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Area of Polar Function

  1. Mar 5, 2008 #1
    1. Problem: Use the spiral r = 5(e^.1θ). Find the area of the region in Quadrant I that is outside the second revolution of the spiral and inside the third revolution.

    2. Relevant equations:

    3. Attempt at solution:

    My problem with finding the area of a polar graph is determining the bounds, so to get the right bounds for this graph do I set the equation equal to zero? I am really at a loss as to how to set up the bounds. Any hints would be helpful.
  2. jcsd
  3. Mar 5, 2008 #2
    So... the second revolution is between which values of [itex]\theta[/itex] and the third is between which values of [itex]\theta[/itex]? [itex]\theta = 0[/itex] is the start of the first.
  4. Mar 5, 2008 #3
    2nd between -2pi and 2pi and the third between -4pi and 4pi?
  5. Mar 5, 2008 #4
    One full revolution is [itex]2\pi[/itex] radians. If revolution 1 is for [itex]0 < \theta < 2\pi[/itex] and the second revolution starts at [itex]\theta = 2\pi[/itex] radians and spans [itex]2\pi[/itex] subsequent radians, the second revolution has what range of [itex]\theta[/itex]?.
  6. Mar 6, 2008 #5
    [itex]2\pi < \theta < 4\pi[/itex]
    and the third would be [itex]4\pi < \theta < 6\pi[/itex].
  7. Mar 6, 2008 #6
  8. Mar 6, 2008 #7
    So the bounds would be between [itex]2\pi < \theta < 4\pi[/itex]?
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