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Line Integrals (yayyy )

  1. Jan 1, 2009 #1
    Line Integrals (yayyy!!)

    1. The problem statement, all variables and given/known data
    [​IMG]


    Okay, so I have already done it using the surface integral; now I need to compute the 3 individual line integrals.


    By definition, the integral (I will call it I since I am that creative) is given by:

    [itex]I=\oint v\cdot\, dl[/itex]

    [itex]v=<xy, 2yz, 3xz>[/itex]

    [itex]dl=<dx, dy, dz>[/itex]

    [itex]\Rightarrow v\dot dl=xydx+2yzdy+3xzdz[/itex]

    So the three individual integrals are:

    along the base of the triangle,

    [itex]I_1=\int_o^2 xydx+2yzdy+3xzdz [/itex]

    along the height of the triangle,

    [itex]I_2=\int_o^2 xydx+2yzdy+3xzdz[/itex]

    Now the hypoteneuse is what is getting me (I think),

    I need to again evaluate [itex]\int xydx+2yzdy+3xzdz[/itex]

    Are my bounds just from 0--->(2-y) ?

    Casey
     
  2. jcsd
  3. Jan 1, 2009 #2
    Re: Line Integrals (yayyy!!)

    I am looking at my work and now I think that I really do not know what I am doing.

    I do not understand what these integrals even mean.

    There are three lines i, ii, and iii that correspond to the base, height and hyp, respectively.

    When I integrate along i, I am integrating v dot dl from y=0 to y=2 right?
     
  4. Jan 1, 2009 #3
    Re: Line Integrals (yayyy!!)

    Don't you have to parameterize the curve?
     
  5. Jan 1, 2009 #4

    gabbagabbahey

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    Re: Line Integrals (yayyy!!)

    Let's start with the base (i): What can you say about the values of x and z all the way along the base? ...that should simplify your vector function a little. Now, what variable changes along the base? Clearly that will be your integration variable for that section of the path.

    The same applies to the height (ii).

    The hypotenuse is a little trickier: what can you say about the values of x and y along the hypoteneuse? :wink: ....So what is the relationship between dx and dy?.... So dl=___?
     
  6. Jan 2, 2009 #5
    Re: Line Integrals (yayyy!!)

    This sounds familiar.

    I think I confused myself. I forgot that the triangle is (more or less) the domain of v ...NOT the function that is being integrated. Silly me.

    So... let me review my Calc notes to see how we parameterize the curves. I realize that it is probably a simple problem, but I rather do it the hard way.
     
  7. Jan 2, 2009 #6

    gabbagabbahey

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    Re: Line Integrals (yayyy!!)

    I wouldn't bother parameterizing such a simple curve. The hypotenuse is just a straight line segment that can be written either as [itex]z=m_1y+b_1[/itex] or [itex]y=m_2z+b_2[/itex] with y and z both going from 0 to 2. Hence, [itex]dz=m_1y[/itex] and [itex]dy=m_2z[/itex]. In this way, you can write both your vector function and your dl in terms of a single variable (y or z) and its differential and then integrate from 0 to*2.
     
  8. Jan 2, 2009 #7

    HallsofIvy

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    Re: Line Integrals (yayyy!!)

    Along the base line z= 0 (and, of course, dx= dz= 0) so the integral of 2zy dy is easy.

    Along the vertical x= 0 (and dx= dy= 0) so the integral of 3xz dz is easy.

    Along the hypotenuse of the triangle y+ z= 2 so dy+ dz= 0 and dx=0.
    y= 2- z, x= 0 and dy= -dz so 2zy dy+ 3xz dz= 2z(2-z)(-dz)+ 3(0)zdz= (4z- 2z2)(-dz)= (2z2- 4z)dz. Integrate that from z= 0 to z= 2.
     
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