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Evaluating double integrals to find area

  1. Nov 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the area of the region R bounded by the parabola y = 2x2−2
    and the line y = 2x + 2 by sketching R, and evaluating the area
    integral

    A = (doubleintegral) R dxdy.

    3. The attempt at a solution

    i found the points of intersection which were -1 and 2 and sketched the region. but im really stuck on evaluating it. If it was dydx i would have no problem but since its dydx i keep ending up with x's in my answer!
     
  2. jcsd
  3. Nov 20, 2009 #2

    rock.freak667

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    What did you put as the limits for the integrals?
     
  4. Nov 20, 2009 #3

    tiny-tim

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    Welcome to PF!

    Hi footboot! Welcome to PF! :smile:

    (have an integral: ∫ and try using the X2 tag just above the Reply box :wink:)

    Show us exactly what you've done. :smile:
     
  5. Nov 20, 2009 #4
    im sorry its quite hard ot get used to writing maths on the internet!
    The limits i used were 2 and -1 on the left integral and 2x+2 and 2x2 -2 on the right integral.
    I integrated with respect to x which just gave x, then filled in the values and got
    (2x+2)-(2x2-2) which gave -2x2 +2x +4
    I then integrated with respect to y and filled in 2 and -1 for y but all the x's from the first parrt are still there so I get an answer of x2+x+10. Which is obviously not right as I presume the answer should have no variables in it?
     
  6. Nov 20, 2009 #5

    tiny-tim

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    Hi footboot! :smile:
    oh i see now …

    you've integrated in the wrong order!

    integrate wrt y first, that gives you a function of x, then integrate wrt x and that gives you a number. :smile:

    (alternatively, if you integrate wrt x first, you must do it between limits which are a function of y)
     
  7. Nov 20, 2009 #6
    In the question it says dxdy so doesnt this mean i have to integrate wrt to x first? And if i do have to do it that way how would I find the limits as a function of y?
    Thanks for your help!
     
  8. Nov 20, 2009 #7

    tiny-tim

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    Without seeing the exact question, it's difficult to say.

    dxdy = dydx, and the order of integration doesn't matter, so you can do it either way.

    However, of course, if the question insists on integrating wrt x first, then it's checking whether you know how to get the limits (which you don't, yet! :wink:), and of course you must do it that way.

    So, what are the limits for x (they are functions of y) if you integrate wrt x first? :smile:

    (in other words, for a fixed value of y, what are the limits of x?)
     
  9. Nov 20, 2009 #8
    In lectures we have been doing examples of functions where dxdy is not equal to dydx so thats why i thought i couldnt do it here. I did integrate wrt y first anyway and came out with an answer of 9 which looks about right from the graph iv drawn.
    By changing the limits for x did you mean change the original functions for the parabola and line? because I did that first and i came out with quite complicated terms for the limits! So im really hoping 9 is right
    Thanks for all your help!
     
  10. Nov 20, 2009 #9

    tiny-tim

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    Unless you show what you've done, I can't say whether it's right. :redface:
     
  11. Nov 20, 2009 #10
    I integrated wrt to y and got y, filled in 2x+2 and 2x2-2 into y and got -2x2 +2x +4. Then integrated with respect to x and got -2x3/3 + x2 + 4x. filled in 2 and -1 to this function and ended up with a value of 9. Which is hopefully right as iv been spending hours on this question! Also I was wondering I did a different
    integration area question earlier and got a negative answer even though the function was above the x axis would this indicate that I went wrong somewhere in the question?
     
  12. Nov 20, 2009 #11

    tiny-tim

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    Yes, that looks ok. :smile:

    (if you got a negative area in another problem, then yes, that's obviously wrong …

    I'd guess you got the two limits the wrong way round :wink:)
     
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