Answer
$D=$ {$(x,y) | 0 \leq x \leq \sqrt {1-y^2}$}
Work Step by Step
As we are given that $G(x,y)=\sqrt x+\sqrt {1-x^2-y^2}$
The function $G(x,y)=\sqrt x+\sqrt {1-x^2-y^2}$
represents a squared root function which cannot be less than $0$.
Thus,
$ x \geq 0 ; 1-x^2-y^2 \geq 0$
or, $x^2+y^2 \leq 1$
or, $x^2 \leq 1-y^2$
or, $x \leq \sqrt {1-y^2}$
This means that $x \geq 0,x\leq \sqrt {1-y^2}$
Hence, Domain: $D=$ {$(x,y) | 0 \leq x \leq \sqrt {1-y^2}$}