1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Exact differentials (thermody)

  1. Jun 29, 2008 #1
    1. The problem statement, all variables and given/known data

    This isnt exactly HW problem , but a "simple" worked problem that is supposed to illustrate exact differentials....

    Consider a differential dZ =2xy⋅dx+x^2dy integrated
    on two paths where Path I is x=y and Path II is x^2=y.

    The integrations are from (x,y)=(0,0) to (x,y)=(1,1).

    Then it works the problem, and says this:

    delta Z sub-i = 2*integral from 1 to 0 xy*dx + integral from 1 to 0 x^2 dy

    = 2 integral from 1 to 0 x^2dx + integral from 1 to 0 y^2dy = 1



    2. Relevant equations
    Why are they integrating twice, to arrive at x=1???

    If this doesnt make any sense, I will upload the pdf file, Im have trouble writing out all the symbols using basic script.

    3. The attempt at a solution
     
  2. jcsd
  3. Jun 29, 2008 #2

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's hard to read what you wrote. Let me TeX it:
    is that what you meant to write? I don't understand why you put your integrations from 1 to 0 (instead of 0 to 1). The integrals you wrote down don't seem to make sense to me. Are you sure these are the correct ones? The first two ones are good, but the last two ones (using the line y=x^2) are wrong.

    And why do you ask about integrating "twice"? What twice are you referring to? They simply want you to show that the integral around two different paths gives the same answer.
     
  4. Jun 29, 2008 #3

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member



    For the first path: take from the origin to (x=1,y=0) and then from (x=1,y=0) to (x=1,y=1).

    These are two segments. On the first segment (origin to (1,0), you have x varies from 0 to 1, y is constant at y=0 and dy=0. So set dy=0, y=1 and integrate x from 0 to 1.

    For the second part of the segment (1,0) to (1,1), x is held constant at x=1 and y varies from 0 to 1. So set x=1, dx=0 and integrate y from 0 to 1.

    Add the two results from the two segments to get your final answer.

    Now, repeat for the second path which is along the parabola y=x^2. You may reexpress everything in terms of x. Replace y by x^2 and then replace dy by 2x dx. Integrate your expression over x from 0 to 1. Your final answer will give the same result as previously which will be equal to one.
     
  5. Jun 29, 2008 #4
    Thanks alot for helping me with this.
    your right, it should be from 0 to 1, my mistake.
    The 2nd integrations are exactly as professor wrote them (I uploaded pdf file--its on pg 19).
    This is why Im asking about integrating twice:
    If we integrate xy with respect to x, we get x^2 right?
    But then after we have integrated it (to get x^2) the integral sign should dissapear shouldnt it?

    But in order for the equation to = 1, you have to integrate a second time, specifically getting 2*(2/3) + (1/3) = 1

    Isnt that integrating twice? Obviously im stumbling on something but i dont know what it is.
    Thanks for the help!
     
  6. Jun 29, 2008 #5
    sorry i meant to say 2*(1/3) + (1/3) = 1
     
  7. Jun 29, 2008 #6
    (here is pdf file that I pulling this from)
     

    Attached Files:

  8. Jun 29, 2008 #7

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    They don't make any sense to me! Are they supposed to be along the parabola y =x^2??
    It depends!!!!!! It depends of the path you are following!!
    If you are going along the straight line (0,0) to (1,0) then y =0 so [itex] \int xy dx =0 [/itex]
    If you are integrating along the straight line from (1,0) to (1,1), then x is held constant so dx=0 and the integral is again zero. If you integrate along the parabola y=x^2 from (0,0) to 91,1) then you have
    [tex] \int_0^1 dx ~ x^3 = 1/4 [/tex]

    So you must specify the path you are following! And you don't integrate twice.



    You know, I just realized that the second expression would make sense if the integral was along the straight line y=x instead of the parabola y=x^2! Are you sure the prof did not say y=x??
     
  9. Jun 29, 2008 #8
    yes, he explains 2 paths. Path 1 is y=x.
    But how can the above equation = 1 unless you integrate twice?

    So first you go from xy to x^2
    then (2nd time) you go from x^2 to (1/3)x^3

    So, we're going from 0 to 1, so from the above equation you would have 2(1/3) + (1/3) = 1

    But I arrived at 1, by integrating twice.
    (??)
     
  10. Jun 29, 2008 #9
    OK I figured out why i though integrating twice.
    The first part he inserts the equality y=x (for path 1). thats where i thought it was being integrated.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Exact differentials (thermody)
  1. Exact Differentials (Replies: 2)

  2. Exact differential (Replies: 2)

  3. Exact differentials (Replies: 3)

  4. Exact differentials (Replies: 0)

Loading...