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Calculating regions of double integrals

  1. May 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate the follow by first changing the order of integration

    [tex]\int_{x=-1}^{1}\int_{y=x^2}^{2-x^2}dydx[/tex]


    3. The attempt at a solution

    This is the region we're concerned with:

    http://www.wolframalpha.com/input/?i=plot%28y%3Dx^2%2C+y+%3D+2+-+x^2%2C+x%3D+1%2C+x%3D+-1%29

    The new inequalities would be:

    [itex]0 ≤ y ≤ 2 [/itex]

    [itex]\sqrt {-y+2} ≤ x ≤ √y [/itex]

    [tex]\int_{y=0}^{2}\int_{x=sqrt(-y + 2)}^{sqrt(y)}dxdy[/tex]

    Doing this double integral gives me a final answer of zero which I checked on MAPLE. The correct answer is 8/3

    I'm guessing the bounds are incorrect, but I can't figure out why. On the plot y clearly has a lower limit of 0 and an upper limit of 2. The limits of x I just manipulated through the previous inequality.
     
  2. jcsd
  3. May 2, 2012 #2
    Hi. Calculate

    [tex]\int_{y=x^2}^{2-x^2}dy[/tex]

    first. Then integrate in x.

    Regards.
     
  4. May 2, 2012 #3
    Yea, but I think they want me to do it the other way, it's a bit futile, since changing the order complicates things here, but I think it's just for practice to get an understanding of both ways.
     
  5. May 2, 2012 #4
    Look at your integral again - particularly the y limits of integration. You cannot do a change of variables with just one integral in this case.
     
  6. May 2, 2012 #5

    sharks

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    To change the order of integration, basically you need this integral:
    [tex]\iint dxdy[/tex]
    Now, to find the limits of this new double integral, split the region vertically in half and describe each half to determine the limits:

    For the upper quarter region:
    For y fixed, x varies from [itex]x=\sqrt{2-y}[/itex] to x=0
    y varies from y=1 to y=2

    For the lower quarter region:
    For y fixed, x varies from [itex]x=\sqrt{y}[/itex] to x=0
    y varies from y= 0 to y=1

    Hence the entire area of the region can be found by this double integral:
    [tex]I=2 \left( \int^2_1 \int^0_{\sqrt{2-y}} dxdy + \int^1_0 \int^0_{\sqrt{y}} dxdy \right)=2\int^2_1 \int^0_{\sqrt{2-y}} dxdy + 2\int^1_0 \int^0_{\sqrt{y}} dxdy[/tex]
     

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    Last edited: May 2, 2012
  7. May 2, 2012 #6
    That was very insightful thanks for that.

    Although, for the lower quarter region do you mean y varies from y = 1 to y = 0?
     
  8. May 2, 2012 #7

    sharks

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    You should always take the integral limits in the positive direction of the axes.

    If you calculate from y=1 to y=0 and then add up, you will get zero as the sum of the area of the regions, since the area of the upper quarter region will cancel out with the (negative) area of the lower quarter region.
     
  9. May 2, 2012 #8

    SammyS

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    If you solve [itex]\displaystyle y=x^2[/itex] for x you get [itex]\displaystyle x=\pm\sqrt{y\ }\ .[/itex]

    If you solve [itex]\displaystyle y=2-x^2[/itex] for x you get [itex]\displaystyle x=\pm\sqrt{2-y\ }\ .[/itex]
     
  10. May 3, 2012 #9

    sharks

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    Indeed, the integral could also be solved more directly, without splitting the upper and lower regions into quarters, but the limits can be a little confusing if you don't know which half of the curve corresponds to +ve square root and which half corresponds to the -ve square root.

    For the upper half region:
    For y fixed, x varies from x=√(2-y) to x=-√(2-y)
    y varies from y=1 to y=2

    For the lower half region:
    For y fixed, x varies from x=-√y to x=√y
    y varies from y=0 to y=1

    Hence the entire area of the region can be found by this double integral:
    [tex]I= \int^2_1 \int^{-\sqrt{2-y}}_{\sqrt{2-y}} dxdy + \int^1_0 \int^{\sqrt y}_{-\sqrt y} dxdy[/tex]

    Maybe someone can confirm this? I based the limits on the negative signs affecting the gradients of each half of the individual curves.
     
    Last edited: May 3, 2012
  11. May 3, 2012 #10

    SammyS

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    To find an area (which in this case should be positive), the order of the limits of integration should be from a lesser value to a greater value,
     
  12. May 3, 2012 #11

    sharks

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    So, this is the correct integral for the total area:
    [tex]I= \int^2_1 \int^{\sqrt{2-y}}_{-\sqrt{2-y}} dxdy + \int^1_0 \int^{\sqrt y}_{-\sqrt y} dxdy[/tex]
     
    Last edited: May 3, 2012
  13. May 3, 2012 #12

    HallsofIvy

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    Yes, that is correct.
     
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