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Some terminology problems involving Stokes' Theorem

  1. Nov 20, 2012 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    I am some trouble understanding the surfaces required to integrate over in the following questions. I can tackle them, I just don't understand some terminology.
    Q1) A circle C is cut on the surface of the sphere ##x^2 +y^2 +z^2 = 25##
    by the plane ##z=3##. The direction round C is anticlockwise, as perceived from (0,0,5). Evaluate ##\int_{C} z\,dx + x\,dy +x\,dz## via Stokes' Thm

    Q2)Use Stokes' to evaluate $$\iint_{\sum} (∇ \times \vec{F}) \cdot \hat{n}\,dS$$ where ##\vec{F} ## is some vector field and ##\sum## is the surface defined by ##z=4-x^2-y^2, z≥0##.

    Q3)Same as Q2) but ##\sum## is the curved surface of the cylinder ##x^2 +y^2 = 4## for ##z\,\in\,[0,3]##

    3. The attempt at a solution
    Q1) I can do it easily directly, but using Stokes' Thm, I said that the required surface is the portion of the sphere between z = 3 and z =5. I.e the part of the sphere bounded by C. I think this is correct, because it follows from the def of Stokes' Thm. However, in the solutions I have, they say the required surface is a disk of radius 4. Why?

    Q2) So I said the required surface is the whole surface between z =4 and z=0. But in the answers they consider only the boundary of the paraboloid?

    Q3) I said the surface was the curved part of the cylinder (ie the boundary of the circle ##x^2 +y^2 =4##, projected up to z =3) That is ##x=2\cos\theta## and ##y=2\sin\theta## and ##z=z## and the integral in the end is $$\int_{0}^{2\pi} \int_{0}^{3} ...\,dz\,d\theta.$$ But, and I have no idea why this is the case, the solutions say the required surface is the two surfaces ##x^2 +y^2 =4## at ##z=3## and ##z=0## (the two circles). This is not the curved part of the surface? What am i missing?
     
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  3. Nov 20, 2012 #2

    vela

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    You can use any surface Ʃ whose boundary is the circle C. The choice you made is valid, but the disk in the plane z=3 is also a valid choice. Using the latter probably makes the integral easier to evaluate.

    I'm not sure what exactly you mean here. There's the two-dimensional surface you describe between z=0 and z=4; its boundary is the circle of radius 2 in the xy-plane, centered at the origin. What do you mean by "boundary of the paraboloid"? Are you talking about the disk in the xy-plane inside the circle?

    Can you write down some expressions to show what they did? It's not clear to me what you mean.
     
  4. Nov 21, 2012 #3

    CAF123

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    I see. Just to check my parametrisation the other way. I said $$\vec{r}(r,\theta) = r\cos\theta\,\hat{i} +r\sin\theta\,\hat{j} + (3+\sqrt{25-r^2})\,\hat{k},$$ found a normal and dotted this with the curl of the vector field ##\vec{F} = <z,x,-x>##. I get $$\int_{0}^{2\pi} \int_{0}^{4} \frac{2r^2 \sin\theta}{\sqrt{1-r^2}} + r\,dr\,d\theta$$ Is this ok? (the dimensions of the integrand are fine, it just looks difficult to compute)

    Yes,sorry, they took the surface as the disk of radius 2 in xy plane. But Ʃ is the whole surface I thought(by the wording of the question) and so I said ##x = r\cos\theta, y = r\sin\theta, z = 4-r^2## and formed a parametrisation r(r,θ).

    They took the border to be comprised of the two circles ##x^2 +y^2 =4## at ##z=3## and ##z=0##. They then said since ##\vec{F} = <-yz^2,xz^2,2xyz>##, ##\vec{F} = 0## on the circle in xy plane so just considered the circle at z=3. 'curved surface of the cylinder' does imply to me they want just the boundary. However, 'curved' suggests they want the cylindrical bit so I set r =2 and got a parametrization r(θ,z) (See my first post).
    EDIT: Actually, doing it my way also gives the correct answer when I checked it over again. Still, why does it work by considering these two circles?

    Many thanks for your help.
     
    Last edited: Nov 21, 2012
  5. Nov 21, 2012 #4

    vela

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    Is that the correct vector field? It doesn't match what you gave in your original post. If it is correct, your expression looks fine. If you integrate with respect to θ first, it's pretty straightforward to evaluate.

    Because of Stoke's theorem. You can evaluate the surface integral directly, which is what you did, or you can apply Stoke's theorem to evaluate the equivalent line integral, which is what they did in the solutions.
     
  6. Nov 22, 2012 #5

    CAF123

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    Thanks - it all makes sense now. Just a couple more questions: when I parametrized the portion of the sphere between z=3 and z=5, I said ##z= 3+ \sqrt{25-r^2}##. Is this correct? I am not sure because I don't see how it takes into consideration the fact that z is restricted to lie between 3 and 5.(I see only why it need be above 3).
    Also, in Q2, about the paraboloid - I can now get the correct answer by doing it via the line integral over the boundary curve. But when I do it via surface integral, I don't. For the surface integral, i said: ##x=r\cos\theta, y=r\sin\theta, z=4-x^2-y^2 = 4-r^2##,found a normal and dotted this with the curl of F where ##\vec{F} = \langle x^2y, xyz,xz^2 \rangle##. Subbing in my polar representation gives $$\int_{0}^{2\pi} \int_{0}^{2} -2r^3 \cos^2\theta \sin\theta - 2r\sin\theta(4-r^2)^2 + r\sin\theta (4-r^2) - r^2\cos^2\theta\,dr\,d\theta,$$ which when I evaluate I get -8pi/3 rather than -4pi, the correct answer.
     
    Last edited: Nov 22, 2012
  7. Nov 22, 2012 #6

    vela

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    No, it's not correct. It should simply be ##z=\sqrt{25-r^2}##. Since r ranges from 0 to 4, z will automatically be between 3 and 5. It turns out the constant 3 you threw in didn't matter because it drops out when you differentiate.

    I got a different expression for ##(\nabla\times\vec{F})\cdot \hat{n}\,dS##. I think you just have an algebra error somewhere.
     
  8. Nov 22, 2012 #7

    CAF123

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    Hmm.. Can't see my error. Did you get ##\vec{n} = 2x\hat{i} + 2y\hat{j} + \hat{k}## and ##(\nabla \times \vec{F}) \cdot \vec{n} = -2x^2y -2yz^2 + (yz - x^2)##? (I use the ##\text{curl} \vec{F} \cdot \vec{n} dA ## approach)
     
  9. Nov 22, 2012 #8

    vela

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    Yes, but you need ##dS = r\,dr\,d\theta##. I actually calculated ##\hat{n}\,dS##, rather than ##\hat{n}## separately, and got what you have for ##\hat{n}## multiplied by r.
     
  10. Nov 22, 2012 #9

    CAF123

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    Why do I need ##r\,dr\,d\theta##? I just said ##dA = dr\,d\theta##
     
  11. Nov 22, 2012 #10

    vela

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    OK, I'm not sure about the notation you're using. The last calculus book I looked in, which was awhile ago, made a distinction between dA and dS and the vector ##\hat{n}## was the unit normal.

    So let me backtrack a bit and say I found ##\vec{n}## (unnormalized) had another factor of r compared to yours. That's where your error lies, I think.
     
  12. Nov 22, 2012 #11

    CAF123

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    What I have is $$ \vec{dS} = \frac{r_u \times r_v}{|r_u \times r_v|}dS = (r_u \times r_v)\,dA.$$I still don't see why we have that extra r. Thanks.
     
  13. Nov 22, 2012 #12

    vela

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    OK, so we're using the same notation. It seems like you dropped the factor of r when you calculated ##\partial\vec{r}/\partial\theta##.
     
  14. Nov 22, 2012 #13

    CAF123

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    I found the normal using cartesian coords first and then converted to polars afterwards, when I had computed the curl dotted with the normal. So using cartesian gives: ##\vec{n} = 2x\hat{i} + 2y\hat{j} + \hat{k} ## and computing the normal right away using polars I get effectively ## 2xr\hat{i} + 2yr\hat{j} + r\hat{k}## So there is that 'factor of r'. Why do I not get the same result using cartesian?
     
  15. Nov 22, 2012 #14

    vela

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    You do, actually. Using the normal you found using Cartesian coordinates, you would then have ##dA = dx\,dy = r\,dr\,d\theta##. So if you integrated with respect to x and y, it would have worked out, but when you change variables after setting up the integral, you have to remember the factor of r that comes from the Jacobian.
     
  16. Nov 22, 2012 #15

    CAF123

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    I thought the infinitesimal volume element was ##r\,dr\,d\theta## not area.
     
  17. Nov 22, 2012 #16

    vela

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    No, look at the units. Angles are unitless, so only r and dr contribute units of length, which leaves you with length squared, which is an area.
     
  18. Nov 22, 2012 #17

    CAF123

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    This makes sense. There is a volume element in cylindrical coords that is r dr dθ though? Like in spherical, there is r2sin##\phi##dr d##\phi##dθ.
     
  19. Nov 22, 2012 #18

    CAF123

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    Also, if I had computed the normal via polar coords first , I believe I take ##dA = dr\,d\theta## (similarly, if I had a parametrisation r(z,θ) my area would be dz dθ). So only when we transform between different coordinates do we have to compute the Jacobian? Why is dr dθ now considered an area? I am a bit confused.
     
  20. Nov 24, 2012 #19

    CAF123

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    Any ideas?
     
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