Plane region in polar coordinates

[Telemachus]

Homework Statement

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.

 

[Mark44]

Homework Statement

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}}

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.)

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.

I think your set can be described this way: \{(r, \theta) | 0 \le r \le 2sin(\theta), 0 \le \theta \le \pi \}

 

[Telemachus]

Thank you Mark.