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Area enclosed between y^2=4-x and y=x-2

  1. Jun 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Please find attached


    2. Relevant equations
    Definite integrals and area between two curves


    3. The attempt at a solution
    Also find attached.

    The answer in the back of the book says that part c should be 10/3, but that would mean that (8sqrt (2))/3 would need to cancel out in both area 2.1 and area 2.2, whichdoesn't work because areas are always positive.
    20160610_103133.jpg 20160610_103219.jpg
     
  2. jcsd
  3. Jun 9, 2016 #2

    fresh_42

    Staff: Mentor

    Why didn't you calculate the area under the parabola and only subtract the triangle? It seems you used more areas than necessary. But I'm not good at reading on the side.
     
  4. Jun 9, 2016 #3
    Ah yes that is a much better idea thank you. But still, my method should still be getting the same answer but it's not. Could you please explain to me where I'm going wrong?
     
  5. Jun 9, 2016 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Areas may always be positive, but definite integrals are not required to be positive. To calculate an area using a definite integral sometimes requires some interpretation of what the definite integral means in that case.

    http://tutorial.math.lamar.edu/Classes/CalcI/AreaBetweenCurves.aspx

    Going off fresh_42's comment, you can also evaluate the definite integral to calculate the area under a curve using dy instead of dx as the variable of integration.
     
  6. Jun 9, 2016 #5

    fresh_42

    Staff: Mentor

    To be honest: I didn't like the root expressions. First thing I did was to exchange x and y.
     
  7. Jun 9, 2016 #6

    Mark44

    Staff: Mentor

    In other words, using horizontal strips instead of vertical strips.
     
  8. Jun 9, 2016 #7
    Thank you all for your help, but I feel like some of this stuff is a little beyond my current course. We've only just started looking at definite integrals, and have spent even less time with areas between two curves. I highly doubt we're expected to be using horizontal strips instead of vertical ones when we haven't even learned about conics or anything really.

    But again thank you all for your help still
     
  9. Jun 9, 2016 #8

    SteamKing

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    Staff Emeritus
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    You don't need to know anything about conics to switch the variable of integration. Often times, switching is done as a matter of convenience to simplify the integral.
     
  10. Jun 9, 2016 #9
    I think my main problem is that I still don't know how to calculate the area between two curves properly.

    Say I was just given y=x-2 and y=-sqrt(4-x) and had to find the area enclosed between them between x=0 and x=2
    Isn't the correct way of doing this to just set up an integral over the given domain and subtract the function on the bottom from the function in top? So integral(2, 0, (x-2-(-sqrt(4-x)))dx
     
  11. Jun 9, 2016 #10

    fresh_42

    Staff: Mentor

    .... or simply exchange x and y and integrate as you are used to, which is the same.
     
  12. Jun 9, 2016 #11
    Okay thank you I'll just do it that way
     
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