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Luigi0210

  • one year ago

Need a little help setting up an integral:

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  1. Luigi0210
    • one year ago
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    Consider \(\large \int_{C} (x^2+y^3)dx+3xy^2dy \) where C is the parabola \(y=x^2\) from (2,4) to (0,0) followed by the line segment from (0, 0) to (2, 0) and another line from (2,0) to (2, 4) Evaluate the integral using Green's Theorem

  2. Luigi0210
    • one year ago
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    @hartnn

  3. hartnn
    • one year ago
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    could you plot the region ?

  4. hartnn
    • one year ago
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    \(\Large \oint_{C} (L\, dx + M\, dy) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\, dx\, dy\)

  5. myininaya
    • one year ago
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    something really weird happens with that integrand thingy doesn't it?

  6. Luigi0210
    • one year ago
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    Yes, I end up getting 0 when taking the partials, unless I did that wrong... I had set it from y: 0 -> 2x and x: 0 -> 2

  7. myininaya
    • one year ago
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    nah I got the same thing for the integrand

  8. hartnn
    • one year ago
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    oh, thats right, it becomes 0

  9. Luigi0210
    • one year ago
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    That's why I got confused.

  10. myininaya
    • one year ago
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    I never applied green's theorem before... but if I were to read this like a "normal" integral we are looking at nothing

  11. myininaya
    • one year ago
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    you know like when I think about this integral I think about something like: \[\int\limits_{1}^{2}0 dx=0\] this 0 because |dw:1436983532812:dw|

  12. myininaya
    • one year ago
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    but @hartnn as I said I'm not familar with green's theorem are there any conditions that the integrand must satisfy?

  13. hartnn
    • one year ago
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    "If L and M are functions of (x, y) defined on an open region containing D and have continuous partial derivatives there ..." ~Wiki

  14. hartnn
    • one year ago
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    so i guess, there are no conditions other than that you should be able to find partial derivatives, in this case, you could find it...so you can apply

  15. Luigi0210
    • one year ago
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    So using Green's Theorem would technically give me the same answer as if I solved it using line integrals?

  16. hartnn
    • one year ago
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    thats right...can we find the answer using line integral and verify it to be =0 ... ?

  17. Luigi0210
    • one year ago
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    I don't know how to set up the integral as a line integral though, not used to the form :/ Could you help me set it up? :P

  18. hartnn
    • one year ago
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    |dw:1436984142535:dw|

  19. hartnn
    • one year ago
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    |dw:1436984249338:dw|

  20. hartnn
    • one year ago
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    |dw:1436984422686:dw|

  21. hartnn
    • one year ago
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    |dw:1436984491305:dw|

  22. hartnn
    • one year ago
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    we need to find the bounds of the double integration. limits for x : from curve to vertical line so, x = + sqrt y to x =2 limits for y: from 0 to 2

  23. hartnn
    • one year ago
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    hmm.. i was treating this as double integral :P

  24. Luigi0210
    • one year ago
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    Yes, the bounds make sense, just don't know how to handle it when it's split into dx and dy instead of something like \(\large \int_{C} y^ ds \)

  25. Luigi0210
    • one year ago
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    Would it need to be split into two single integrals?

  26. Luigi0210
    • one year ago
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    *ds not s

  27. hartnn
    • one year ago
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    we'll have to use parametric equations for x and y x= f(t) y= g(t)

  28. hartnn
    • one year ago
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    I am aware of the linear parameterization, but thinking how do we include y= x^2 there...

  29. hartnn
    • one year ago
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    x= t y= t^2 dx =dt dy = 2t dt

  30. hartnn
    • one year ago
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    plug all those in the original integral

  31. Luigi0210
    • one year ago
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    \(\large \int [(t)^2+(t^2)^3 (1)+3(t)(t^2)^2)(2t)] dt\)?

  32. hartnn
    • one year ago
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    yes, this integral we need to find along, t = 0 to t =2

  33. hartnn
    • one year ago
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    and i highly doubt you'll get a 0, because all terms are positive...

  34. hartnn
    • one year ago
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    @dan815 @ganeshie8

  35. Luigi0210
    • one year ago
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    I'm looking in the book and it says something about using the line segments in \(r(t)=(1-t)r_{0}+tr_{1} \) where t is from 0 to 1 to make a parametric equation?

  36. ganeshie8
    • one year ago
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    Here curl is 0, that means the given vector field is conservative. Which is same as saying the work done is path independent, consequently any closed loop line integral is 0.

  37. hartnn
    • one year ago
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    so using green's theorem, getting a 0 is indeed correct. we were trying to find the line integral, (not using green's theorem)

  38. hartnn
    • one year ago
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    along the curve, y= x^2, we took, x =t, y = t^2 and we'll get a certain integral value. now along the line x= 0 to x= 2, again try to find the integral. in this case, y =0, dy =0 similarly along the line y= 0 to y=4, again try to find the integral, in this case, x= 2, dx =0 the addition of all these 3 integrals should be = 0

  39. hartnn
    • one year ago
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    if you get 0, you would have verified the answer we got from green's theorem

  40. ganeshie8
    • one year ago
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    |dw:1436985910945:dw| 3 line integrals are needed... looks a bit painful hmm

  41. hartnn
    • one year ago
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    oh yes, i took 0,2 instead of 2,0 :P

  42. hartnn
    • one year ago
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    |dw:1436986201469:dw|

  43. Luigi0210
    • one year ago
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    Thanks guys!

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