How to convert $y>0$ to polar coordinates?

Question

How to convert $y>0$ to polar coordinates?

Given another constraint $x^2+y^2 < 1$.

What I thought is writing $y^2<1-x^2$ so then

$$y > 0 \implies 0<y^2<1-x^2 \implies 0 < y <\sqrt{1-x^2}$$

Additionally $$x^2+y^2<1 \implies 0 < r^2 < 1 \implies 0 < r < 1$$

But I am unsure on how to get $\theta$ bounds.

What I thought is

$$y=r\sin(\theta)$$ $$\iff \arcsin(\frac{y}{r})=\theta$$

which I think should give me the bounds for $\theta$ if I just see what the bounds for $y$ and $r$ imply for $\theta$. However I find it difficult to progress from this because of $0<y<\sqrt{1-x^2}$ being dependent on $x$.

What to do?

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I can progress by writing

$$0<y<\sqrt{1-x^2} \implies x \in [-1,1] \implies 0<y<1$$

which then should enable me to find $\theta$. However, how do I infer $\theta$ if

$$\arcsin(\frac{y}{r})=\theta$$

and $0<y<1$ and $0<r<1$.

Answer 1

Instead of losing yourself in algebra here, draw a diagram and think about how the coordinate systems in question work.

It should become immediately clear that the set you're looking to describe is exactly the (open) upper half-plane, which in the usual polar coordinates is characterized by $0<\theta<\pi$.

The additional constraint $x^2+y^2<1$ (which would be separately represented by $r<1$, of course) does not enter into this.

Answer 2

$ y > 0 $ means the upper half plane so the constraint on $\theta$ is $ 0 < \theta < \pi $. Note the strict inequalities.

To address your comment about algebra: $\sin(\theta) = \frac{y}{r}> 0 \Rightarrow 0 < \theta < \pi$ ? Consider a sketch of $\sin(\theta)$ to see this.