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anonymous

  • 5 years ago

I need to find the flux through a disk of radius 1 centered at (1,1,1) with its normal pointing towards the origin. How do i parameterize this disk?

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  1. anonymous
    • 5 years ago
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    I found that it lies in the plane -x-y-z=3

  2. anonymous
    • 5 years ago
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    dont parametrize it

  3. anonymous
    • 5 years ago
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    should i take the intersection between this plane and a sphere centered at 111

  4. anonymous
    • 5 years ago
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    find a vec tor normal to it, with magnitude of the area of the disk

  5. anonymous
    • 5 years ago
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    -1,-1,-1

  6. anonymous
    • 5 years ago
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    whats the vec tor field??

  7. anonymous
    • 5 years ago
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    f=<z,x,y>

  8. anonymous
    • 5 years ago
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    I cant use divergence theorem right?

  9. anonymous
    • 5 years ago
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    im thinking

  10. anonymous
    • 5 years ago
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    what with 0 volume.

  11. anonymous
    • 5 years ago
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    thats whts confusing me..no i dont think divergence thm is applicabe..old way then

  12. anonymous
    • 5 years ago
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    yeah so i need a parametric surface

  13. anonymous
    • 5 years ago
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    yeah

  14. anonymous
    • 5 years ago
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    thats where im stuck

  15. anonymous
    • 5 years ago
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    thats wht i really dunno..;.parametrizing surfaces

  16. anonymous
    • 5 years ago
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    i can find the intersection between the sphere and a plane but its an ugly integral

  17. anonymous
    • 5 years ago
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    I know theres another way to do it using tangents to the normal but i cant remember

  18. anonymous
    • 5 years ago
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    wht u do

  19. anonymous
    • 5 years ago
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    well the disk lies in the plane -x-y-z=3 and theres a sphere with rad 1 cestered at (1.1.1) of the form <rcos(theta)sin(phi), rsin(theta)sin(phi), rcos(phi)>

  20. anonymous
    • 5 years ago
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    go on.bt watll u do wid d sphere?

  21. anonymous
    • 5 years ago
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    Im trying to figure it out i cant remember exactly

  22. anonymous
    • 5 years ago
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    i know paramet for cylinders and all bt nt this

  23. anonymous
    • 5 years ago
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    Using vectors, generally if t is the parameter then and point P on the circle is given by; LaTeX Code: P = R\\cos(t) \\vec{u} + R\\sin(t) \\;\\;\\vec{n}\\times\\vec{u} + c

  24. anonymous
    • 5 years ago
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    check this out: http://www.physicsforums.com/showthread.php?t=123168

  25. anonymous
    • 5 years ago
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    you need 2 parameters for surfaces

  26. anonymous
    • 5 years ago
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    go down to the post where it writes a parametric eqn for a circle in 3d..our surface is actually a circle at some angle

  27. anonymous
    • 5 years ago
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    read it up

  28. anonymous
    • 5 years ago
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    uve got t and the vector u here as parameters

  29. anonymous
    • 5 years ago
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    u add u the components of i,j,k

  30. anonymous
    • 5 years ago
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    kool ill try the eq on that link

  31. anonymous
    • 5 years ago
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    u get x,y,z in terms of t and u

  32. anonymous
    • 5 years ago
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    i think thatll parametrize it

  33. anonymous
    • 5 years ago
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    U is an orthognal vector not a parameter

  34. anonymous
    • 5 years ago
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    t and r are the parameters

  35. anonymous
    • 5 years ago
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    r is the radius mate...u is the parametric vector from the centre to that point and n is the orthogonal vector

  36. anonymous
    • 5 years ago
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    can you use polar coords?

  37. anonymous
    • 5 years ago
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    you have to integrate over the radius to get a disk

  38. anonymous
    • 5 years ago
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    thats how far i cn get at parameterizing it

  39. anonymous
    • 5 years ago
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    i g2g.. i learnt a lot thru this discusion..thnx..hope u get it

  40. anonymous
    • 5 years ago
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    tnx

  41. anonymous
    • 5 years ago
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    polar is the way to go, I'm not sure what him1618 meant

  42. anonymous
    • 5 years ago
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    its in 3 space so its in cylindrical

  43. anonymous
    • 5 years ago
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    just have z=z be one condition

  44. anonymous
    • 5 years ago
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    if z doesn't vary, well i guess cylindrical still applies, anyway the jacobian is still r

  45. anonymous
    • 5 years ago
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    but z does change the disk is tilted in 3 dimentions

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