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Double Integration, finding Area

  1. Mar 12, 2008 #1
    [SOLVED] Double Integration, finding Area

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

    Find: [tex]\int[/tex][tex]\int_{A} xdxdy[/tex] , where A is the area between [tex]y=x^2[/tex] and [tex]y=2x+8[/tex]

    2. Relevant equations

    The points of intersection of the two functions is at [tex]x=-2[/tex] and at [tex]x=4[/tex]. Attached is a plot with the area asked to find.

    3. The attempt at a solution

    I'm seeing a problem with the x limits of integration changing at x=-2, one of their intersections. I am pretty sure this can be done by summing the 2 areas separately, but the problem asks to solve for the double integral.

    Attached Files:

    Last edited: Mar 12, 2008
  2. jcsd
  3. Mar 12, 2008 #2


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    IF you were to integrate [itex]\int \int x dx dy[/itex], yes, you would have to do it as two separate integrals: as y goes from 0 to 4, x must range between the two sides of the parabola, [itex]x= -\sqrt{y}[/itex] and [itex]x= \sqrt{y}[/itex]. For y between 4 and 16, x ranges from the straight line on the left to the parabola on the right: [itex]x= (1/2)y- 4[/itex] . The integral is given by
    [tex]\int_{y=0}^4\int_{x= -\sqrt{y}}^{\sqrt{y}}x dxdy+ \int_{y= 4}^{16}\int_{x= y/2- 4}^{\sqrt{y}} xdxdy[/tex]

    Are you required to do it in that order? The other order, dydx, would seem to me simpler and more natural. In this case, x must range from -2 to 4 and, for every x, y ranges from [itex]x^2[/itex] to 2x+ 8. In that order, the integral is
    [tex]\int_{x= -2}^4\int_{y= x^2}^{2x+ 8} x dydx[/tex]
  4. Mar 12, 2008 #3
    Yeah, that's what I ended up doing. I often get too caught up in small details that prevent me from even starting the problem. This was pretty straightforward once I reversed the order of integration. Once again, thanks for your thorough help. /SOLVED
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