Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find area of region (double integral)

  1. Dec 26, 2011 #1


    User Avatar
    Gold Member

    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.
    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
    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

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


    User Avatar
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Gold Member

    OK, thanks SammyS.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook