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Double integral bounded by closed parametric curve

  1. Jul 10, 2009 #1
    question:

    how do i find the area under f[x,y] bounded by a closed parametric curve x[t],y[t]? it doesnt look like i can use a change of variables. it seems as though double integrals only with functions where the curve is given explicitly such as y[x] or x[y].
     
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  3. Jul 10, 2009 #2

    tiny-tim

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    Hi okkvlt! :smile:
    I don't understand … what double integral? :confused:

    The area is a single integral … ∫y dx, or ∫y[t] x'[t] dt.
     
  4. Jul 11, 2009 #3
    thats a typo i meant

    how do i find the area under f[x,y] bounded by a closed parametric curve x[t],y[t]? it doesnt look like i can use a change of variables. it seems as though double integrals only work with functions where the curve is given explicitly such as y[x] or x[y].

    what about when the boundary is given by a parametric curve?
     
  5. Jul 11, 2009 #4

    HallsofIvy

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    Use Green's theorem:
    [tex]\int P(x,y)dx+ Q(x,y)dy= \int\int \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}dxdy[/tex]
    Taking Q(x,y)= y so that [itex]\partial Q/\partial y= 1[/itex] and P(x,y)= 0 so that [itex]\partial P/\partial x= 0[/itex] and
    [tex]Area= \int\int dxdy= \int y dx= \int y(t) (dx/dt)dt[/tex]
    just as tiny-tim said! But notice this is a path integral, not an integral over an area.

    For example, if x= cos(t) and y= sin(t), with t going from 0 to [itex]2\pi[/itex] (this is a circle of radius 1 with center at (0,0)), (dx/dt) dt= -sin(t) dt and ydx= -sin^2(t) dt. The area is
    [tex]\int_{t=0}^{2\pi} -sin^2(t)dt= -\frac{1}{2}\int_0^{2\pi}(1- cos(2t))dt[/tex]
    [tex]= (1/2)(2\pi)= \pi[/tex]
     
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