For which of the following function domain is open and bounded region in $\mathbb R^2$
1.$f(x,y)=\sin^{-1}(x-y)$
2.$f(x,y)=4x^2+9y^2$
3.$f(x,y)=\frac{1}{\sqrt {16-x^2-y^2}}$
4.$f(x,y)=\ln(x^2+y^2)$
Solutions:
Argument 1: $f$ is well defined if $x-y\in [-1,1]$,which is correct for $\mathbb R^2$,which is unbounded.Hence,it is incorrect
Argument 2: Since $4x^2+9y^2$ is valid for all $(x,y)\in \mathbb R^2$.Hence,the domain of $f$ is $\mathbb R^2$,which is both open and closed,but it is unbounded.Hence,option (2) is incorrect.
Argument 3: $f$ is well defined if $16-x^2-y^2> 0$ i.e., $16> x^2+y^2 $,this region is open in $\mathbb R^2$ and can be bounded by the circle $x^2+y^2=17$.Hence,$16> x^2+y^2 $ is open and bounded in $\mathbb R^2$.Hence,option (3) is correct.
Argument : $f$ is well defined for $\mathbb R^2$-{(0,0)},which is unbounded in $\mathbb R^2$.Hence,option (4) is incorrect.
Please check my arguments...whether they are correct or not.
If there are some other arguments please do share...
thank you!!