Answer
$D=$ {$(x,y) | x \geq 0 ; x^2+y^2 \leq 1$}
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 is defined for positive numbers greater than zero.
Thus,
$ x \geq 0 ; 1-x^2-y^2 \geq 0$
or, $x^2+y^2 \leq 1$
