Cartesian vs. Polar Coordinates in sets

Question

The set $D_{2}$ is defined as $\{(x, y) \in \mathbb{R}^2 ~ \big| ~ x^2 + y^2 \leq 1\}$. This is in Cartesian coordinates.

If I wanted to represent the set in terms of polar coordinates, I would write: $\{(r, \theta) \in \mathbb{R}^2 ~ \big| ~ 0 \leq r \leq 1 , 0 \leq \theta \leq 2\pi \}$.

It looks as though I could replace $r, \theta$ with $x, y$ and get an entirely different set (a rectangle in the plane). Is my notation correct, how do I specify that the second set refers to polar coordinates?

Answer 1

The essentially correct answers here don't seem to answer this question of yours:

how do I specify that the second set refers to polar coordinates?

That is indeed a problem, since both the cartesian and polar representations of a point are pairs of real numbers.

I don't know if there's a standard way other than the context of a sentence and the names chosen for the coordinates to specify that $(r, \theta)$ is meant to be a polar representation.

The unit disk would be $$ \{ (r, \theta) | r \le 1 \} $$

with no need to restrict $\theta$ or to say that $r$ is nonnegative.

Answer 2

$\{(r, \theta) \in \mathbb{R}^2 ~ \big| ~ 0 \leq r \leq 1 , 0 \leq \theta \leq 2\pi \}$ is indeed a rectangle. It does not matter how you call the variables.

$\{(r \cdot \cos(\theta), r \cdot \sin(\theta)) \in \mathbb{R}^2 ~ \big| ~ 0 \leq r \leq 1 , 0 \leq \theta \leq 2\pi \}$ is a set of points defined by polar coordinates.

Answer 3

As you defined them, the two sets do not represent the same points in R2. The first is a disc, the second a rectangle.

Answer 4

Define the polar transform as $$\gamma: \Bbb R^2 \to \Bbb R^2 \\ (r,\theta)\mapsto(r\cos(\theta),r\sin(\theta))$$

Then what you want is the image of your set under $\gamma$. So let $D = \{(r, \theta) \in \mathbb{R}^2 \mid 0 \leq r \leq 1 \,\wedge\, 0 \leq \theta \leq 2\pi \}$. Then your region of interest is $\gamma(D)$.

You could of course write $\gamma(D)$ in set notation if you like as $$\{(x,y)\in\Bbb R^2\mid (x,y) = \gamma(r,\theta)\,\wedge\, 0 \leq r \leq 1 \,\wedge\, 0 \leq \theta \leq 2\pi\}$$ or $$\{(x,y)\in\Bbb R^2\mid (x,y) = \gamma(r,\theta)\,\wedge\, (r,\theta)\in D\}$$

but $\gamma(D)$ gets the point across nicer I think.

Answer 5

You have a sheet of paper on which you have drawn an $x$- and a $y$-axis. This sheet is your working space $X$. You can address individual points $p\in X$ using their $x$- and $y$-coordinates. In other words: The chosen setup defines two coordinate functions $x:\>X\to{\mathbb R}$ and $y:\>X\to{\mathbb R}$, such that $p\mapsto\bigl(x(p),y(p)\bigr)$ represents $X$ bijectively as a "copy of ${\mathbb R}^2\>$".

In some situations it is preferable to replace the coordinate functions $x$ and $y$ by new coordinate functions $r:\>X\to{\mathbb R}_{\geq0}$ and $\theta:\>X\to{\mathbb R}/(2\pi{\mathbb Z})$, called polar coordinates. These have some misgivings, but anyway. Your "working space", inhabited by curves and shapes, is still $X$, but there is an "abstract" auxiliary $(r,\theta)$-plane hidden away somewhere. The "old" cartesian coordinates $(x,y)$ of the points $p\in X$ and their new polar coordinates are related by formulas like $$x=r\cos\theta,\quad y=r\sin\theta,$$ resp., $$r=\sqrt{x^2+y^2},\quad \theta={\rm arg}(x,y)\ ,$$ whereby ${\rm arg}(x,y)=\arctan{y\over x}$ if $x>0$, and similar formulas hold in other parts of the plane.