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Evaluate the Following Line Integral Part 2

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\displaystyle \int_c zdx+xdy+ydz[/itex] where C is given by [itex]t^2\vec i +t^3 \vec j +t^2 \vec k[/itex]

    Can this [itex]\displaystyle \int_c zdx+xdy+ydz[/itex] be written as [itex]\displaystyle \int_c z\vec i+x \vec j+y \vec k[/itex]?

    I believe I need to evalute the integral [itex]\displaystyle \int_c \vec F( \vec r (t)) d\vec r(t)[/itex]....?

    How do I proceed?
     
  2. jcsd
  3. Nov 1, 2011 #2

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    Hi again bugatti79! :smile:

    It should be:
    [itex]\int_c (z\vec i+x \vec j+y \vec k) \cdot d\vec r(t)[/itex]


    What would [itex]d\vec r(t)[/itex] be?
     
  4. Nov 1, 2011 #3
    How did you establish this?

    From what you have shown I rearranged to find [itex]d\vec r = \vec i dx+\vec j dy+ \vec k dz[/itex]...........?

    No, that wouldnt be right as dr is a function of t.....

    [itex]d\vec r = d(t^2 \vec i +t^3\vec j+ t^2\vec k)[/itex]
     
    Last edited: Nov 1, 2011
  5. Nov 1, 2011 #4

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    What you have is a dot product:
    [itex]zdx+xdy+ydz=(z,x,y) \cdot (dx,dy,dz)[/itex]

    The vector (z,x,y) corresponds to your function [itex]\vec F(\vec r)[/itex]

    The vector (dx,dy,dz) corresponds to your [itex]d\vec r(t)[/itex]

    The vector (x,y,z) corresponds to your [itex]\vec r(t)[/itex]



    Yes, the latter.
    Can you work this out as a derivative?
     
  6. Nov 1, 2011 #5
    [itex]\displaystyle \int_c (z\vec i+x \vec j+y \vec k) \cdot d\vec r(t)=\int_c (t^2 \vec i+t^2 \vec j+t^3 \vec k) \cdot (2t \vec i + 3t^2 \vec j + 2t \vec k)dt = \int_{0}^{1} (2t^3 +3t^4+2t^4) dt[/itex].......?
     
  7. Nov 1, 2011 #6

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    Yep!
    (That was easy, was it not? :wink:)
     
  8. Nov 1, 2011 #7
    Cheers! There will be more trickier ones along the way. :-) Thanks
     
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