Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

SBurchette Group Title

Stokes' Theorem I'm trying to evaluate the line integral around the curve obtained by intersecting the cylinder x^2+z^2=1 with the plane y=3. The vector field is F=<3xz, e^(xz), 2xy> and the curve is oriented counterclockwise when viewed from the right.

  • one year ago
  • one year ago

  • This Question is Closed
  1. SBurchette Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I parameterized the surface needed to use stokes' thm as r(u,v)=<u*cos(v), 3, u*sin(v)> where 0 <=u<=1 and 0<=v<=2*pi. The curl of F and the cross product of the partials of r(u,v) are quite long, but I got an answer of 6*pi which is far from the answer in the solution key.

    • one year ago
  2. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    F(r) . r' right?

    • one year ago
  3. SBurchette Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    For the line integral it is, but we are supposed to use Stokes' theorem

    • one year ago
  4. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i can never keep the names straight

    • one year ago
  5. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    what is stokes thrm by chance?

    • one year ago
  6. slaaibak Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Flux is F.n

    • one year ago
  7. slaaibak Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    The line integral in space is equal to the flux of the surface

    • one year ago
  8. SBurchette Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\int\limits_{C} F.dr = \int\limits \int\limits_{S} curl(F).dS\]

    • one year ago
  9. slaaibak Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    what was your normal vector? <0,y,0? ?

    • one year ago
  10. amistre64 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    oy, its way to late for me to be doing anything remotely sane with that :/

    • one year ago
  11. malevolence19 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    If the plane intersects the cylinder you know the cross section's shadow on the x-y plane is a circle right? So you need only to integrate around the circle? And since it's a surface integral r if FIXED at r=1 not to be integrated over. So integrate over z and phi (the polar angle).

    • one year ago
  12. malevolence19 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I believe.

    • one year ago
  13. malevolence19 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Especially with y=3, its not even slanted so it will DEFINITELY only give you a circle of constant radius.

    • one year ago
  14. SBurchette Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    The formula I have been following states that flux integral over a paramterized surface is \[\int\limits \int\limits_{D} F(r).(r_u \times r_v) dA\] This, in effect, eliminates having to compute the normal directly.

    • one year ago
  15. malevolence19 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    And since you're only integrating with scalars you don't need to worry about integrating over a "position" vector which would NOT give you a circle if the plain was slanted. And you still compute the normal directly, you just don't normalize it.

    • one year ago
  16. malevolence19 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    And I guess it would be the x-z plane not x-y.

    • one year ago
  17. SBurchette Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Integrating along the curve would likely be more simple, but we are supposed to use Stokes' theorem instead of directly computing the line integral

    • one year ago
  18. slaaibak Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    but the normal is quite obvious though, isn't it? It just points in the y-direction. so <0,1,0> is fine

    • one year ago
  19. SBurchette Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, I meant we don't compute the nomalized normal, the wording gets tricky ;)

    • one year ago
  20. malevolence19 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    And I misread, since you want to calculate the line integral you only need to integrate over phi. So: |dw:1354845022728:dw| So: \[\Phi(\phi)=(- \cos(\phi),3,\sin(\phi))\] That is your parameterization for that circle. So \[0 \le \phi \le 2 \pi\] And then you have: \[\oint \vec{F} \cdot d \vec{l}=\int\limits_0^{2 \pi} \vec{F}(\vec{\Phi}(\phi)) \cdot \vec{\Phi}'(\phi)d \phi\]

    • one year ago
  21. slaaibak Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    wait, I did this a long while ago... but if you just use xyz coordinates, and calculate the curl of the vector field, then dot it with <0,1,0> and then use parametrizations, it's not really tedious at all? since you're only dealing with the j component?

    • one year ago
  22. SBurchette Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    That's what we need if we were doing the line integral directly, but to use stokes thm, we have to make it into a flux integral. and @slaaibak I didn't have too much trouble actually doing it, but the solution was incorrect and I can;t figure out where I went wrong

    • one year ago
  23. malevolence19 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    A flux integral of the curl of F.

    • one year ago
  24. SBurchette Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I just realized that I was checking the wrong section of the manual... so the answer I got actually was correct. Thanks for the help guys, I apologize for my oversight,.

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.