Answer
Domain = $\{(x,y)|\quad 0\leq y\leq\sqrt{25-x^{2}}\}$
(see image)
Work Step by Step
Because of the restrictions for square roots, f is defined when
$ y\geq 0\qquad$and$\qquad 25-x^{2}-y^{2}\geq 0.$
The first condition allows only points above or on the axis.
The second condition:
The $\mathbb{R}^{2}$ plane is divided into two regions by the circle around the origin,
$x^{2}+y^{2}=5^{5}$
(the circle itself is included)
(0,0) satisfies the condition so the region defined is the disk of radius 5 around the origin.
We combine the 2 conditions
$y^{2}\leq 25-x^{2},\qquad y\geq 0$
$-\sqrt{25-x^{2}}\leq y\leq\sqrt{25-x^{2}},\qquad y\geq 0\quad\Rightarrow\quad 0\leq y\leq\sqrt{25-x^{2}}$
which is half of a disk around the origin, radius 5, above the x-axis, border lines included.
Domain = $\{(x,y)|\quad 0\leq y\leq\sqrt{25-x^{2}}\}$
(see image)
