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Evaluate flux across surface

  1. Jan 18, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    http://s1.ipicture.ru/uploads/20120119/gW4x6Mxi.jpg

    The attempt at a solution
    I drew the graph. It's a vertical cylinder, with its axis along the z-axis, with the top cut off by the plane z=3 but the cylinder goes down to infinity. The radius of the cylinder is 1.

    [tex]\phi(x,y,z)=x^2+y^2-1[/tex]
    [tex]∇\vec{\phi}=2x\vec{i}+2y\vec{j}[/tex]
    Since the direction of unit normal vector is downward, [itex]-∇\vec{\phi}=-2x\vec{i}-2y\vec{j}[/itex]
    [tex]\hat{n}=\frac{-∇\vec{\phi}}{|-∇\vec{\phi}|}[/tex]
    [tex]\hat{n}=\frac{-x-y}{\sqrt{x^2+y^2}}[/tex]
    [tex]Flux=\int\int\frac{-x^2-y^2}{\sqrt{x^2+y^2}}\,.d \sigma=-\sqrt{x+y}\,.d \sigma[/tex]
    Projecting the surface S onto the yz-plane (i cannot project onto the xy-plane as the cylinder starts from z=0 and goes down to infinity):
    [tex]x=\sqrt{1-y^2}[/tex]
    [tex]x_y=-\frac{y}{\sqrt{1-y^2}}[/tex]
    [tex]x_z=0[/tex]
    [tex]d\sigma=\frac{1}{\sqrt{1-y^2}}\,.dzdy[/tex]
    [tex]S.A.=2\int\int \frac{1}{\sqrt{1-y^2}}\,.dzdy[/tex]
    The limits are -∞≤z≤3 and -1≤y≤1 but i think i messed up somewhere above, as these limits look suspicious.

    When i visualize this thin slice of the surface at the plane z=3, it seems to me that if i project the thin slice onto the yz-plane, the slice would become invisible, as its thickness is zero. Am i right? But since we're dealing with flux here, then this slice does not have a thickness of zero. Correct?
     
    Last edited: Jan 18, 2012
  2. jcsd
  3. Jan 18, 2012 #2

    lanedance

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    can you explain a bit?

    isn't the surface a disk radius one centred on the z axis and in the z=3 plane, giving [itex] \hat{n}=(0,0,-1)^T[/itex]
     
  4. Jan 18, 2012 #3

    sharks

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    No, it's in the plane z=3 and i don't understand how you got [itex] \hat{n}[/itex] directly?
    From my notes, i must always follow this method:
    [tex]\hat{n}=\frac{∇\vec{\phi}}{|∇\vec{\phi}|}[/tex]
     
  5. Jan 18, 2012 #4

    lanedance

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    yeah was editing as you replied
     
  6. Jan 18, 2012 #5

    lanedance

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    i'm not sure what you mean about a cylinder, isn't it just the disk as described?
     
  7. Jan 18, 2012 #6

    lanedance

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    then only the z component of the field will contribute to the integral
     
  8. Jan 18, 2012 #7

    sharks

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    OK, i think i understand. For flat surfaces, the outward unit normal is +1 is the direction is along the positive side of the axis, otherwise it's negative.
    So, for example, for plane x=2, [itex]\hat{n}=(1,0,0)^T[/itex]
    And, for plane y=-3, [itex]\hat{n}=(0,-1,0)^T[/itex]
    Are these correct?

    So, for flat surfaces, there is no need to use this formula:
    [tex]\hat{n}=\frac{∇\vec{\phi}}{|∇\vec{\phi}|}[/tex]
    Correct?

    But if i do use the formula, then i don't get -1. Instead i get this:
    [tex]\hat{n}=\frac{-x-y}{\sqrt{x^2+y^2}}[/tex]
    Using [itex]x^2+y^2=1[/itex], i can simply:
    [tex]\hat{n}=\frac{-x-y}{\sqrt{x^2+y^2}}=-x-y[/tex]
    But i don't know how it becomes -1.
     
    Last edited: Jan 18, 2012
  9. Jan 18, 2012 #8

    lanedance

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    well the normal can be in the positive or negative direction depending on the surface orientation you choose, and that gets important when you start looking at stokes theorem, but yeah they are valid normals

    the problem is the function phi you defined, when phi=0, it gives the unit cylinder, NOT the plane z=3.

    and though the method is a good one, it's not really required as the surface normal is constant and easily to calculate

    if you really want to use it the method, you should start with
    phi=z-3
     
  10. Jan 18, 2012 #9

    lanedance

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    just to clarify, you might want to re-read the question, the surface is z=3, the boundary is given by x^2+y^2=1 (the unit circle)
     
  11. Jan 19, 2012 #10

    sharks

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    Yes, i did a mistake with [itex]\phi[/itex].
    [tex]\phi(x,y,z)=z-3[/tex][tex]∇\vec{\phi}=\vec{k}[/tex][tex]-∇\vec{\phi}=-\vec{k}[/tex][tex]\hat{n}=-\vec{k}[/tex][tex]Flux=\int\int \vec{F}.\hat{n}\,.d \sigma=\int\int-z\,.d \sigma[/tex]
    Now, i'm not sure, but i think the only plane where i can project the surface z=3 is the xy-plane. Am i right?
    So, projecting on the xy-plane:
    [tex]z_x=0[/tex][tex]z_y=0[/tex][tex]\sigma=\sqrt{0+0+1}\,.dxdy[/tex][tex]S.A.=\int\int(-z)\,.dxdy[/tex][tex]S.A.=\int\int(-3)\,.dxdy[/tex]
    Transforming to polar coordinates:[tex]S.A.=\int\int(-3)\,.dxdy=\int^{2\pi}_0\int^1_0(-3)\,.rdrd\theta=-3\pi[/tex]
    The answer is wrong.
     
    Last edited: Jan 19, 2012
  12. Jan 19, 2012 #11

    lanedance

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    why is it wrong?
     
  13. Jan 19, 2012 #12

    sharks

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    Thanks for answering, lanedance. I've been waiting for confirmation the whole day.

    Well, the answer in my notes is: [itex]\frac{3\pi}{2}[/itex]
    Furthermore, the answer in my notes is positive, whereas the answer i got is negative.
     
  14. Jan 19, 2012 #13

    lanedance

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  15. Jan 19, 2012 #14

    sharks

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    I checked out the example in your link and there is this last part that i can't understand from the website:

    At this point we can acknowledge that D is a disk of radius 1 and this double integral is nothing more than the double integral that will give the area of the region D so there is no reason to compute the integral. Here is the value of the surface integral.
    eq0065M.gif

    My question is: How can he directly know that the answer is [itex]\pi[/itex] without evaluating the integral?
    At this point, i'm having a hard time to fully understand, as i'm still learning the fundamentals in this chapter, and my notes are apparently wrong.
     
  16. Jan 19, 2012 #15

    lanedance

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    So the full surface vector integral is
    [tex]
    \int\int \vec{F}\cdot\vec{dS}
    [/tex]

    the surface differential is equivalent to an infintesimal area on the surface in the direction of the normal
    [tex]
    \vec{dS}
    =\hat{n}dA
    [/tex]

    Thus the integral becomes
    [tex]
    \int\int \vec{F}\cdot\hat{n}dA
    [/tex]

    In our case the dot product is equal to -z, and z=3 on the surface
    [tex]
    \vec{F}\cdot\hat{n}=-z=-3
    [/tex]

    So the integral becomes
    [tex]
    \int\int \vec{F}\cdot\hat{n}dA
    =-3\int\int dA
    [/tex]

    Now what is the area of our surface, a circle of radius 1? you could integrate, as you did with the change to polar coords, or just use the area [itex]\pi r^2=p\i[/itex]
    [tex]
    \int\int \vec{F}\cdot\hat{n}dA
    =\int\int dA
    [/tex]
     
  17. Jan 20, 2012 #16

    sharks

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    I see it now.
    [tex]\int\int \vec{F}\cdot\hat{n}dA=-3 \int\int dA[/tex]
    But since [itex]\int\int dA={\pi}{r}^2[/itex] where r=1
    [tex]\int\int \vec{F}\cdot\hat{n}dA=-3 \int\int dA=-3\pi[/tex]
    Thanks for the help, lanedance.
     
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