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

enter image description here

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?

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Note that as you go from $0$ to $2\pi$, the bit from $\pi$ to $2\pi$ retraces what you have between $0$ and $\pi$. – Omnomnomnom Dec 10 '13 at 19:34
@Omnomnomnom If I plot the bit just from $\pi$ to $2\pi$, I get the entire graph:… – user1251385 Dec 10 '13 at 19:47
Apparently all had the same idea to answer. – Luis Valerin Dec 10 '13 at 19:48

$$r = \cos \theta$$

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

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

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

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

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

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(Where $r=\sqrt{x^2+y^2}$ and $\cos\theta=\frac x{\sqrt{x^2+y^2}}$, using the translation from polar to rectangular coordinates) – abiessu Dec 10 '13 at 19:38

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.)

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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.

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Use the polar-rectangular coordinates passing:

$$x=r\cos\theta\;,\;\;y=r\sin\theta\implies r=\sqrt{x^2+y^2}\;,\;\;\cos\theta=\frac xr=\frac x{\sqrt{x^2+y^2}}$$


$$r=\cos\theta\iff \sqrt{x^2+y^2}=\frac x{\sqrt{x^2+y^2}}\iff$$

$$\iff x^2+y^2=x\iff \left(x-\frac12\right)^2+y^2=\frac14$$

and there you have a beautiful circle centered at $\;(1/2\,,\,0)\;$ and radius $\;1/2\;$

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In Cartesian coordinates $(\cos \alpha, \sin \alpha)$ is a circle. What you have is $$(\cos \alpha \cos \alpha, \sin \alpha \cos \alpha) = \frac 1 2 (1-\cos 2\alpha, \sin 2\alpha)$$

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To see why the graph displays a function, ignoring the origin for the moment, note that the "vertical line test" you may be thinking of is inappropriate for polar functions. Each value of $\theta$ specifies a line through the origin, and it is this line which should only pass through one point on the graph. However this test is still inappropriate, because there are multiple values of $\theta$ which specify the same line through the origin (specifically, $\theta+\pi n$ for every integer $n$), so it is possible for a graph of a polar function to have many values passing through the same line through the origin.

In this particular case, every nonvertical line through the origin passes through two points of the graph: one on the origin, and the other elsewhere. It is the other point that the equation $r = \cos \theta$ specifies. On the vertical line, $\theta = \pi/2+\pi n$, we have $\cos\theta=0$, so this gives the origin.

As for why this gives a circle, the other answers have given algebraic proofs for this. I'm partial to the one Thomas and Luis gave, which multiplies by $r$ first, to give $r^2 = r\cos \theta$, myself. ;)

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here your circle is $x^2 + y^2 = x$, in polar coordinates we take $x = r \cos \theta, y = r \sin \theta$ on substituting in the equation you will get $r = 0$ or $r = \cos \theta$. Thus we should vary $r$ from $0$ to $\cos \theta$.

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