# Plane region in polar coordinates

• Aug 27th 2010, 02:11 PM
Ulysses
Plane region in polar coordinates
Hi there. I must express the next region in polar coordinates:
$\displaystyle \{x\in{R^2:x^2+y^2\leq{2y}}\}$

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

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

Then, polar coordinates form:

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

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

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

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.
• Aug 27th 2010, 02:34 PM
mr fantastic
Quote:

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

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

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

Then, polar coordinates form:

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

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

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

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.

$\displaystyle 0 \leq \rho \leq 2 \sin \theta$ and $\displaystyle 0 \leq \theta \leq 2 \pi$.
• Aug 27th 2010, 02:35 PM
Ulysses
Thanks a lot!