I'm not sure if my way of solving this task is correct. Perhaps I've done some "math-grammar" mistakes, or some calculating errors? Especially speaking of assumptions 2. Would anyone double-check my calculations?
Given function $f(x,y) = \sqrt{\pi-3\arcsin\frac{x+y}{\sqrt{3}}} + \ln(1-y^2-x^2)$ find the domain of this function.
Assumption 1: $$-1 \leq \frac{x+y}{\sqrt{3}} \leq 1$$ $$-\sqrt{3} \leq x+y \leq \sqrt{3}$$ $$\Rightarrow -\sqrt{3} -x\leq y \quad \land \quad y \leq -x + \sqrt{3}$$ $$\Longrightarrow y \geq -x -\sqrt{3} \leq \quad \land \quad y \leq -x + \sqrt{3}$$
Assumption 2: $$\pi - 3\arcsin{\frac{x+y}{\sqrt{3}}} \geq 0$$ $$- 3\arcsin{\frac{x+y}{\sqrt{3}}} \geq -\pi$$ $$3\arcsin{\frac{x+y}{\sqrt{3}}} \leq \pi$$ $$\arcsin{\frac{x+y}{\sqrt{3}}} \leq \frac{\pi}{3}$$ $$\Rightarrow \sin{\frac{\pi}{3}} \leq \frac{x+y}{\sqrt{3}}$$ $$\frac{\sqrt{3}}{2} \leq \frac{x+y}{\sqrt{3}}$$ $$\frac{3}{2} \leq x+y$$ $$y \geq -x + \frac{3}{2}$$
Assumption 3: $$1 - y^2 - x^2 > 0$$ $$- y^2 - x^2 > - 1$$ $$y^2 + x^2 < 1$$
Hence: $$\Longrightarrow D = \Bigg( (x,y) \in \mathbb{R}^2 : (y^2 + x^2 < 1), (y \geq - x + \frac{3}{2} ), (y \leq -x + \sqrt{3}) \Bigg) $$
Is this a good enough answer? Do you see some obvious mistakes? Thanks.