Why does $r = \cos \theta$ produce a circle?

I am trying to do a double integral over the following region in polar coordinates:

I know that the limits of integration are:

$$\theta = -\frac{\pi}{2} \quad \to \quad \theta = \frac{\pi}{2} \\ r = 0 \quad \to \quad r = \cos \theta$$

However, I don't understand how $r = 0 \quad \to \quad r = \cos \theta$ works. Cosine is a function (not just a relation) meaning that it has only one value of $r$ for every value of $\theta$. However, it seems like the graph $r = \cos \theta$ has two values of $r$ for every value of $\theta$. Why does $r = \cos \theta$ produce a circle?

Solution 1:

$$r = \cos \theta$$

$$\sqrt{x^2 + y^2} = \frac{x}{\sqrt{x^2 + y^2}}$$

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

$$x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}$$

$$\left(x - \frac{1}{2}\right)^2 + (y - 0)^2 = \left(\frac{1}{2}\right)^2$$

This is the equation for a circle in Cartesian coordinates $(x, y)$ with center $({1 \over 2}, 0)$ and radius $1 \over 2$.

Solution 2:

A bit different then the other answers: $$\begin{align} r &= \cos(\theta) \quad &\Rightarrow \\ r^2 &= r\cos(\theta) \quad&\Rightarrow\\ x^2 + y^2 &= x. \end{align} $$ (And then finish as in the other answers.)

Solution 3:

No, there aren't several values of $r$ associated with each value of $\theta$. Remember that the rays are being issued from the origin, not from the center of the circle.

Solution 4:

It is not clear to see actually, because $r=\cos \theta$ is in polar coordinates. If you make a change of coordinates, to rectangular for example this is most clear. Take $x=r\cos \theta$ and $y=r\sin \theta$ then you have $r^2=r\cos\theta$ or $$x^2+y^2=x$$ or $$(x-\frac{1}{2})^2+y^2=\frac{1}{4}$$ which describes a circle.