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Line integral

  1. Mar 3, 2010 #1
    what is the different between line integral and surface integral?
    If we parameterize curve by x=t , y=t , what is the range of t ? Is it 0<= t <=1? why?
  2. jcsd
  3. Mar 3, 2010 #2


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    A "line integral" is along a one-dimensional line and a "surface integral" is over a two dimensional surface.

    There are no bounds. if you are talking about a portion of a curve, say from [itex](x_0, y_0)[/itex] to [itex](x_1, y_1)[/itex], the range of t is from whatever value of t gives [itex]x_0[/itex] and [itex]y_0[/itex] to whatever value of t gives [itex]x_1[/itex] and [itex]y_1[/itex].
  4. Mar 3, 2010 #3
    Erm.. ok..
    Let say, integral ( x^2 + y + z)ds , the line segment of curve is from (0,0,0) to (1,2,1).
    We parameterize curve C by x=t , y=2t , z=t.
    As you said, t should be between 0 and 2.
    but according to my reference book, 0<=t<=1.
  5. Mar 5, 2010 #4
    HallsofIvy said that the range of t is such that your parametrization will agree with your end points. (Of course, you first need to check that your parametrization does describe a line, in this case it does).

    With your paramterization, its 0<=t<=1. But I could choose:

    x=t/2, y=t, z=t/2

    Which still describe a line with the same orientation, instead now t must range from 0 to 2 to give the desired segment.
  6. Mar 5, 2010 #5


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    When t=1, x,y,z is the final endpoint. Try plugging t=1 into the forumulas for x,y,z to convince yourself of this.
  7. Mar 5, 2010 #6


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    No, I did NOT say that! I said " if you are talking about a portion of a curve, say from (x0, y0) to (x0, y0) , the range of t is from whatever value of t gives x0 and y0 to whatever value of t gives x1 and y1" . Since this is in three dimensions, we need to include z0 and z1 also.

    We must have x(t)= t= 0, y(t)= 2t= 0, and z(t)= t= 0. Obviously t= 0 satisfies all of those.

    We must also have x(t)= t= 1, y(t)= 2t= 2, and z(t)= t= 1. Obviously t= 1, NOT t= 2, satisfies all of those.

  8. Apr 15, 2010 #7
    HallsofIvy, ya.. i missunderstand already.. :p
    Well, now i understand it.. thanks to elibj123,Redbelly98, HallsofIvy for helping..
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