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Plane region in polar coordinates

  1. Aug 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi there. I must express the next region in polar coordinates:
    [tex]\{x\in{R^2:x^2+y^2\leq{2y}}\}[/tex]


    So, this is what I did to visualize the region:
    Completing the square we get:

    [tex]x^2+y^2-2y\leq{0}\Rightarrow{x^2+(y-1)^2\leq{1}}[/tex]

    Then, polar coordinates form:

    [tex]f(x)=\begin{Bmatrix} x=\rho \cos\theta \\y=\rho \sin\theta \end{matrix}[/tex]

    So I got
    [tex]\rho^2=2y\Rightarrow{\rho=\displaystyle\frac{2y}{\rho}}\Rightarrow{\rho=2\sin\theta}[/tex]

    [tex]f(x)=\begin{Bmatrix} x=2\sin\theta \cos\theta \\y=\sin^2\theta \end{matrix}[/tex]

    Now, how do I express the region with polar coordinates? this is the inside of the circle, I've just get the expression for the boundary. How do I include the inside of it?

    Bye there. Thanks for posting.
     
  2. jcsd
  3. Aug 27, 2010 #2

    Mark44

    Staff: Mentor

    Convert this inequality to polar form, which gives you r2 - 2rsin(theta) <= 0, or
    r(r - 2sin(theta)) <= 0.

    I believe this is equivalent to r <= 2sin(theta). (I'm being cautious here because I haven't thought through the ramifications of r being negative and r - 2sin(theta) being positive and vice-versa. It's much more straightforward if you're dealing with an equation.)
    I think your set can be described this way: [tex]\{(r, \theta) | 0 \le r \le 2sin(\theta), 0 \le \theta \le \pi \}[/tex]
     
  4. Aug 27, 2010 #3
    Thank you Mark.
     
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