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Surface Integral (and Greens theorem) confusion

  1. Sep 24, 2010 #1
    Hello all,


    [tex] \int\int r. da [/tex]

    over the whole surface of the cylinder bound by

    [tex]x^{2} + y^{2} = 1, z=0 [/tex] and [tex] z=3. [/tex]

    [tex]\vec{r} = x \hat{x} + y \hat{y} + z \hat{z} [/tex]

    Sorry for the awkward formatting, this site is giving me trouble.
    it seems to me that since I have 3 dimensions I'm asked to use I must resort to Green's theorem,
    and I will eventually have a triple integral with x=0 to 1, y=0 to 1 and z=0 to 3.

    What's my tau (volume unit) though? I have the area condition which only includes x and y, and the z condition on its own...

    Last edited: Sep 24, 2010
  2. jcsd
  3. Sep 24, 2010 #2
    Or maybe I need to parametrize the equation... that would work, but I still don't know how r relates to that...
  4. Sep 24, 2010 #3
    Don't you need to rewrite
    [tex]\vec{r} = x \hat{x} + y \hat{y} + z \hat{z} [/tex]

    [tex]\vec{r} = x \cdot \frac{x}{\|x\|} + y \frac{y}{\|y\|} + z \cdot \frac{z}{\|z\|} = x \cdot \frac{x}{\sqrt{x \cdot x}} + y \frac{y}{\sqrt{y\cdot y}} + z \cdot \frac{z}{\sqrt{z \cdot z}} = |x|+|y| + |z|[/tex]
    Last edited: Sep 24, 2010
  5. Sep 24, 2010 #4


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    Green's theorem?? Is this a line integral around a closed curve in the plane?

    Those limits would describe a rectangular block, which a cylinder isn't.

    Why do you want the unit of volume? You are given a surface integral, not a volume integral.

    You need to parameterize the top, bottom, and lateral surfaces in terms of two parameters and work each piece separately. Since you are using an area vector, presumably you have an orientation given for the surface. I'll give you a hint to get you started. For the lateral surface, try parameterizing it in terms of z and θ.
  6. Sep 24, 2010 #5
    Sorry I meant to say the divergence theorem, which is why I would need the unit of volume.

    My problem is, I don't see, regardless of what I do, how I'm going to do the dot product of something with 2 dimensions (da) with something with 3 dimensions (r)...

    If I parametrize to cos(theta) and sin(theta) (radii are 1), I don't see how I've solved my problem..

    thanks for the help by the way!
  7. Sep 24, 2010 #6
  8. Sep 24, 2010 #7


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    The da in your problem is the area vector. It is the unit normal multiplied by the scalar area element. How to approach this problem depends on the intent of your textbook author for that section. You can work the problem directly by parameterizing the surfaces and using the fact that if the surface is parameterized by R(u,v), then

    [tex]d\vec a = \pm \vec R_u \times \vec R_v[/tex]

    where the sign is chosen to agree with the orientation. Or you can apply the divergence theorem. Some texts ask you to work both sides of the divergence theorem as an exercise for practice. Which leads to the obvious question: Have you written out what the divergence theorem says for this particular problem? You might be surprised...
  9. Sep 24, 2010 #8


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    I don't have time to work through that example right now, but I would point out to you that your question involves a flux integral and the vector da is not the same as the scalar da.
  10. Sep 26, 2010 #9
    I've run into another problem. When parametrizing everything, I obviously need my surface to be in cylindrical coordinates, but I can't very well parametrize r = x+y+z in cylindrical because there is no axis...

    What do I do? I can't do the dot product between two things parametrized differently =(
  11. Sep 26, 2010 #10


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    I have several comments for you.

    First, r is not the scalar x + y + z. r is the position vector to (x,y,z) which would be written as xi+yj+zk or <x, y, z> depending on your preference. And in a problem such as this which might involve cylindrical coordinates it would be better to use R instead of r to avoid confusion with the r of cylindrical coordinates.

    Second, I gave you the hint above to parameterize the lateral surface using z and θ. Have you expressed x, y, and z in terms of those for that surface? Then you can use the other formula I gave you for

    d\vec a = \pm \vec R_u \times \vec R_v\ dudv

    Did you understand all that?

    Third, I asked you whether you have actually written what the divergence theorem says for this problem. Have you done that? What does it give you?
    Last edited: Sep 27, 2010
  12. Sep 27, 2010 #11

    so, if I make R equal to the parametrization of my surface then,
    R= [tex]cos \theta \hat{i} + sin \theta \hat{j} + z \hat{k} [/tex] since the axis of the cylinder is along z.

    Next, following

    [tex]\int\int \bar{r}.d\bar{a} = \int\int \bar{r} .\bar{n} da[/tex]

    Doing the cross product of R w.r.t. theta and z gives:

    [tex]- cos \theta \hat{i}- sin \theta \hat{j}[/tex]

    Now is where I hit my problem...

    I realize r is a vector and not a scalar, that is exactly what I'm confused about.

    I need to do the dot product r.n. Where r=xi + yk+zk, and n is what I already stated in cylindrical coordinates. But, I can't put r into cylindrical coordinates since there is no axis, so I don't know how to proceed with the dot product.

    I haven't done anything with the divergence theorem... I don't know how I would come up with a dV since I am only given the shape in 2-D or 1-D pieces.
  13. Sep 27, 2010 #12


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    Good so far. That correctly parameterizes the lateral surface as a function of θ and z.

    Do you mean to say the cross product of rθ and rz? At this point you need to check the orientation of the surface. You haven't actually said it yet but I assume the surface is probably oriented outward. Does the above vector agree with that orientation or do you need its opposite?

    I don't know what you mean "there is no axis". You have the substitution x = cos(θ), y = sin(θ), z = z that you used before. Use the same substitution to put r into cylindrical coordinates.

    Now you need to look again at the general formula for flux integrals:

    [tex]\int\int_S \vec F \cdot d\vec S =\pm\int\int_{(u,v)} \vec F(u,v)\cdot\vec r_u \times \vec r_v\ dudv[/tex]

    Here your u and v are θ and z, your F is r, and the sign is chosen so that the direction of the cross product agrees with you orientation. So what does this give you for your problem?

    That's a shame because the divergence theorem is the obvious easy way to do this problem. Your three surfaces enclose a volume. So I will ask you once more to write down what the divergence theorem says for your particular problem.
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