Answer
${(x,y)∈R^{2}|x^{2}+y^{2}\leq 4}$ and $-1\leq x\leq 1$
Work Step by Step
Given: $f(x,y)=\sqrt {4-x^{2}-y^{2}}+\sqrt {1-x^{2}}$
Since we can only take the square root of non-negatives quantities, we want
$4-x^{2}-y^{2}\geq 0$
This implies
$x^{2}+y^{2}\leq 4$
and
$1-x^{2}\geq 0$
This implies
$x^{2}\leq 1$ or $-1\leq x\leq 1$
Therefore, the domain for given function is
${(x,y)∈R^{2}|x^{2}+y^{2}\leq 4}$ and $-1\leq x\leq 1$
Sketch the graph for the domain as shown below:
