## Calculus: Early Transcendentals 8th Edition

$D=$ {$(x,y) | x \geq 0 ; x^2+y^2 \leq 1$}
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, $x^2+y^2 \leq 1$