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Work Integral

  1. Oct 12, 2015 #1
    1. The problem statement, all variables and given/known data
    Evaluate the Work Integral,
    ## I = \int_\Gamma [ (\frac{y} {x^2 + y^2} + 1) dx - \frac{x} {x^2+y^2}dy]##
    between the points (5, 30/pi) and (2,8/pi), using the potential function. Pesent your answer in exact form.

    2. Relevant equations
    ##W = \int F . dr##
    ##\int_\Gamma (Pdx+Qdy)##

    3. The attempt at a solution
    Usually, we are given the curve along the line. This question we have not.

    (5 - 2) and (30/pi - 8/pi) is the region between the points is it not?

    Or do I rearange the given formula I, for y?

    Do we even use the line integral formula, or use Green's Theorem instead?


    Thanks in advance.
     
    Last edited: Oct 12, 2015
  2. jcsd
  3. Oct 12, 2015 #2

    HallsofIvy

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    You can't use "Green's theorem" because that requires an integral around a closed path. However, this problem said "using the potential function" which certainly implies that there is a potential function! That is, that there exist a function, F(x,y), such that [itex]\frac{\partial F}{\partial x}= \frac{y}{x^2+ y^2}+ 1[/itex] and [itex]\frac{\partial F}{\partial y}= -\frac{x}{x^2+ y^2}[/itex]. Find that function and evaluate at the endpoints. (That is, the "work" is the change in potential between the two points.)
     
  4. Oct 12, 2015 #3

    Ok so I've obtained the partial derivatives,
    ##\frac{\partial F}{\partial x}= \frac{-2yx}{(x^2+ y^2)^2}##
    ##\frac{\partial F}{\partial y}= \frac{2yx}{(x^2+ y^2)^2}##

    Now, do I integrate from the points,

    ##I =\int^5_2 [\frac{-2yx}{(x^2+ y^2)^2}+ \frac{2yx}{(x^2+ y^2)^2}] = [\frac{-2y5}{(5^2+ y^2)^2}+ \frac{2y5}{(5^2+ y^2)^2}] - [\frac{-2y2}{(2^2+ y^2)^2}+ \frac{2y2}{(2^2+ y^2)^2}] = 0 ##

    Would that not just equal zero, even with the y values substituted in place? Making this a conservative system? I've most definitely done this the wrong way, haven't I?.
     
    Last edited: Oct 12, 2015
  5. Oct 12, 2015 #4

    LCKurtz

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    Yes, that is wrong. You are given ##F_x = \frac y {x^2+y^2}+1## and ##F_x =\frac{-x}{x^2+y^2}##. You have differentiated them and called them the same things. To find a potential function you must find an ##F(x,y)## whose partial derivatives agree with what is given. This means you need to find anti-partial derivatives. Integrate, not differentiate. And your answer will be of the form ##F(x,y) = ...## Surely your text must have examples for finding potential functions.
     
  6. Oct 13, 2015 #5
    I've practiced very little with line integrals. I've done some further research and I think I have come up with a solution. Im very new to latex so I written a brief summary.

    After doing the partial derivatives, they equated to each other. I think this shows that we can find a potential function. A check. Even though it is implied in the question

    I then found the potential function by integrating and differentiating with respect to y. Obtaining,

    ##f(x,y)=\tan^-1 \frac{x}{y} + x ##

    Inputting the given points, obtaining the answer,

    ##W = f(2,8/pi) - f(5,30/pi)##
    ## = tan^-1\frac{2}{8/pi} + 2 - tan^-1\frac{5}{30/pi} - 5##
    ## = tan^-1\frac{pi}{4} - tan^-1\frac{pi}{6} - 3 ##
     
    Last edited: Oct 13, 2015
  7. Oct 13, 2015 #6

    Ray Vickson

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    This is all correct.
     
  8. Oct 15, 2015 #7
    Yes! Thankyou everyone for the kind and helpful guidance. :-)
     
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