group Title

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

9. wio Group Title

So to use stokes theorem, you need to parametrize the surface.

10. wio Group Title

They don't want you to set up the line integral, but I can help you do that if you want.

11. ?12 Group Title

The surface is the triangle?

12. ?12 Group Title

I want to be able to do it using stoke's theorem/how they want it

13. wio Group Title

Here is stokes theorem: $\iint_S\mathbf{F}\cdot d\mathbf{S}=\int_{\partial S}\nabla \times \mathbf{F}\cdot d\mathbf{r}$

14. wio Group Title

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?

15. ?12 Group Title

No. I guess I was making a mistake. So how would I parametrize the surface? After that would the surface integral require a normal?

16. ?12 Group Title

I'm getting line integrals and surface and green's theorem andstoke's theorem confused since we are barely covering them.

17. wio Group Title

You can easily find two parallel vectors for the triangle right?

18. ?12 Group Title

the yz-plane?

19. wio Group Title

No, you do it by finding a vector between two points.

20. ?12 Group Title

Between two points. So would I have to use points not contained in the triangle?

21. ?12 Group Title

?

22. wio Group Title

oh wow, you're right.

23. wio Group Title

yeah so we want the bounds on the yz plane.

24. ?12 Group Title

okay so since I know the yz plane is parallel to the trianle would I be able to choose any 2 points there?

25. ?12 Group Title

oh so it is a projection onto the yz plane

26. ?12 Group Title

of the triangle, right?

27. wio Group Title

|dw:1362708042969:dw|

28. ?12 Group Title

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?

29. wio Group Title

You could parametrize it as $\Phi(u,v) = (6,u,v),\quad 0\le u \le 5,\quad u\le v \le 5$

30. ?12 Group Title

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?

31. wio Group Title

In this case $$\mathbf{F}$$ doesn't need to be messed with, you only take the curl to do the line integral part.

32. ?12 Group Title

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?

33. wio Group Title

How many times do I have to say you don't take the curl?

34. ?12 Group Title

Sorry I keep looking up at the equation. Okay. What would be the next step in my process?

35. wio Group Title

$\iint_S \mathbf{F}\cdot d\mathbf{S}= \iint_D\mathbf{F}\circ\Phi(u,v)\cdot \frac{\partial \Phi}{\partial u}\times \frac{\partial \Phi}{\partial v}dudv$

36. wio Group Title

$\mathbf{F}\circ \Phi(u,v) = \mathbf{F}(x(u,v),y(u,v), z(u,v))$

37. wio Group Title

$$D$$ corresponds to the bounds of $$u,v$$.

38. ?12 Group Title

okay so it is the vector field f in terms of u and v dotted with the normal of the parametrization?

39. wio Group Title

Yes.

40. ?12 Group Title

Okay I'm going to work it out really quick.

41. ?12 Group Title

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?

42. wio Group Title

You have to consider the orientation.

43. ?12 Group Title

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?

44. wio Group Title

|dw:1362709722289:dw|

45. wio Group Title

Yes it should be negative.

46. ?12 Group Title

the normal?

47. wio Group Title

yeah.

48. ?12 Group Title

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?

49. wio Group Title

Do you know what the answer is?

50. ?12 Group Title

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?