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Finding the Area Enclosed by a Polar Curve

  1. Mar 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the area enclosed by the polar curve
    r = 2 e^(0.9theta)

    on the interval 0 <= theta <= 1/8
    and the straight line segment between its ends.

    2. Relevant equations

    arclength = eq0006MP.gif

    3. The attempt at a solution

    I need help finding the boundaries for this problem. Is it from 0 to pi? I'm not sure how exactly would I go about finding the limits of integration(boundaries) since it starts from the origin and ends at the x-axis.

    http://img5.imageshack.us/img5/7910/whatareloi.jpg [Broken]

    Attached Files:

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 14, 2009 #2


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    The integration boundaries for theta are given.

    [Hint: first find the area between the curve and the x-axis, then find the area under the straight line segment between its ends and subtract appropriately.]
  4. Mar 14, 2009 #3
    are you talking about the inequality from 0 to 1/8? so my boundaries of int. are a=0, b=1/8?
  5. Mar 14, 2009 #4


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    Note that the dtheta is outside the square root and don't forget the Jacobian factor when doing an integral in polar coordinates.
  6. Mar 14, 2009 #5
    integrating from 0 to 1/8 is giving me .355, which isn't right =-\
  7. Mar 14, 2009 #6
    Why did you include an arclength formula in your OP when the question asks for area?
  8. Mar 14, 2009 #7
    i thought it meant arclength, but i guess it is area
    so it should be int [a->b] of (1/2(r)^2)dtheta

    but I still don't know how to set up this problem correctly
  9. Mar 14, 2009 #8
    I figured it out, it's Integral[0->1/8] (1/2*(2 e^(0.9theta))^2)dTheta

    thanks everyone
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