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Independence of path of line integral

  1. Jan 16, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    http://s2.ipicture.ru/uploads/20120117/ReWSCD1f.jpg

    The attempt at a solution
    [tex]\frac{\partial P}{\partial y}=\frac{2y}{x^3}[/tex]
    [tex]\frac{\partial Q}{\partial x}=\frac{2y}{x^3}[/tex]
    [tex]\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}[/tex]

    According to my notes: Both functions are continuous, except at x=0.
    My tentative explanation: This is due to the denominator in [itex]\frac{2y}{x^3}[/itex] which will make the fraction equal to infinity at x=0. Therefore the functions won't be continuous at x=0. Is this explanation correct?

    Apparently, at this stage, i am free to choose whatever path i please. I'm going to start from path y=1 (1,1) to (2,1) and then move up the path x=2 (2,1) to (2,2).

    Then, i do direct substitution:
    [tex]I=\int^2_1\frac{11}{x^3}\,dx+\int^2_1 -\frac{5}{4}\,ydy[/tex]
    Is this correct?
    After i did the integration, i get the final answer: [itex]-\frac{87}{8}[/itex] but it's wrong.
     
  2. jcsd
  3. Jan 16, 2012 #2
    I think if your line integral is path independent, you can use the Fundamental Theorem of Calculus as your vector field is just a gradient field of a scalar function.
     
  4. Jan 16, 2012 #3

    sharks

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    I don't understand what you're talking about.
     
  5. Jan 16, 2012 #4

    sharks

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    The correct answer is: -9/8

    For the path i chose in the first post, the graph is like this:
    http://s2.ipicture.ru/uploads/20120117/K2c405Y4.jpg

    I think you're wrong about the path going "under" the origin, as the line x=0 is the y-axis. So the graph is not allowed to cross the y-axis but can go under the x-axis. Correct?
     
  6. Jan 17, 2012 #5

    sharks

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    If the answer is 9/4, then what is wrong with my calculation from post #1?
    Is this expression correct?
    [tex]I=\int^2_1\frac{11}{x^3}\,dx+\int^2_1 -\frac{5}{4}\,ydy[/tex]


    And no one answered this:
     
  7. Jan 17, 2012 #6

    sharks

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    Can anyone help?
     
  8. Jan 17, 2012 #7

    ehild

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    It is correct, and the result is 9/4. How did you get different result? Show details.

    ehild
     
  9. Jan 17, 2012 #8

    sharks

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    Hi ehild!:smile:
    [tex]\int^2_1\frac{11}{x^3}\,dx=\int^2_111x^{-3}\,dx=-\frac{11}{2x^2}=\frac{33}{8}[/tex]
    [tex]\int^2_1 -\frac{5}{4}\,ydy=-\frac{5y^2}{8}=-\frac{15}{8}[/tex]
    [tex]\frac{33}{8}-\frac{15}{8}=\frac{9}{4}[/tex]
    Indeed, the answer is [itex]\frac{9}{4}[/itex]. However, in my notes, the answer is [itex]-\frac{9}{8}[/itex] for this problem. It must be a mistake in my notes.
     
    Last edited: Jan 17, 2012
  10. Jan 18, 2012 #9

    ehild

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    Your notes are wrong then. It happens... :smile:

    And the functions are not continuous at x=0, because of the x in the denominator.

    ehild
     
  11. Jan 18, 2012 #10

    sharks

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    To confirm my suspicions on this:

    For example, if [tex]\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}=\frac{2x}{y^2}[/tex]
    Then, the functions are not continuous at y=0.

    And if [tex]\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}=\frac{2x}{x+y^2}[/tex]
    Then, the functions are not continuous at (0,0).

    Are my conclusions correct?
     
  12. Jan 18, 2012 #11

    ehild

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    I do not get you. What functions are you talking about? Of course, the functions P and Q are not continuous where the denominator is zero (x=0). So you need to use such path for integration which does not include x=0.

    ehild
     
  13. Jan 18, 2012 #12

    sharks

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    In my post #12 above, i just gave 2 random examples, as i wanted to understand that it's really the denominator from [itex]\frac{\partial P}{\partial y}[/itex] and [itex]\frac{\partial Q}{\partial x}[/itex] which determine whether the functions P and Q are continuous or not. Since there is also x^3 and x^2 in the denominators from the original problem. I'm sorry if this seems a bit confusing.
     
  14. Jan 20, 2012 #13

    sharks

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    Ah OK! Thanks for this clarification, sheriff89.:smile:
     
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