Why does the function $r = \theta$ graph a spiral? If $\theta$ denotes an angle (in radians I assume), and $r$ means the distance from the origin, then why does $r = \theta$ make a perfect spiral?  I'm not that advanced in math (only in geometry and Algebra II), and I saw a 3Blue1Brown video that involved spirals and prime numbers, and I got curious. When I graphed $r=\theta$ in Desmos, I saw the spiral and asked "Why?" 
Does anyone know why this happens?
 A: So, if we graph $r = f( \theta )$, as we often do, what does this mean? This means, for each angle $\theta$ (where $\theta$ is a real number), its distance from the origin (or radius) $r$ is given by $f(\theta)$. This is often easiest to see if you take a particular function $f$ and graph a bunch of points $(r, \theta)$ with increasingly higher values of $\theta$. Animations are also nice. Some are in this video. You can play with some demos yourself on Wolfram here.
In this sense, what does $r = \theta$ represent? Basically, for any given angle $\theta$, its distance from the origin is equal in value to $\theta$. 


*

*An angle of $\theta = 0$? The radius $r$ is $0$.

*An angle of $\theta = \pi/4$? The radius $r$ is $pi/4$.

*An angle of $\theta = \pi$? The radius $r$ is $\pi$.

*An angle of $\theta = 2\pi$? The radius $r$ is $2\pi$.

*An angle of $\theta = 10^{10}$? The radius $r$ is $10^{10}$.


All you're doing is graphing the points $(\theta,\theta)$, really (again, $r=\theta$). It's sort of like graphing $y=x$, which gives you the points $(x,x)$ -- now just with a polar flair.

A: If you draw a straight line from the origin, it would be $\theta=\theta_{0}+2k\pi$. Such line intersects with $r=\theta$ at $(r,\theta)=(\theta_{0}+2k\pi,\theta_{0}+2k\pi)$. Note that the gaps between these points, which are the gaps between $r$ values, are the same ($2\pi$), which makes it a perfect spiral.
A: The crucial point here is that we trace out over all $\theta\ge0$*, not just one revolution's worth of values of $\theta$. Every time we complete a revolution, $\theta$ has increased by $2\pi$ radians (or $360^\circ$ if that's more comfortable); this is therefore also added to $r$. Each line through the origin contains infinitely many equally spaced values of $\theta$, so the resulting curve intersects every such line infinitely many times. A curve that achieves this has to be an outward spiral.
*Subtle addendum: what if you try to also include the $\theta<0$ values? We can't have $r<0$, right? Well, we can interpret a negative value of $r$ by rotating the positive counterpart through a half-revolution.
A: $\theta$ isn’t undefined, it is the independent variable in the equation. You get to control $\theta$, and then the equation determines $r$. That produces a point on the graph of the equation. As $\theta$ increases from $0$, the point traces out the graph.
Any graph of the form $r=f(\theta)$ where $f$ is strictly increasing (assuming $r,\theta\geq 0$) [or strictly decreasing] is a spiral.
That’s just because the distance $r$ of a point on the graph from the origin is increasing [decreasing] as you walk around the origin counterclockwise (\theta increases). Isn’t that the very definition of a spiral?
Edit: I see now you were asking why it is a perfect spiral. That’s just because you are moving outward at a constant rate with respect to the angle. If you know calculus, it is because $\frac{dr}{d\theta}$ is constant.
Any graph of the form $r=c\theta$ has the same property. The constant $c$ determines how squeezed ($0<|c|\leq 1$) or stretched ($|c|\geq 1$) the spiral is. 
A: On a

ω(x′,y′)=ω(ρ,φ)=a₀(ρ)+∑_{n=1}^{∞}[a_{n}(ρ)cos nφ+b_{n}(ρ)sin nφ]

si je prends
ω(x′,y′)=ω(ρ,φ)=a₀(ρ)=√(1-ρ²)

tel que
T(ζ,η)=(1/(2π))∬_{S}ω(x′,y′)exp[i(ζx′+ηy′)]dx′dy′

et
S:((x²)/(a²))+((y²)/(b²))≤1;z=0,


(ζa,ηb) = (λcosχ,λsinχ); λ∈[0,∞[,χ∈[0,2π],
(x,y) = (arcosθ,brsinθ); r∈[0,1],θ∈[0,2π],
(x′,y′) = (aρcosφ,bρsinφ); ρ∈[0,1],φ∈[0,2π],

