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Finding average value

  1. Nov 20, 2009 #1
    Find the average value of y on the region D that is bounded above by the parabola y = 4 - x2 and below by the x-axis.

    Does this question makes sense?

    I know the formula for the average values function :

    1/(A(x)) * doubleIntegralOf F(x,y)DA


    But its says above the parabola and below the x-axis, does that contradict each other?

    There is no explanation in the answer but here it is :

    The area of the region is 64/3 and the double integral of y over the region is 256/15. So
    y_avg = 4/5.

    Can you explain and show me how to do this ?
     
  2. jcsd
  3. Nov 20, 2009 #2
    "bounded above" means that the parabola is the upper limit.
     
  4. Nov 20, 2009 #3
    so is it just this :

    1/2[tex]\int^{2}_{0} 4 - x^2[/tex]

    And how about the double integral of it?
     
  5. Nov 20, 2009 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The parabola, y= 4- x2, intersects the x-axis at (-2, 0) and (2, 0). On the entire region, bounded by the parabola and the x-axis, x ranges from -2 to 2. For each x, y ranges between the x-axis and the parabola- from y= 0 to y= 4- x2. Those are the limits of integration:
    [tex]\int_{x=-2}^2 \int_{y= 0}^{4- x^2} y dydx[/tex]
    Of course, because of the symmetry, you can integrate from x=0 to x= 2 and double.

    Notice that the double integral for the area is just
    [tex]\int_{x=-2}^2 \int_{y= 0}^{4- x^2} dydx[/itex] and becomes
    [tex]\int_{x=-2}^2 \left[y\right]_{y=0}^{4-x^2}= \int_{x=-2}^2 (4- x^2)dx[/tex]
    [tex]= 2\int_0^2 4- x^2 dx[/tex]
     
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