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Area of region between circle and curve

  1. Jan 6, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    http://s2.ipicture.ru/uploads/20120107/67Ag24Qb.jpg

    The attempt at a solution
    So, i plotted the graphs of the circle and the curve:
    http://s2.ipicture.ru/uploads/20120107/x32KTV6y.jpg

    The shaded area is what i need to find. My plan to solve this problem is to find the area with respect to the y-axis of the curve minus the right-half of the circle.

    Area of left-side of curve:
    For y fixed, x varies from x=0 to x=4-y^2
    y varies from y=-2 to y=2

    I did double integral with the limits above, w.r.t. dxdy and got 32/3

    Area of right-half of circle:
    For y fixed, x varies from x=0 to x=√(4-y^2)
    y varies from y=-2 to y=2

    Again, i've done double integral and ended up with 2pi

    I don't know which of these 2 integrals are wrong, as the difference doesn't give me the correct answer, although it's close.
     
    Last edited: Jan 6, 2012
  2. jcsd
  3. Jan 6, 2012 #2

    Mark44

    Staff: Mentor

    Why don't you use just a single integral? Determining an iterated integral seems more complicated than is necessary. Also, I would exploit the symmetry in your curves by calculating the area of the upper half and then doubling it.

    I would use horizontal slices (of thickness Δy) and of length xparabola - xcircle, and integrate from y = 0 to y = 2.
     
  4. Jan 6, 2012 #3

    ehild

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    The parabola and the circle intersect at x=0 and at x=1. I think the shaded area should extend from x=1 to x=4, while y varies between-√3 and √3.

    ehild
     
  5. Jan 6, 2012 #4

    SammyS

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    Notice where the circle & parabola intersect.

    attachment.php?attachmentid=42520&stc=1&d=1325900049.gif
    (from WolframAlpha)​
     

    Attached Files:

  6. Jan 7, 2012 #5

    sharks

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    I obviously had the wrong limits. I had drawn the graph roughly in my copybook so it wasn't clear at first. I used the correct limits as advised and got the answer. Thanks for the help.
     
  7. Jan 7, 2012 #6

    SammyS

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    Sometimes a picture is worth a thousand words! :smile:
     
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