Function:
$$x^2-\sin^{-1}y=\frac \pi2$$
A simple WolframAlpha query outputs that its domain is:
$$D=\{(x,y)\in R^2:y\in[-1,1]\}$$
I understand $y$ is restricted due to $\sin^{-1}y$, but I have trouble understanding why $x$ is unrestricted. The following logic makes me believe $x$ should be restricted:
Since $x=\pm\sqrt{\frac\pi2+\sin^{-1}y}$, and $\sin^{-1}y\in[-\frac\pi2,\frac\pi2]$, so, $\color{blue}{x\in[-\sqrt{\pi},\sqrt{\pi}]}$.
I don't exactly see the fault in my reasoning. Can anyone elaborate the reason for this anomaly?
Note: on that WA page, the "Input interpretation" box clearly says that it is finding the domain of $x^2-\sin^{-1}y=\frac \pi2$ and NOT $f(x,y)=x^2-\sin^{-1}y$ as was initially proposed by @HansLundmark.
UPDATE: added additional notes from discussion in comments. Corrected final answer thanks to @ziprovich.
The function $sin^{-1}$ is defined on $[-1,1]$, so $y\in [-1,1]$. The function $F(x,y)=x^2-sin^{-1}y$ is defined on $\mathbb R \times [-1,1]$, but if you want to find the domain in which the equation $F(x,y)=\frac{\pi}{2}$ has solutions, then $x^2\in [0,\pi]$, hence $x\in [-\sqrt \pi, \sqrt \pi]$.
Note $x^2=a\geq 0 \Rightarrow x=\pm \sqrt a$.