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Homework Help: Polar Area question

  1. Jun 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the area of the region between the inner and outer loop of the limicon r=2cos(x)-1


    2. Relevant equations
    A=(2(1/2)small circle)-(2(1/2)large circle)


    3. The attempt at a solution
    I don't even know where to start with this question because I can't figure out the formula for the inner circle to plug into the Area formula. The hint I have received from the teacher is "Use symmetry: the area of the shaded region is twice the area in half of the little loop subtracted from twice the area in half of the big loop. The integration is much neater if you choose a good way to compute those two half areas." I have included the teachers hint into the equation above to try and help my understand what he's looking for but I'm still scratching my head. Any help would be greatly appreciated and thanks in advance!
     
  2. jcsd
  3. Jun 24, 2010 #2

    Mark44

    Staff: Mentor

    I hope you have sketched the limaçon. An easy way to do this is to sketch a graph of y = 2cos(x) - 1, and then use this to sketch a graph of your polar curve r = 2cos(theta) - 1.

    The graph of y = 2cos(x) - 1 starts off at (0, 1), drops down to cross the x-axis at about pi/3 and remains below the x-axis until about 5pi/3 or so. There is a low point at (pi, -3).

    On the polar curve you have the point (1, 0), and r becomes zero at about pi/3. After that, even though theta is still positive, the r values are negative, so the points go on the negative side of the rays. For example, the point (pi, -3) on the Cartesian graph corresponds to the polar point (-3, pi).

    There is only one formula for the polar curve. There is not a separate formula for the inner loop and the larger loop. Every point on the limaçon is obtained from the equation r = 2cos(theta) - 1. What you need to find are the values of theta that correspond to the start and end of the inner loop, and the values of theta that correspond to the start and end of the outer loop (upper half of each).
     
  4. Jun 24, 2010 #3
    I was just working thru what you wrote and noticed I wrote the equation as 2cos(x)-1, but it is actually 2cos(theta)-1. I must have been writing too fast and not thinking. I have calculated the values where 2cos(theta)-1 = 0 and I get theta = (pi/3) and (5pi/3) as the limits of integration. Wouldn't the start and end points of the inner and outer loop be the same?
     
  5. Jun 24, 2010 #4

    Mark44

    Staff: Mentor

    Yes, I noticed that. I was careful to distinguish between the Cartesian equation and the polar equation.
    Slow down - you shouldn't be thinking about limits of integration just yet.
    The curve crosses over itself at the pole, but the values for theta are different.

    On your sketch of the polar function, label a few points with their theta values. For example, the rightmost point of the inner loop should be labelled as theta = 0. What is theta at the leftmost point of the inner loop the first time the curve hits that point?

    Try to follow the curve as theta increases from 0 to 2pi. That might give you some better insight as to what the limits of integration are. Keep in mind that you want only the area of the upper (or lower) half of the outer loop minus the area of the upper (or lower) half of the inner loop, and you will double that result.
     
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