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Area b/w curves given by parametric eq's

  1. Jul 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the area between the curves:
    x = r(theta-sin(theta)), y = r(1-cos(theta))

    2. The attempt at a solution

    Usually I would just change the parametric equations into a single equation by solving for theta and substituting back into one of the equations. But that results in a nasty expression that I can't integrate. Basically neither my text nor my professor ever explained how to find the area between two parametric curves and I don't know how to approach the problem. Should I subtract the integrals with respect to theta, using the end points for theta where x=y?
     
  2. jcsd
  3. Jul 29, 2009 #2
    Trying to solve for x=y I end up with the equation,

    theta - sin(theta) = 1 - cos(theta)

    Which is only equal at theta = 0... so how am I supposed to pick points to evaluate the definite integral? :confused: Intuitively I feel like 2pi and 0 are the same thing but 2pi doesn't satisfy the equation, right?
     
  4. Jul 29, 2009 #3
    integrating x(theta) with respect to theta gives me,

    r*[(theta)^2/2 + cos(theta)]

    and integrating y with respect to theta gives me,

    r*[theta-sin(theta)]

    If I was to choose the limits of integration as 2pi and 0 and evaluated the integral of x minus the integral of y would I get the correct answer?
     
  5. Jul 29, 2009 #4

    Mark44

    Staff: Mentor

    This problem seems odd to me in that you're given x and y, both in terms of r and theta. The usual way that problems like this are presented is in two different equations in r and theta.

    Something you might try is to convert the x and y into their polar counterparts: x = rcos(theta) and y = rsin(theta). If you did that you would have two equations in r and theta that you could graph.

    Be careful about solving for intersection points in polar equations. It can happen that the two curves intersect, but have different coordinates at the point of intersection.
     
  6. Jul 29, 2009 #5
    I was under the impression that r was a constant. The question is written on my paper exactly as I typed it up here, so no more info was given. This is for a calc 2 class so the solution should be relatively straight forward. If I set x = r*cos(theta)=r*(theta-sin(theta)) the r's cancel and i'm unsure how that would help? We did recently cover polar coordinates so I suppose it would make sense for them to come into play.
     
  7. Jul 29, 2009 #6

    zcd

    User Avatar

    If I remember correctly this was how the Stewart textbook gave the parametric equations of a cycloid. To find area just use [tex]\int_{a}^{b}y\,dx[/tex], where y=r(1-cost) and dx=r(1-cost)dt.

    You'll end up with [tex]r^{2}\int_{a}^{b} \cos^{2}(t)-2\cos(t)+1 \,dt[/tex]
     
  8. Jul 29, 2009 #7
    We are using the Stewart textbook so I bet thats what it is. I'm going to find that section and try to figure it out now. Thanks!
     
  9. Jul 29, 2009 #8
    Would my limits of integration be 2pi and 0? The book says a cycloid is essentially the line traced out by a point on a circle as it rolls, so it would have a period of 2pi regardless of the radius right?
     
  10. Jul 29, 2009 #9

    zcd

    User Avatar

    Yes that's right. And for the record, two parametric equations form one curve, so you're not finding the area between the two parametric equations, you're finding the area beneath the curve given by the equations.
     
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