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Center of Quadrant

  1. May 17, 2007 #1
    Hi. this is probably a really dumb question but here goes.
    I know the radius of a circle, say 6 cm.
    i need to know what is the (x,y) coordinate of the CENTER of the Northeast quadrant (first quadrant).
    is it just 3,3? or do i need to use some angle and trig to figure this out.

    thank you in advance
     
  2. jcsd
  3. May 17, 2007 #2

    HallsofIvy

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    What you are calling the "center" is technically the "centroid"- the point at which the "center of mass" would be if the figure were a thin plane of constant density. From formulas for center of mass, you want determine that the x and y coordinates of the centroid [itex](\overline{x},\overline{y})[/itex] are given by:
    [itex]Area*\overline{x}= \int\int x dA[/itex] and
    [itex]Area*\overline{y}= \int\int y dA[/itex] where "dA" is the differential of area for the figure. For a quadrant of a circle of radius A, we would have
    [tex](1/4)\pi R^2 \overline{x}= \int_{r=0}^R\int_{\theta=0}^{\pi/2}(r cos \theta)(r dr d\theta)[/tex]\
    [tex]= \left(\int_{r=0}^R r^2 dr\right)\left(\int_{\theta= 0}^{\pi/2} cos\theta d\theta\right)[/tex]
    [tex]= \left((1/3)R^3\right)(1)[/tex]
    so
    [tex]\overline{x}= \frac{4}{3\pi}R[/itex]
    By symmetry, [itex]\overline{y}= \overline{x}[/itex]
     
  4. May 17, 2007 #3
    wow. thank you!
     
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