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Area of a plane region

  1. Oct 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the area of the region in the plane where, r2sin(2theta) >2(sq.root3) , r^2 < 4

    2. Relevant equations


    3. The attempt at a solution
    To try to visualize the problem a little better I converted from r2sin(2theta) to 2xy. However I'm confused after this, since I don't know what the upper limit to integrate is. Also, in the context of the question what does r^2 < 4 mean? Thanks very much. :)
     
  2. jcsd
  3. Oct 15, 2015 #2
    Recall that [itex]r = \sqrt{x^2 + y^2}[/itex], so the inequality [itex]r^2 < 4[/itex] covers all points within a circle of radius 2 about the origin in the xy-plane. Sketch the graph of this bounding circle, and the graph of the bounding curve [itex]2xy = 2\sqrt{3}[/itex] (a rectangular hyperbola). Then shade in the regions that satisfy the inequality. Use the boundaries of those regions to define your integrals.
     
  4. Oct 17, 2015 #3

    HallsofIvy

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    [itex]r^2< 4[/itex], in polar coordinates, is the same as "r< 2" since r is not negative. That is the interior of a circle, centered at the origin, with radius 2
     
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