$\begingroup$Find and sketch the domain of the following function $f(x,y) = \arcsin(x^2 + y^2 -2) $.
I have a difficulty in finding this domain, I know that domain $\arcsin$ is [-1,1] but then what shall I do?
$\endgroup$2 Answers
$\begingroup$So the whole bunch $x^{2}+y^{2}-2$ is in $[-1,1]$, you need to think of composition, that is, as long as the range of the function $(x,y)\rightarrow x^{2}+y^{2}-2$ is the domain of $\sin(\cdot)$, then it goes through.
Now the domain of the composite map is then $\{(x,y): -1\leq x^{2}+y^{2}-2\leq 1\}=\{(x,y): 1^{2}\leq x^{2}+y^{2}\leq(\sqrt{3})^{2}\}$, the graph is a doughnut of bounded by inner circle with radius $1$ and outer circle with radius $\sqrt{3}$, both circles have center $(0,0)$.
$\endgroup$$\begingroup$The expression $\arcsin(...)$ is defined whenever $(...)$ is in $[-1,1]$, which means your expression is defined whenever $$x^2+y^2-2\in[-1,1]$$
Now you need to figure out which values of $x,y$ satisfy that condition.
Hint:
For any positive number $R$, what does the set $\{(x,y)| x^2+y^2=R\}$ look like?
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