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?
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.
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],