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Homework Help: Find the area of the region inside both polar graphs

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data
    r1= 1+sin(theta)
    r2= 5sin(theta)

    2. Relevant equations
    see above?

    3. The attempt at a solution
    totally stumped. usually i would set the two curves equal to each other, but i have no idea how to do that. using my ti-89's solve function just gives me a weird answer using arcsin as well.
  2. jcsd
  3. Nov 10, 2009 #2


    Staff: Mentor

    Set the two r values equal to each other -- you can't set two equations equal to each other -- which gives you 1 + sin([itex]\theta[/itex]) = 5sin([itex]\theta[/itex]).

    The graph of r = 5sin([itex]\theta[/itex]) is a circle of radius 2.5, whose center is at (0, 2.5) in rectangular coordinates. The other curve is a cardioid, a kind of heart-shaped curve. These curves intersect at two points.
    Last edited: Nov 11, 2009
  4. Nov 11, 2009 #3
    but how do I find the numerical values of these two points?
  5. Nov 11, 2009 #4


    Staff: Mentor

    Solve the equation 1 + sin([itex]\theta[/itex]) = 5sin([itex]\theta[/itex]). That will give you one of the values of [itex]\theta[/itex]. And because sin([itex]\pi - \theta[/itex]) = sin([itex]\theta[/itex]), that will give you the other one.
  6. Nov 11, 2009 #5
    is there a way to solve it manually? my ti-89 gives me 6.28....(2pi) and 2.8889, which im not sure what that converts to in radians.
  7. Nov 11, 2009 #6
    which on second thought, doesnt make sense.


  8. Nov 11, 2009 #7


    Staff: Mentor

    Absolutely, there's a way to solve it manually. Nothing I have done required the use of a calculator of any kind.

    2.8889 is in radians, but both answers you show are WAY off. I really think you should put your calculator away.

    What exactly did you do to solve 1 + sin([itex]\theta[/itex]) = 5sin([itex]\theta[/itex])? Show me your steps. No calculator.
  9. Nov 11, 2009 #8
    Just woke up. A little groggy but I'll give it a shot.
    theta=x out of laziness

    .....now what?
    no where out of the "standard" points on the unit circle (pi/3,pi/2,pi/4, etc) does the y value equal 1/4

    unless I can use the half/double angle formula somehow?

    I REALLY gotta go over my trig.
  10. Nov 11, 2009 #9


    Staff: Mentor

    OK, good so far.
    sin(x) = 1/4 ==> x = arcsin(1/4) = sin-1(1/4). Now you can use a calculator to get an approximate value for x, which is about 14.5 degrees, or about .253 radians. The exact value is The other angle is pi - x, or about 165.5 degrees.

    Now, to find the area that is inside both curves, you should probably set this up as a polar integral. You can make life slightly easier by using the symmetry here, noting that both curves are symmetric about the y axis. Draw a graph of the two curves so you can figure out what your integrand needs to be. Also, because the boundary changes at the points of intersection, you'll need two integrals.
  11. Nov 11, 2009 #10
    Got it. The actual finding the area part was easy, just blanked out at the sinx=1/4 part. thanks for you help
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