I am seeking clarity / intuition on the meaning of the condition $y \geq x$ when performing a change of variables from Cartesian to Cylindrical Coordinates for volume integration.
In the context of Cylindrical Coordinates, variables $x$ and $y$ are expressed in terms of $r$ and $\theta$, where $x = r \cos(\theta)$ and $y = r \sin(\theta)$. However, I have uncertainties about the implication of the condition $y \geq x$ in this context.
Could someone provide intuition / an analytical development / a detailed explanation of what $y \geq x$ signifies in Cylindrical Coordinates during a variable transformation for volume integration?
I appreciate any guidance or clarification that can be offered on this matter.
Thank you in advance for your assistance!
When we consider Cylindrical Coordinates & check only $x$ & $y$ , we are basically using Polar Co-ordinates with $z$ having no restriction.
Hence we can consider only the $X\text{-}Y$ Plane , which we want to convert to $r\text{-}\theta$ Plane , where we want Interval $0 \le \theta \le 2\pi$ & $r \ge 0$.
Consider the Points in the $1^{st}$ Quadrant , at various Positions of the Blue line.
Coordinates are $x$ & $y$ which get converted to $r=\sqrt{x^2+y^2}$ & $\theta=\tan^{-1}y/x$
It will have $0 \le \theta \le \pi/4$
Blue line cannot occur in $2^{nd}$ Quadrant.
In $3^{rd}$ Quadrant , $\theta$ is between $5\pi/4$ & $3\pi/2$.
In $4^{th}$ Quadrant , $\theta$ is between $3\pi/4$ & $2\pi$.
Consider the Points in the $1^{st}$ Quadrant , at various Positions of the Purple line.
Coordinates are $x$ & $y$ which get converted to $r=\sqrt{x^2+y^2}$ & $\theta=\tan^{-1}y/x$
It will have $\pi/4 \le \theta \le \pi/2$
In $2^{nd}$ Quadrant , $\theta$ is between $\pi/2$ & $\pi$.
In $3^{rd}$ Quadrant , $\theta$ is between $\pi$ & $5\pi/4$.
Purple line cannot occur in $4^{th}$ Quadrant.
Over-all , there is no restriction on $r$ here.
To match $y<x$ , $\theta$ should be between $0$ & $\pi/4$ or between $5\pi/4$ & $2\pi$.
To match $y>x$ , $\theta$ should be between $\pi/4$ & $5\pi/4$. [[ This is what OP wants ]]
When we want the alternate Interval $-\pi \le \theta \le +\pi$ , we can make suitable changes to the Intervals.
We should then get :
$\frac{-3\pi}{4} \le \theta \le \frac{+\pi}{4}$ for $y<x$
$\frac{\pi}{4} \le \theta \le +\pi$ & $-\pi \le \theta \le \frac{-3\pi}{4}$ for $y>x$ [[ This is what OP wants ]]