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Use Green's theorem to calculate work done ?

  1. Aug 10, 2008 #1
    Hey guys! I have been on the forum for about a week or so and have compiled a lot of information and techniques to help me understand calculus, so i really appreciate everyone's help!

    I am a soon-to-be freshman in college and am taking a summer class, calculus II (took calc I in HS). This is our last week of class after our final exam so my professor is taking this time to give us a preview of what we will be learning in the fall semester in Calc III (since this is the same professor). Every Tuesday class our professor gives us a few problems from future sections and asks us to "see what we can come up with" and to work together to find solutions. The following Tuesday he asks us to discuss the problems as a class, seeing which ones of us know our stuff =P

    Basically, i want to ask you guys what you think about these problems as i do them along before i have my discussion. I really want to make a lasting impression on my professor by "knowing my stuff" -to show him i can do it! All's i need is a little help! Would you guys mind giving me some help?

    We are using the textbook Calculus 8th edition by Larson, Hostetler and Edwards and the problems come from the book.

    The problem is on pg 1096 in chapter 15.4 in the text, number 24. It reads:

    Use green's theorem to calculate the work done by the force F on a particle that is moving counterclockwise around the closed path C
    It gives:
    F(x,y)=(3x^2+y)i + 4xy^2j
    and gives:
    C: boundary of the region lying between the graphs of y=(sqrt x) and y=0, and x=9

    I looked at similar problems in the same section and came up with the following for this one:

    work= integral (with C at bottom) of 3x^2+y dx + 4xy^2 dy
    =Integral (with R at bottom) of the integral of ?
    ..this is where i get lost, kind of confused as to what the C is and how to integrate with it. Also, what's R?

    This doesn't seem that bad of a problem, i just think i'm missing something since it seems too easy?

    Any further help would be greatly appreciated. Thanks guys!! =/
  2. jcsd
  3. Aug 10, 2008 #2


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    Homework Helper

    Start by stating Green's theorem and the conditions under which it holds. Hint: The region bounded by C is crucial here.
  4. Aug 10, 2008 #3
    how do i use it though? Ive seen it done for similar problems but i cant follow their work.
  5. Aug 10, 2008 #4


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    Science Advisor

    If you are not going to take the advice given, why post here?

    What is Green's theorem?
  6. Aug 10, 2008 #5
    Let C be a positively oriented, piecewise smooth curve and let D be the region bounded by C, and let P and Q be functions of (x,y) defined on a region containing D, then:

    [tex]\int[/tex][tex]\int_{D}[/tex][tex]\left(\frac{dQ}{dx}[/tex]-[tex]\frac{dP}{dy}\right)[/tex]=[tex]\oint_{C}[/tex] Qdx+Pdy
    Last edited: Aug 10, 2008
  7. Aug 10, 2008 #6


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    Homework Helper

    That isn't enough. You also need to know how to use and more importantly know when it applies.
  8. Aug 10, 2008 #7


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    I think Halls was talking to the OP.
  9. Aug 11, 2008 #8
    Yeah but she hadn't replied, so I figured I'd edge her on a bit? I took a vector calculus class, and Green's Theorem was the last thing we learned, and it wasn't taught very clearly, so I'm interested in being able to better understand it. I was hoping she would answer so I could see what everyone said, but when she hadn't answered, I did it for her to prevent the thread from fizzling away. From what I can understand, Green's Theorem applies when you can use a line integral over a curve, or a double integral over a region, and sometimes one method is easier than the other.
  10. Aug 11, 2008 #9


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    Green's was the last thing you learnt? That's strange, because Green is usually taught before Stokes which would definitely be covered in a vector calc class.
  11. Aug 11, 2008 #10
    Yup we did not cover Stoke's Theorem. Our class was a little slow, you could say. And the professor was a little ineffective at times...again this is why I'm interested in this thread.
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