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Math Help - Plane region in polar coordinates

  1. #1
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    Plane region in polar coordinates

    Hi there. I must express the next region in polar coordinates:
    \{x\in{R^2:x^2+y^2\leq{2y}}\}


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

    x^2+y^2-2y\leq{0}\Rightarrow{x^2+(y-1)^2\leq{1}}

    Then, polar coordinates form:

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

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

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

    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.
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  2. #2
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    Quote Originally Posted by Ulysses View Post
    Hi there. I must express the next region in polar coordinates:
    \{x\in{R^2:x^2+y^2\leq{2y}}\}


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

    x^2+y^2-2y\leq{0}\Rightarrow{x^2+(y-1)^2\leq{1}}

    Then, polar coordinates form:

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

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

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

    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.
    0 \leq \rho \leq 2 \sin \theta and 0 \leq \theta \leq 2 \pi.
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  3. #3
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    Thanks a lot!
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