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Homework Help: Integral what has gone wrong here?

  1. Feb 27, 2006 #1
    I am still struggling my way through line integrals, and this here is one where I do not understand what has gone wrong - does anyone see what it is ( I really want to understand all of this)?

    [tex] \int_{0}^{1}{(y*e^{xy} + 2x + y)dx + (x*e^{xy} + x )dy } [/tex]

    the curve joins (0,0) to (1,1)

    where x= t and y=t (0<= t <= 1)

    so, I said that

    [tex] \frac {dx} {dt} = \frac {dy} {dt} = 1 [/tex]

    therefore:
    [tex] \int_{0}^{1} {( t*e^{t^2} + 2t + t + t*e^t^2 + t)dt} = \int_{0}^{1}{ (2t*e^{t^2} + 4t) dt} = e^1 + 1 [/tex]

    However, the solution is supposed to be -2pi
     
    Last edited: Feb 28, 2006
  2. jcsd
  3. Feb 27, 2006 #2
    What do you mean by [tex]e^{t2}[/tex]? Is it [tex]e^{2t}[/tex] or [tex]e^{t^2}[/tex]?

    I guess it comes from the first expression, viz. [tex]ye^xy[/tex]. Is it [tex]y^2e^x[/tex] or [tex]ye^{xy}[/tex]?

    Never mind, going through your calculation, I can see what you meant.

    Hmmm... I obtain the same thing as you. Are you sure about that parametrization?
     
    Last edited: Feb 27, 2006
  4. Feb 28, 2006 #3
    I am so sorry about that.....I fixed it.

    That's the parametrization the question stated:

    (....) the curve joins (0,0) to (1,1)

    where x= t and y=t (0<= t <= 1)

    ( I just double checked this)
     
    Last edited: Feb 28, 2006
  5. Feb 28, 2006 #4
    does it mabe help that the field is conservative?
     
    Last edited: Feb 28, 2006
  6. Feb 28, 2006 #5

    HallsofIvy

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    A bit. That of course is the reason the path did not need to be given!

    And it verifies that the correct answer is e+ 1.
     
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