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Evaluate the Line Integral

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Let C be the (positively oriented) boundary of the first quadrant of the unit disk. Use the definition of the line integral to find ∫(xy)dx+(x+y)dy


    2. Relevant equations
    x=rcos(x)
    y=rsin(x)
    dx=-sin(x)
    dy=cos(y)
    0≤ t ≤ ∏/2

    3. The attempt at a solution
    ∫-cos(t)sin^2(t)+cos^2(t)+sin(t)cos(t) dt from 0 to ∏/2
    Then I finished out the integral and was left with ∏/4-5/6 which is incorrect. Could it possibly have to do with the r or my limits of integration?
     
  2. jcsd
  3. Dec 5, 2011 #2
    BTW I can do this using Greens theorem
     
  4. Dec 5, 2011 #3

    Dick

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    No, the setup looks ok. I think you just did the integral wrong. I don't get the number you got as a answer.
     
  5. Dec 5, 2011 #4
    I used wolfram to see where I mesed up and when it gave me the integral it gave the answer pi/4-1/6 which is also incorrect. Also another weird thing is I did the integral again and I got the same thing wolfram did unfortunatley the back of the book disagrees it says the answer is pi/4-1/3. Which I believe because I can solve using greens so you got anything else
     
  6. Dec 5, 2011 #5

    Dick

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    You really need to show more of your work before we can help. Yes, Green's theorem gives you pi/4-1/3. So far you've only given the line integral over the arc. What do you get for the line integral over the x=0 and y=0 parts of the region?
     
  7. Dec 6, 2011 #6
    Yah sorry about not showing too much work but that would require a ton of typing and I am lazy. Other than that, I think I have solved it. I found that the line integral over x=0 would cancel out. As for y=0, I determined it to be -1/2 so when I add -1/2 to pi/4+1/6. I get pi/4-1/3 which is correct. Let me show my work at least for the part I just did x=0 and y=t and dx = 0 and dy = 1 so I get ∫t dt from 1 to 0. which gives a -1/2
     
  8. Dec 6, 2011 #7

    Dick

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    That's it all right.
     
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