## Calculus 8th Edition

$y=2+\sqrt{4-x^2}$ With domain: $-2\leq x \leq2$
We need to solve for $y$: $x^2+(y-2)^2=4$ $(y-2)^2=4-x^2$ $y-2=\pm\sqrt{4-x^2}$ $y=2\pm\sqrt{4-x^2}$ Since we need the top half of the circle, we use: $y=2+\sqrt{4-x^2}$ With domain: $-2\leq x \leq2$