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Equation of Graph in Polar Coordinates

  1. Mar 10, 2009 #1
    1. The question was find the area between the curves using DOUBLE Integrals

    Area between:
    r = sin theta
    r = cos theta

    well to draw them i made them into cartesian form by

    r^2 = rsin theta
    r^2 = rcos theta


    x^2 + y^2 = y

    x^2 + y^2 = x

    completing square

    1) x^2 + (y - 1/2)^2 = 1/4
    2) (x - 1/2)^2 + y^2 = 1/4

    these are two circles

    their intersection or bounded region that we need to find the area is like a disc...

    I know if i use the double integral

    Integral of (Integral of 1) dA OVER D where D is the intersection bounded area

    i get the area i need....

    but I dont know how to define that region

    if someone can help me define the region it will be helpful

    I know THETA will change from 0 to pi/2

    but R will change from 0 to some equation of that area but i dont know how to find that equation!!

  2. jcsd
  3. Mar 11, 2009 #2


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    Hi Salman

    If you look at r = sin theta in polar coordinates, I assume you mean the the polar equation is:

    r = (sin(theta), theta)

    which is defintely not a circle, similar for the cos
  4. Mar 11, 2009 #3

    in polar coordinates
    r = sin (theta)

    is NOT equal to
    r^2 = rsin (theta) = x^2 + y^2 = y

  5. Mar 11, 2009 #4


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    Hi salman, sorry jumped the mark on that one, think I know what you are doing now...

    so you have 2 circles radius 1/2 at centres (0,1/2) and (1/2,0), and need to calculate the area between them

    If you draw a picture where do the circles intersect? at the orgin, and at a point along y = x (or equivalently theta = pi/4) due to the symmetry of the situation

    think about the bounds for r in each half of the area, both upper & lower will correspond to one of the circles,

    I would try and write it as sum of 2 integrals, one from theta 0 to pi/4, and theta pi/4 to pi/2, with appropriate r bounds. So do the r integral first & define lower & up bounds in terms of theta for each segment...
  6. Mar 12, 2009 #5
    so in other words i can just do

    theta limits 0 to pi/4
    r limits r = sin (theta) to the intersection point

    add that to

    theta limits pi/4 to pi/2
    r limits intersection point to r = cos (theta)

    would that be the proper limits?
  7. Mar 12, 2009 #6


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    lookin good - so the first half is

    [tex]\int^{\pi/4}_0 d\theta \int^{sin{\theta}}_{0}dr (?)[/tex]

    what is your integrand for area? do you know what a jacobian is?
  8. Mar 14, 2009 #7
    yea got it ty...it was also on my quiz yesterday so it was great i asked here :)

    r*dr*d(theta) :)
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