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This integral is destroying my life

  1. Mar 4, 2008 #1
    1. The problem statement, all variables and given/known data
    Evaluate Int{(y+yz cos(xyz))dx + (x^2+xzcos(xyz)dy + (z + xycos(xyz)dz}
    along the ellipse:
    x = 2cost
    y = 3sint
    z = 1
    (its a line integral problem)

    2. Relevant equations

    3. The attempt at a solution
    I have tried the following things:
    First, i tried plugging x,y,and z into the integral, finding dx, dy, and dz in terms of the parameter t. However, doing so i get an expression that cannot be solved using elementary functions:

    Int{(3sint+3sint cos(2cost*3sint))dx + ((2cost)^2+2cost*cos(2cost*3sintz)dy + (1 + 2cost*3sint*cos(2cost*3sint)dz}

    dx = -2sint dt
    dy = 3cost dt
    dz = 0

    gives me the following:
    Int{(3sint+3sint cos(2cost*3sint))(-2sint dt) + ((2cost)^2+2cost*cos(2cost*3sintz)(3cost dt)}

    the above cannot be solved (according to my calculator) with elementary functions...

    I also tried to simply eliminate t and to work from there using the following expression for the elipse:

    x^2/4 + y^2/9 = 1

    However, doing so I cannot get a solvable integral either because it leaves me with too few equations for the 3 unknowns ( i need to use the parameter t somehow, is what I am taking in from this). Is there anything else that I can try to solve this integral? absolutely any help would be greatly appreciated!
  2. jcsd
  3. Mar 4, 2008 #2


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    It's a line integral around a closed curve. [yzcos(xyz),xzcos(xyz),xycos(xyz)] is the gradient of a function. What function? What does that tell you?
  4. Mar 4, 2008 #3
    I'd look at the limit of integration and look at the cosine terms.
  5. Mar 4, 2008 #4
    many of the terms do end up disappearing because of the limits, but i'm still left with several terms that i cant figure out how to solve. I'm not sure how to use the gradient in this case to help me -- but the function that that is the gradient of is f = sin(xyz). Im not sure what this tells me... After i did a bunch of integration i'm left with

    -2pi + 2Int(cos(6cost))dt from 0 to 2pi -- i dont think this integral will go to zero however.
  6. Mar 4, 2008 #5


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    You can THROW AWAY all of the terms that come from a gradient. The integral of a gradient around a closed curve is zero, provided the function is well defined over the whole domain. This means you don't have any nasty stuff like cos(cos(t)). You should be able to handle the rest directly.
    Last edited: Mar 4, 2008
  7. Mar 4, 2008 #6
    I haven't ran into this in the book and I'm not sure why this is so. If you could explain or point me to a place where I can read about this on my own I would appreciate that. Also, I'm not sure where I can throw away these terms -- in the initial integral? and if so, what about the dx, dy, and dzs?

    this is the first integral:
    Int{(y+yz cos(xyz))dx + (x^2+xzcos(xyz)dy + (z + xycos(xyz)dz}

    can i just say
    Int{ydx + x^2dy + zdz}?

    thank you for the help
  8. Mar 4, 2008 #7


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    If they didn't give you that then the problem doesn't seem really fair. It's a nasty contour integral to try to work directly. Yes, you can replace the first integral with the second. The topics to look for are 'conservative forces' or 'conservative vector fields'. There's some wikipedia stuff, like http://en.wikipedia.org/wiki/Conservative_vector_field

    You can stop letting the integral destroy your life now.
  9. Mar 4, 2008 #8
    Ahh, that will come in a few more sections. Thanks a lot for the help!
  10. Mar 4, 2008 #9


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    If you want to think about it this way, the integral around a closed curve of d(sin(xyz)) is zero. Maybe they told you that?
  11. Mar 4, 2008 #10
    The book didn't talk about it yet, but the article was helpful. Thank you for directing me to it and for helping me
  12. Mar 5, 2008 #11
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