Finding the Area Enclosed by a Polar Curve

  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 = [​IMG]

    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.

    [​IMG]
     
    Last edited: Mar 14, 2009
  2. jcsd
  3. CompuChip

    CompuChip 4,299
    Science Advisor
    Homework Helper

    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. are you talking about the inequality from 0 to 1/8? so my boundaries of int. are a=0, b=1/8?
     
  5. CompuChip

    CompuChip 4,299
    Science Advisor
    Homework Helper

    Yes.
    Note that the dtheta is outside the square root and don't forget the Jacobian factor when doing an integral in polar coordinates.
     
  6. integrating from 0 to 1/8 is giving me .355, which isn't right =-\
     
  7. Why did you include an arclength formula in your OP when the question asks for area?
     
  8. 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. I figured it out, it's Integral[0->1/8] (1/2*(2 e^(0.9theta))^2)dTheta

    thanks everyone
     
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