1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Possible application of Stoke's Theorem

  1. Nov 10, 2012 #1

    CAF123

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    A circle C is cut on the surface of the hemisphere [itex] x^2 + y^2 + z^2 = 1, z ≥ 0 [/itex] by the cylinder [itex] x^2 + y^2 = y [/itex]. Evaluate [tex] \int_{C} -y^2\,dx + y^2\,dy + z^2\,dz, [/tex] where the direction round C is such that the point (0,0,1) is directed into the first octant.

    3. The attempt at a solution
    So completing the square gives a cylinder of centre (0,1/2) and radius 1/2. Using Stoke's thm, I identified the vector field F to be [itex] -y^2 i + y^2j + z^2 k [/itex] and took the curl of it to give [itex] 2y \,k [/itex] I believe everything is right up to here.

    I am confused about what the surface is here that is bounded by C. I realise that to compute [itex] d\vec{S} = \frac{\vec{r_u} ×\vec{r_v}}{|\vec{r_u} × \vec{r_v}|} dS [/itex], I have to find a suitable parametrisation of some surface. I found where the cylinder and sphere intersected : [itex] y + z^2 = 1 => z = \sqrt{1-y}\,\,\text{since}\,\,z≥0 [/itex] .So then my parametrisation would be [itex] r(x,y) = xi + yj + \sqrt{1-y}k [/itex], from which I could then compute two tangent vectors and a normal.

    I am not sure if my parametrisation is correct. Can anyone offer any advice?
    Many thanks
     
  2. jcsd
  3. Nov 10, 2012 #2
    The part of the hemisphere that is bounded by the circle.
     
  4. Nov 10, 2012 #3

    CAF123

    User Avatar
    Gold Member

    So this is the required S? The part of the hemisphere bounded by the circle is the part of the hemisphere in the region where y≥0 and [itex] x \in\,[-1,1] [/itex] So my parametrisation is correct? Why don't we care about the other half of the hemisphere in the region y≤0?
     
  5. Nov 10, 2012 #4
    Stock's theorem allows to take any surface with the boundary C with the suitable orientation.
     
  6. Nov 10, 2012 #5

    CAF123

    User Avatar
    Gold Member

    I am just stuck on the parametrisation. I think what I have done is the parametrisation of the cylinder. What I said now was that r(x,y) = xi + yj + [itex] \sqrt{1-x^2-y^2} [/itex]k and my parameter domain being the circle formed by the intersection of the cylinder with the sphere.

    I have found a unit normal from the above r(x,y), but now I need to find the limits of integration. Can you help here?
     
  7. Nov 10, 2012 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What does that mean? (0,0,1) is a vector pointing in the z direction or else it is a point which doesn't point anywhere.

    Is there any reason you have to use Stoke's theorem? My suggestion would be to work the line integral directly. The first step would be to write ##x^2+y^2 = y## in polar (cylindrical) coordinates, which is very easy. Then write your curve parameterized by ##\theta##. Then just plug and chug.
     
  8. Nov 10, 2012 #7

    CAF123

    User Avatar
    Gold Member

    Hi LCKurtz,
    Yes, I was initially confused by the meaning of the question, but I took it to mean the traversal of C would be anticlockwise so that as you approach the point (0,0,1) you head into the first octant.

    You mention parametrizing a curve in terms of θ. What curve is this?
    This is actually a set question, so given that we have just done Stoke's thm, I think we may have to use it. However, I will do it both ways to confirm my answers.

    If doing it by Stoke's Thm, I had the previous parametrization as in my previous post. I then found the (unit) normal, but I am struggling with the parameter domain. I think it is the circle cut from the sphere when the cylinder intersects, but I am not sure about the limits here. Any advice?
    Thanks.
     
  9. Nov 10, 2012 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Did you express the circle (cylinder) in cylindrical (polar) coordinates? That will give your ##r,\theta## limits once you see how the circle plots. You need to do that. Then whether you do the line or surface integral, use ##\theta## for the parameter for the line integral and ##r,\theta## if you do a surface integral. Then the limits will be just like polar coordinates.
     
  10. Nov 10, 2012 #9

    CAF123

    User Avatar
    Gold Member

    So, [itex] x^2 + y^2 = y => r^2 = r\sin\theta => r = \sin\theta [/itex] . I think that we would have [itex] r \in\,[0, \sin\theta] [/itex] and [itex] \theta \in\,[0,2\pi] [/itex] right?
    These are my limits. Then taking the curve in question, letting [itex] x = r\cos\theta [/itex] and [itex] y = \frac{1}{2} + r\sin\theta [/itex], allows me to compute dx, dy. Is [itex] z = \sqrt{1-y} = \sqrt{(1-(1/2 + r\sin\theta)} [/itex]here? That's for the line integral.
     
  11. Nov 10, 2012 #10

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No for ##\theta##. Plot it to get the correct ##\theta## limits.
    It's easier but I have to run. Maybe more this evening.
     
  12. Nov 10, 2012 #11

    CAF123

    User Avatar
    Gold Member

    Ah, sorry, made a careless mistake and confused the axes. It should be [itex] \theta \in\,[0,\pi] [/itex]. There is an easier way? In terms of what?
    Thanks
     
  13. Nov 10, 2012 #12

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    OK. So ##r=\sin\theta## on the surface of the cylinder and therefore on the curve of intersection between the cylinder and the sphere. Now you know a point (x,y,z) in cylindrical coordinates is ##\vec R=(r\cos\theta,r\sin\theta,z)##. And ##z## must be on the sphere so ##z=\sqrt{1-(x^2+y^2)}=\sqrt{1-r^2}##. So ##\vec R =(r\cos\theta,r\sin\theta,\sqrt{1-r^2})##. Now you can let ##r,\theta## vary over the interior of the circle if you want to parameterize that portion of the sphere, or if you just want the boundary curve of intersection use that ##r=\sin\theta## to parameterize that curve as a function of just ##\theta##.

    [Edit]:Made a correction
     
    Last edited: Nov 11, 2012
  14. Nov 11, 2012 #13

    CAF123

    User Avatar
    Gold Member

    How did you get [itex] \sqrt{1-(x^2 + y^2)} = 1-r? [/itex] I only see a simplification to [itex]\sqrt{1 - r^2}[/itex].

    So assuming [itex] z = \sqrt{1 - r^2}[/itex], i have [tex] \int_{0}^{\pi} -\sin^4\theta (-\frac{1}{2}\cos^2\theta) + \sin^4\theta (\sin 2\theta) + \cos^2\theta (-\sin\theta)\, d\theta, [/tex] for the line integral.
    (One question: Why do we choose to parametrize the curve in terms of θ? is it because r stays constant on the given C?)

    For Stokes' thm appraoch, I should be able to get a normal vector from the parametrization [itex] \vec{r}(r, \theta) = < r\cos \theta, r\sin \theta, \sqrt{1-r^2}>,[/itex] dot this with the curl of the given vector field and then do a double integral [tex] \int_{0}^{\pi} \int_{0}^{\sin\theta} ... dr\,d\theta.[/tex]
     
    Last edited: Nov 11, 2012
  15. Nov 11, 2012 #14

    CAF123

    User Avatar
    Gold Member

    So in evaluating the integral above (the line integral) I get pi/32 + 2/3. By evaluating : [tex] \int_{0}^{\pi} \int_{0}^{\sin\theta} 2(r\sin\theta)r^2\,dr\,d\theta [/tex] i get 8/15. [itex] 2(r\sin\theta)r^2 [/itex] I got by dotting curl(F)with a normal vector found by taking [itex] r_r \times r_\theta [/itex] of the parametrization r(r,θ) = rcosθi + rsinθj + √(1-r^2)k.
    I don't know where I went wrong.
     
  16. Nov 11, 2012 #15

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, of course. Typing in a hurry :blushing: I will edit it if I'm not too late.

    ##r## isn't constant on the curve, it is a function of ##\theta##: ##r = \sin \theta##. Much easier to get everything in terms of ##\theta## rather than ##r##.
    I will look at it a bit later today.
     
  17. Nov 11, 2012 #16

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Me neither without seeing the steps. But you are close. The integrand for the surface integral should be ##2r^2\sin\theta## after simplification. You are off by a factor of ##r##.
     
  18. Nov 11, 2012 #17

    CAF123

    User Avatar
    Gold Member

    Why [itex] r^2 [/itex] and not [itex] r^3 [/itex]?

    EDIT: nevermind - I realise why. Thanks. Is my line integral correct?
     
  19. Nov 11, 2012 #18

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I don't know. But you can check it by seeing if you get the same answer as the surface integral.
     
  20. Nov 11, 2012 #19

    CAF123

    User Avatar
    Gold Member

    I get pi/4 for the surface integral. This is not what i get for the line integral. Using [itex] x = (\sin \theta) \cos\theta, y = (\sin \theta)\sin \theta , z = \sqrt{1 - sin^2 \theta} = \cos \theta [/itex] and then subsequently finding the derivatives [itex] dx, dy, dz [/itex] and subbing into the form in the question, i get the line integral expression in my last couple of posts. I may have made an error in the integration, but I can't see any yet. Is my method right?
     
  21. Nov 11, 2012 #20

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Again, I haven't checked all your substitutions. But one problem I see is on your statement$$
    z = \sqrt{1 - sin^2 \theta} = \cos \theta$$That last equality isn't correct if ##\pi/2 < \theta<\pi## because ##\cos\theta## is negative there. The correct identity is$$
    \sqrt{1 - sin^2 \theta} = |\cos \theta|$$and you need to take care of that when you integrate. And yes, ##\frac \pi 4## is the correct answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Possible application of Stoke's Theorem
Loading...