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Stokes Theorem cone oriented downwards

  1. Jul 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Verify stokes theorem where F(xyz) = -yi+xj-2k and s is the cone [itex]z^2 = x^2 + y^2[/itex] , 0≤ Z ≤ 4 oriented downwards

    2. Relevant equations

    [itex]\oint_{c} F.dr = \int\int_{s} (curlF).dS [/itex]

    3. The attempt at a solution

    Firstly the image of the widest part of the cone on the xy plane is the circle ofradius 4
    we parametize this circle using the parameters x=4cost , y=4sint , z=0 and also dx=-4sintdt , dy=4costdt

    so we get
    [itex] \oint_{c} F.dr = \int^{2\pi}_{0} -ydx + xdy -2dz [/itex]
    [itex] \oint_{c} F.dr = \int^{2\pi}_{0} (-4sint)(-4sint) + (4cost)(4cost)dt [/itex]
    [itex] = 16 \int^{2\pi}_{0} (sin^2 t) + (cos^2 t)dt [/itex]
    [itex] = 16 \int^{2\pi}_{0} dt = 16(2\pi- 0) = 32\pi[/itex]

    next curlF = 2k

    so [itex]= \int\int_{s} (curlF).dS = \int\int_{D} (2).dA [/itex]
    =[itex] 2\int^{4}_{0}\int^{2\pi}_{0} rdtdr = 2\int^{4}_{0} 2\pi r dr [/itex]
    =[itex] 2(\pi r^2)^{4}_{0} = 2(16\pi) =32\pi [/itex]

    ok so i got the same answer both ways which verifies Stoke but the thing im not sure about is the question says the cone is oriented downward so should i have reversed the limits as in 0≥r≥-4 and because it is downward orientated should i have put a minus infront of each side? but then the would have canceled each other out anyways?
     
  2. jcsd
  3. Jul 28, 2012 #2

    LCKurtz

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    First mistake. The bounding circle of the cone is not z = 0, it is z = 4, up on the cone.

    Nitpick, should still be ##\int_C## for the second integral since it isn't expressed in terms of dt yet.

    More than a nitpick here. At this point you must think about orientation. If the normal is downward and you use the right hand rule, what direction do you go around the circle? And the limits must reflect that.

    That's a vector ##d\vec S## and you must choose it to be oriented correctly. Is it? How do you calculate ##d\vec S## and check its orientation?

    No, not that way. See above.

     
  4. Jul 29, 2012 #3
    Hi LCkurtz,
    If the normal is pointed downward and i use the righthand rule i am going clockwise around the circle. So my limits reflecting this should be 2∏ → 0? but then if i use these limits my answer becomes [itex]-32\pi[/itex]?? but it should be just [itex]32\pi[/itex] ???
     
  5. Jul 29, 2012 #4

    LCKurtz

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    You didn't respond about the orientation of ##d\vec S\ ##that I highlighted in red. If you had, you might not have had to ask that last question. In your equation $$
    \int\int_{s} (curlF).dS = \int\int_{D} (2).dA$$you haven't made clear how you got the orientation of ##d\vec S## or how you got from ##d\vec S## to ##dA##. You have a dot in front of dA but you can't dot scalars. How do you know both sides don't give you ##-32\pi##? (Disclaimer: I haven't yet checked the calculation details since there may be no problem once you understand proper orientation.)
     
  6. Jul 29, 2012 #5
    Sorry the dot shouldnt be there.
    Using the right hand rule my limits should be -c or 2∏ to 0. dS is calculated by the normal or [itex]\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2 } dA[/itex] where x=rcost,y=rsint,z=r

    Thanks for the help LCkurtz
     
  7. Jul 29, 2012 #6

    LCKurtz

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    Yes, the limits for the line integral should be ##-C## or ##2\pi## to ##0##. But we are talking about the surface integral now. And yes, ##d\vec S## is the normal surface area vector. You must calculate it and make sure it agrees with the orientation. That square root formula you gave is not a vector and is not the correct formula.

    Either look in your text for the correct formula for ##d\vec S## or read my post #13 here:
    https://www.physicsforums.com/showthread.php?t=611873
     
  8. Jul 30, 2012 #7
    Thanks that clears up alot for me.
    Thanks a million for all your help so far but can i ask another question?
    if i was dealing with a hemisphere that was orientated in the postive y or x axis would [itex] \hat{n} dS = \left\langle -\frac{ \partial f}{\partial x}, 1 ,-\frac{ \partial f}{\partial z}, \right\rangle dz dx [/itex] say if it was oriented in the positive y direction? and set my limits for z and x accordingly, well i'd have to paramaterize it but is my thinking right?
     
  9. Jul 30, 2012 #8
    Or if i use [itex] d\vec S =\hat n dS =\pm\frac{\vec R_u\times\vec R_v}{|\vec R_u\times\vec R_v|}|\vec R_u\times \vec R_v|dudv=\pm{\vec R_u\times\vec R_v}dudv [/itex] it doesnt matter which axis it is orientated towards. i just need to choose my ± accordingly?
     
  10. Jul 30, 2012 #9

    LCKurtz

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    Before changing the subject in this thread, how about posting how you did the surface integral, showing how you got from ##d\vec S## to ##dA##. Did you get ##-32\pi## like you got in the line integral? Then I will know if you really understand it.
     
  11. Jul 30, 2012 #10

    LCKurtz

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    You don't have to learn any such formulas. If ##y = f(x,z)## your parameterization is$$
    \vec R(x,z) =\langle x, f(x,z), z)\rangle$$So ##\vec R_x = \langle 1, f_x,0\rangle## and ##\vec R_z = \langle 0, f_z, 1\rangle##. The cross product is$$
    \vec R_x \times \vec R_z = \langle f_x, -1, f_z\rangle$$
    Notice that this is opposite the vector you have, but, yes, you would think about the sign of the ##y## component when determining the orientation, and choose the ##\pm## sign accordingly when calculating ##d\vec S##.
     
  12. Aug 1, 2012 #11
    Hi LCkurtz,I got from dS⃗ to dA

    dS = [itex]\sqrt{(\frac{dz}{dx})^2 + (\frac{dz}{dy})^2 + 1 } dxdy [/itex]

    x=rcost,y=rsint and sub these values in i get rdrdt or rdA?

    thats how i did it.
    but im still abit confused, i need to look at again.
     
  13. Aug 1, 2012 #12

    LCKurtz

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    So you have a parameterization$$
    \vec R(r,t) = \langle r\cos t, r\sin t, r\rangle$$


    That argument still doesn't address the orientation because what you are calling dS is not a vector.

    I have highlighted in red some of your earlier observations. You know curl F and you have a parameterization. Why don't you actually try using the formula I gave you in that post #13 reference I gave you in post #6 above? Now your parameters are r and t instead of u and v or x and y. And don't forget to think about the orientation choice of ##\pm## when you do it.
     
    Last edited: Aug 1, 2012
  14. Aug 2, 2012 #13
    right so [itex] \vec R(r,t) = \langle r\cos t, r\sin t, r\rangle [/itex]

    [itex] \vec R_{r} = \langle cos t, sin t, 1\rangle [/itex] and [itex] \vec R_{t} = \langle -rsin t, cos t, 0\rangle [/itex]

    [itex]\vec R_r \times \vec R_t = \langle -rcost, -rsint, r\rangle [/itex] = [itex] =\hat n dS [/itex] im going to take this as + positive because although my cone is orientated downwards im still going around it it anticlock wise. like wise my limits will be from 0 to ∏
    so then i get [itex]\int^{4}_{0}\int^{2\pi}_{0} \langle 0,0,2\rangle .\langle -rcost, -rsint, r\rangle dtdr[/itex]
    = [itex]\int^{4}_{0}\int^{2\pi}_{0} 2r dtdr[/itex] =32[itex]\pi[/itex]
     
    Last edited: Aug 2, 2012
  15. Aug 2, 2012 #14

    LCKurtz

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    No. You know that's wrong because the line integral gives ##-32\pi##. Forget the anticlockwise and the line integral. You are doing the surface integral. The vector [itex]\pm \langle -rcost, -rsint, r\rangle [/itex] must agree with the orientation direction. The ##z## component is ##r## which is positive, hence upward. So you must choose the minus sign and use [itex]-\langle -rcost, -rsint, r\rangle = \langle rcost, rsint, -r\rangle[/itex] to agree with the given downward orientation. That will give you ##-32\pi##, in agreement with the circuit integral. If you look back on how we have done the flux surface integral, you will see it really isn't that difficult. I hope you will study what we have done and learn from it.
     
  16. Aug 2, 2012 #15
    Thank you for your patience! i think i see now. I was thinking the normal was off the disk at the base of the cone in the positive direction but it should have been orientated in the same direction as the cone. I'll do a few more question. hopefully things will go a bit smoother.

    Actually i have one more question if it was a cylinder instead of a cone that was open at both ends i would have to get the line integrals of both ends and add add them together to equal the surface integral. right?
     
  17. Aug 2, 2012 #16

    LCKurtz

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    That isn't a surface bounded by a closed curve. But imagine cutting it lengthwise to unroll it, but don't actually unroll it, just separate it very slightly. Now it is a surface bounded by a closed curve. If the cylinder was oriented outward you can use the right hand rule to determine the orientation of the boundary. Notice that the top and bottom circles are oriented in opposite directions and the slit is oriented once in each direction so can be ignored. So you can work such problems but you need to be careful with the orientation of things. But that shouldn't bother you much since you are now expert at orientation.:cool:
     
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