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Flux Integrals

  1. Dec 7, 2014 #1

    dwn

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

    Attached Image

    2. Relevant equations
    this is not a simple plane curve or a close plane curve so I use the formula:
    ∫ F ⋅ dr/dt dt

    3. The attempt at a solution
    From the point (0,0) to (2,4)
    Direction Vector v(t) = <2-0, 4-0>
    Parametric Equation:
    r(t) = (2t + 0) i + (4t + 0) j
    r'(t) = 2i + 4j

    ∴ F(x(t),y(t)) = (4t)2i + 2(2t)(4t)j = (16t2)i + (16t2)j
    F(x(t),y(t)) ⋅ r'(t) =( (16t2)i + (16t2)j ) ⋅ (2i + 4j) = 32t2 + 64t2

    I feel my math is correct which leads me to believe my approach is incorrect. The final answer is supposed to be 32 evaluated from 0 to 2.

    But if I finished my integral : ∫ 96t2dt = 32t3

    Can someone please help me.
     

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    Last edited: Dec 7, 2014
  2. jcsd
  3. Dec 7, 2014 #2

    SteamKing

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    The parameter t should run from 0 to 1, rather than 0 to 2. Plug these limits into your flux integral and everything should work out.

    Try the next path.
     
  4. Dec 7, 2014 #3

    dwn

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    Thank you for your help. I had considered that but I'm still not confident on how interval is defined. Would you mind explaining that to me? All of the examples I see in the book simply set the interval from 0 to 1. Is this always the case?
     
  5. Dec 7, 2014 #4

    SteamKing

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    It doesn't have to be, but the interval from 0 to 1 is usually the simplest to evaluate.

    The path over which you were evaluating the flux integral was y = 2x, where 0 ≤ x ≤ 2. By choosing the parameterization r(t) = 2t i + 4t j for this path, it is important to check and make sure that by plugging in the limits of t into the parametric equation, you obtain the same values of x and y for the limits of the actual path expression, hence the interval 0 ≤ t ≤ 1 when plugged into r(t) gives (0, 0) and (2, 4) for the end points of the path, which are identical to y = 2x for 0 ≤ x ≤ 2.

    You could have specified r(t) = t i + 2t j, for 0 ≤ t ≤ 2, and obtained the same evaluation for the flux integral after making the appropriate integration.
     
  6. Dec 8, 2014 #5

    HallsofIvy

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    There are an infinite number of ways to set up parametric equations for a curve. Myself, since in both of these y is a function of x, I would have just used x itself as parameter. In the first, y= 2x so that [itex]F(x, y)= y^2\vec{i}+ 2xy\vec{j}= (2x)^2\vec{i}+ 2x(2x)\vec{j}=4x^2\vec{i}+ 4x^2\vec{j}[/itex]. Since y= 2x, dy= 2dx and [itex]d\vec{r}= dx\vec{i}+ dy\vec{j}= dx\vec{i}+ 2dx\vec{j}= (\vec{i}+ 2\vec{j})dx[/itex]. The integral becomes
    [itex]\int_0^2 (4x^2\vec{i}+ 4x^2\vec{j})\cdot(\vec{i}+ 2\vec{j})dx= \int_0^2 (4x^2+ 8x^2) dx= 12\int_0^2 x^2 dx[/itex].
    The limits of integration are 0 and 2, of course, because the variable of integration is x and x goes from 2 to 4.

    Similarly, the second integral is over the curve [itex]y= x^2[/itex] so that [itex]F(x, y)= y^2\vec{i}+ 2xy\vec{j}= (x^2)^2\vec{i}+ 2x(x^2)\vec{j}= x^4\vec{i}+ 2x^3\vec{j}[/itex]. Further, since [itex]y= x^2[/itex], [itex]dy= 2xdx[/itex] so that [itex]d\vec{r}= dx\vec{i}+ dy\vec{j}= dx\vec{i}+ 2xdx\vec{j}= (\vec{i}+ 2x\vec{j})dx[/itex]. The integral becomes
    [itex]\int_0^2 (x^4\vec{i}+ 4x^2\vec{j})\cdot (\vec{i}+ 2x\vec{j})dx= \int_0^2( x^4+ 8x^3) dx[/itex].

    Again, the limits of integration are 0 to 2 because x is the variable of integration and x goes from 0 to 2.
     
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