What is the cartesian equivalent of $r=0$?

Question

The question I have asks to convert

$$\int_{\pi/4}^{\pi/2} \int_{0}^{2/\sin{\theta}} r^{3/2} dr d\theta$$

I think I understand how to do most of it:

$r=\frac{2}{\sin{\theta}}\implies r\sin\theta=2 \implies y=2$

$\theta=\frac{\pi}{4} \implies y=x$

$\theta=\frac{\pi}{2} \implies x=0$

I just want to know what happens with the lower bound for $r$, does this automatically mean $x=0, y=0$?

I graphed the lines $y=2, y=x$

And I can see that the integral converted to cartesian becomes

$\int_{0}^{2}\int_{x}^{2} (x^2+y^2)^{1/4} dydx$

or $\int_{0}^{2}\int_{0}^{y} (x^2+y^2)^{1/4} dxdy$

Essentially my question is how do I convert the lower bound $r=0$ into something cartesian that I can use? Do I just need to graph all the other lines first and see what they give? Does $r=0$ even give any important information?

Answer 1

Yes, a lower bound of $r=0$ means $x=y=0$.

The singular point in polar coordinates where $r=0$ (where the "radius", or our distance from the origin is zero) is the origin. (You can only be a distance of zero away from something, if you are at that something.)

Answer 2

$$ r^2= x^2+y^2=0 \to (x=0,y=0)$$

because $r^2$ is sum of two positive quantities, represents the origin or pole in polar coordinates, a singular point with no derivative (directional) definition. $x$ and $y$ must both be zero when going through the origin.