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Lineintegrals-greens theorem

  1. Jul 12, 2012 #1
    1. The problem statement, all variables and given/known data

    show 3 different line integrals that can be used to calculate the area enclosed by a curve. use greens theorem to derive each formulae shown

    2. Relevant equations



    3. The attempt at a solution

    So i show 3 line integrals
    1) [itex]\int_{c}xy^4 ds[/itex] where c is the circle [itex]x^{2} + y^{2} = 16[/itex]

    2) [itex]4x^4 ds[/itex] where c is the circle [itex]x^{2} + y^{2} = 16[/itex]

    3) [itex]2y^2 ds[/itex] where c is the circle [itex]y = x^{2}[/itex]

    i just made these up?

    greens theorm states [itex]\oint[Mdx+Ndy] = \int\int_{R} (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) dxd y[/itex]

    where should i go from here? are my line integrals right? maybe i should use something line integrals that make it easier to use greens theorem on?
     
  2. jcsd
  3. Jul 12, 2012 #2
    You're trying to jump into this too fast. You want to calculate an area--there should be some geometric way to find the area a curve encloses by a line integral (that is, they're not asking you to "make up" line integrals but to come up with three different geometric approaches to calculating an area).

    The way you verify that your formulas are correct is by converting your line integral to an area integral. If you know what the region is, how do you find the area of a region with an area integral?
     
  4. Jul 12, 2012 #3
    Hi Muphrid,
    So they are asking for say [itex]\pi r^2[/itex] the are of a circle? [itex]\pi ab[/itex] for a elipse or say a segment of a circle?
     
  5. Jul 12, 2012 #4
    Getting there. If you wanted to find the area between a curve and the x-axis, you'd break it back down into a series of small rectangles conceptually and build the integral from there. Here, you have a general curve (edit: and you want to find the area it encloses by itself). Can you break it down into a series of small geometric objects with an area you can easily compute?
     
  6. Jul 12, 2012 #5
    it would be [itex]\int^{a}_{b}[/itex] f(x) dx and in the case of a circle f(x) = [itex]\sqrt{x^2 + r^2}[/itex] ?
     
  7. Jul 12, 2012 #6

    HallsofIvy

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    I don't believe this problem is asking for a specific example, as the circle of radius 4, here, but three different integrals that would give the area for a general closed curve.

    Your statement of Green's theorem is correct and note that the area inside a closed curve, as a double integral is just [itex]\int\int dydx[/itex]. So to use Green's theorem, we want to find two functions M(x,y) and N(x,y) such that [itex]\partial N/\partial x- \partial M/\partial y= 1[/itex]. One possibility that leaps to mind immediately is N= 0, M= -y. Another is N= x, M= 0. Can you find another?
     
  8. Jul 16, 2012 #7
    hi HallsofIvy,
    So in that case another example would be N=2x, M=y, right?

    so my 3 different line integrals in this case would be
    1. [itex]\oint -ydx [/itex]
    2. 3 [itex]\oint xdy[/itex]
    3. [itex]\oint ydx + 2xdy[/itex]

    to use Greens Theorem to derive each of the 3 formula i just sub in the N and M values into the double integral side as in for the first one [itex]\int\int - \frac{\partial -y}{\partial y} dxdy[/itex] which is [itex]\int\int dxdy[/itex] =area
     
  9. Jul 16, 2012 #8
    This is the basic idea, yes. What HallsofIvy has pointed out is the direct, grunt-work part of the problem. I was hoping to get you see the geometric side. Each choice corresponds to taking the curve and integrating it in a different way.

    N=0, M=-y: you take the curve and draw rectangles down to the x-axis, integrating around it counterclockwise. For a curve entirely above the x-axis, this gives a negative result, so you multiply by -1 to get the correct answer.

    N=x, M=0: you do as above, except you draw rectangles over to the y-axis. Here, no extra negative sign is required.

    But yes, you can generate any other means of integrating this as long as it fits the constraint that the curl is 1. What I had in mind was doing the integral as a series of triangles from the origin, for instance.
     
  10. Jul 16, 2012 #9
    Thanks Murhrid,
    So i've now got 3 different line integrals. i've used greens theorem on them and shown that it gives the area, but is that right? the question asks to use Greens formula to derive each of the formula...
     
  11. Jul 16, 2012 #10
    Once you have verified that [itex]\partial N/\partial x - \partial M/\partial y = 1[/itex], the surface integral obviously gives the area, and you're done. I don't think there's any more work you need to show to argue this.
     
  12. Jul 16, 2012 #11
    thanks a million
     
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