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Area of a graph with polar coordinates

  1. Jul 2, 2008 #1
    I'm trying to plot the graph r = sin 3t and find its area.

    This is how far I've gotten:

    The graph looks like a plane propellor with one propellor pointing downward, and two pointing up-left / up-right, with the length of each equal to 1. Now to get the area...

    I have to figure out where the graph reaches 0... which I'm not sure how to do (I solve sin3t = 0 and get t = 0, pi, 2pi ?? )...
    or...
    I can just get the area from 0 to 2pi... so i'll try that:

    A = intergral of (1/2 (sin 3t)^2 dt) from 0 to 2pi

    the solution is:
    1/2( t/2 - 1/12 sin(6t) )

    If I plug in 0 or 2pi into this and solve, the last term becomes 0 and I get a weird answer... what do I do?!?
     
  2. jcsd
  3. Jul 2, 2008 #2
    [tex]\frac 1 2\int_a^b r^2d\theta[/tex]

    [tex]\sin\theta=0[/tex]

    [tex]\theta=0, \ \pi, \ 2\pi[/tex]

    So the first limit starts and ends from [tex][0,\pi][/tex] for [tex]\sin\theta[/tex], but you want [tex]\sin 3\theta[/tex], so [tex]3\theta=...[/tex]

    Solve for [tex]\theta[/tex], that would be your first limit of integration, and just multiply by a "constant" that gets you the whole area.
     
    Last edited: Jul 2, 2008
  4. Jul 2, 2008 #3
    can you elaborate on what you wrote a bit?

    sorry i'm just not sure how you got that theta = 0, pi, 2pi/3... since this last one does not equal 0 for sin(theta)

    also 3*theta = inverse sin( 0, pi, etc.) = 0
    therefore, 2pi/3 works here since it becomes 2pi and sin2pi = 0...
     
  5. Jul 2, 2008 #4
    Sorry! I was fixing latex and brain froze a bit and accidently put down 2pi/3. Should be 2pi as you said.

    Anyways, you just want the starting point (obviously 0) and the ending point which it becomes 0 again for the first petal. You only need sint=0, which is 0-pi. Now find it for sin3t.
     
    Last edited: Jul 2, 2008
  6. Jul 3, 2008 #5
    hey i got it! sweet!
    appreciate your help :)
     
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