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Green's Theorem well, sort of.

  1. Jan 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Evaluate [tex]\displaystyle \int_C y^2dx + x^2dy[/tex] for the path C: the boundary of the region lying between the graphs of [tex]\displaystyle y=x[/tex] and [tex]\displaystyle y=\frac{x^2}{4}[/tex].


    2. Relevant equations
    The catch is that you can't use Green's Theorem.


    3. The attempt at a solution
    I think you can break C into two other curves, [tex]C_1:r(t)=t\textbf{i}+t\textbf{j}[/tex] for [tex]0 \leq t \leq 4[/tex] and [tex]C_2:r(t)=(8-t)\textbf{i}+\frac{(8-t)^2}{4}\textbf{j}[/tex] for [tex]4 \leq t \leq 8[/tex].

    I believe my error is somewhere below:

    [tex]
    2\int_{C_1}t^2dt+\frac{-1}{2} \int_{C_2}(8-t)^3((8-t)+1)dt \Rightarrow 2\int_0^4t^2dt+\frac{-1}{2} \int_4^8(8-t)^3((8-t)+1)dt
    [/tex]

    I get that expression equal to [tex]\frac{-1376}{15}[/tex], which I know is incorrect, as I used Green's Theorem, and got [tex]\frac{32}{15}[/tex] which also coincides with the answer key I have.
     
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  3. Jan 10, 2009 #2

    Dick

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    Shouldn't your second integrand be something like (8-t)^4/16+(8-t)^3/2? With only a (-1) coefficient outside?
     
    Last edited: Jan 10, 2009
  4. Jan 10, 2009 #3
    Yes, that's what I had, but I factored it just a little bit in the above expression.
     
  5. Jan 10, 2009 #4

    Dick

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    Then you factored it wrong. The integral of 2*t^2 from 0 to 4 minus the integral of (8-t)^4/16+(8-t)^3/2 from 4 to 8 is -32/15.
     
  6. Jan 10, 2009 #5
    I realized my mistake, I had been thinking that I needed to get positive 32/15, but it all depends on the direction of the curve, which in my case was clockwise, and is usually taken to be counterclockwise. Is that correct?
     
  7. Jan 10, 2009 #6

    Dick

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    Yes, but how did you get -1376/15?
     
  8. Jan 10, 2009 #7
    I forgot a factor of 1/8 that should be in the second expression when I factored, thanks for the help!
     
  9. Jan 10, 2009 #8

    Dick

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    I thought it was something like that. You set up the problem just fine. Very welcome.
     
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