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Find area of region (double integral)

  1. Dec 26, 2011 #1

    sharks

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    The problem statement, all variables and given/known data
    Find area of region bounded by
    [tex]x^2 + y^2 = 4[/tex]above y=1

    The attempt at a solution

    OK, so i drew the graph.
    http://s1.ipicture.ru/uploads/20111226/oR99IdxJ.jpg
    The red part of the graph is the area that i need to find.

    Put y = 1 in the equation of circle;
    x^2 = 3
    x = -√3 and +√3

    Now, from what i understood from struggling with this problem, i must always describe the region area in the following way:

    For x fixed, x varies between -√3 to √3
    Then, y varies between 1 and √(4 - x^2)

    Since this problem is found at the end of the section, "Double Integrals - Polar Coordinates", i'm going to transform to polar coordinates, instead of solving it directly:

    x^2 +y^2 = 4
    r^2 = 4
    r = -2 or +2

    Considering r= +2, since the graph is symmetric about the y-axis, i will evaluate the area of the positive half found in the first quadrant and then later when i get the answer, i will multiply the latter by 2 to get the final answer.

    θ varies between 0 and ∏/2

    So, rewriting the double integrals:
    [tex]\int\int dydx = \int\int rdrd\theta[/tex]
    And using the following limits:

    For r:
    y = 1
    rsinθ=1
    So,r = 1/sinθ

    (1/sinθ)≤r≤2 and 0≤θ≤(∏/2)

    For the inner integral, i get the answer:
    2 - 1/[2sin^2(θ)]
    which i integrate again w.r.t.θ
    2θ + (1/2)cotθ
    Applying the limits, i get:

    which is incorrect. I don't know where i made the mistake/s.
     
  2. jcsd
  3. Dec 26, 2011 #2

    SammyS

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    Your integration limits for θ are incorrect. Look at your figure.
     
  4. Dec 26, 2011 #3

    sharks

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    OK, so now i'm using the limits:
    (1/sinθ)≤r≤2 and (∏/6)≤θ≤(∏/2)

    I get the answer: 2∏/3 - √3/2

    To get the whole area, by symmetry, multiply by 2:

    The final answer: 4∏/3 - √3 (correct!)

    However, i have some doubts about the method:

    1. Could i also have used y as fixed, instead of x, to find the area? Just wondering. As in evaluating the area (not through polar coordinates) i've used both methods.

    2. Are these 2 expressions equivalent?
    [tex]dydx = rdrd\theta[/tex]
    [tex]dxdy = rdrd\theta[/tex]
    In my notes, they are not used in any specific order, so i'm wondering, however i checked in a calculus book and it states:
    [tex]dxdy = rdrd\theta[/tex]
     
    Last edited: Dec 26, 2011
  5. Dec 26, 2011 #4

    SammyS

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    Yes, dxdy = dydx .

    But as you've learned recently, the limits of integration need to be changed to make sure that you integrate over the correct region in both cases.

    BTW: rdrdθ = rdθdr .
     
  6. Dec 26, 2011 #5

    sharks

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    OK, thanks SammyS.
     
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