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Double integrals help

  1. Jul 29, 2014 #1
    1. The problem statement, all variables and given/known data

    Set up the double integral over the region ##y=x+3; y=x^2+1##

    2. Relevant equations



    3. The attempt at a solution

    finding the intersections you get the double integral

    ##\int_{1}^{5}\int_{-1}^{2}dxdy =12 ##


    but why is that not the same as

    ##\int_{1}^{5}\int_{-1}^{\sqrt{y-1}} dxdy =\frac{28}{3}##
     
  2. jcsd
  3. Jul 29, 2014 #2

    LCKurtz

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    That would be the integral over a rectangle.

    That's wrong too. Have you drawn the graphs?
     
  4. Jul 29, 2014 #3
    yeah I drew it, its the line y=x+3 slicing through the parabola x^2 moved up one unit on the y-axis

    ##\int_{1}^{5}\int_{-\sqrt{y-1}}^{\sqrt{y-1}} dxdy ## is the other one I had, but no sure if that is correct either
     
  5. Jul 29, 2014 #4

    LCKurtz

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    It isn't. If you do x first it must go from ##x_{left}## to ##x_{right}##. x on the left is a two piece formula.

    Is there some reason you don't do this problem the more natural and easier dydx order?
     
  6. Jul 29, 2014 #5
    probably out of habit of typically slicing with a y value then seeing how x changes

    ##\displaystyle\int_{-1}^{2}\int_{x^2+1}^{x+3} dydx##

    okay I see it now, if I do x first theres an issue where the left intersection point occurs, below that point the parabola is the left and right x and above that the left is the line and the right is the parabola.

    thanks!
     
  7. Jul 29, 2014 #6

    LCKurtz

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    Yes! That's why you need two integrals to do it it dxdy order. That's the sort of thing you should always look at when deciding how to set up a double integral. Now can you get the same answer both ways?
     
  8. Jul 29, 2014 #7

    ##\displaystyle \int_{1}^{2} \int_{-\sqrt{y-1}}^{\sqrt{y-1}}dxdy + \int_{2}^{5} \int_{y-3}^{\sqrt{y-1}}dxdy = \frac{4}{3} + \frac{19}{6} = \frac{9}{2}##
     
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