# Polar equation of $y = 2$

Maybe I do not understand what is going on here but I cannot get the right answer.

$$y = 2$$

$$y^2 + x^2 = r^2$$

$$4 + 0^2 = r^2$$

$$r = 2$$

$$y = r \sin \theta$$

$$1 = \sin \theta$$

$$\theta = \pi/2$$

This is wrong but I do not see anything wrong with my logic.

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Just because $y=2$, it does not mean that $x =0$. In particular, you want that, for any $x$, $y=2$, but $x$ is not $0$. Since you already know that

$$y=2$$

and that to change to polar coordinates, you can use

$$y= \rho \sin \theta$$,

we plug in $y=2$, which gives

$$2= \rho \sin \theta$$

or

$$2 \csc \theta= \rho$$

Note that we don't consider the variable $x$ and $$x= \rho \cos \theta$$, since it suffices to use only the $y$-equation above, because it fully describes the dependence between $\theta$ and $\rho$.

In particular note the radius can't be always $2$, else you would get a circle! Everytime you state something such as $x=0$ or $r=2$, think the implications it carries, and it might help you to spot the flaw.

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Where does csc come from and what does p represent differently than r? – user138246 Jul 1 '12 at 0:40
$\csc \theta$ is the cosecant function, which is simply $\frac{1}{\sin \theta}$. $\rho$ is rho, the greek $r$ which is usually used in polar coordinates for the $\rho$adius. =) – Pedro Tamaroff Jul 1 '12 at 0:41
So all you did was divide everything by sin? – user138246 Jul 1 '12 at 0:43
In polar coordinates, one wirtes $\rho$ as a function of $\theta$, that is $\rho = \rho(\theta)$ just as we write $y$ as a function of $x$ in cartesian coordinates, $y =y(x)$. That's why I wrote the equation explicitly. – Pedro Tamaroff Jul 1 '12 at 0:44

Remember that $y = r \sin(\theta)$. Hence, $$r \sin(\theta) = 2$$ is the equation in polar coordinates of the straight line $y=2$.

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Polar equation is

$$y = r \sin \theta = 2 = constant$$

Using this equation you computed correctly only one point.

$$x= 0, y=2 ;\, \theta = \pi/2, r = 2$$

$y$ is same for all $\theta$ and $x$ is a variable.

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