Answer
The area of the surface $z$ is
$$
A(S) =4 \pi .
$$
Work Step by Step
$$
f(x,y,z)=x^{2}+y^{2}+z^{2}=4 ,
$$
when $z=1$
we get
$$
x^{2}+y^{2}=3.
$$
so,
$$
D=\left\{ (x, y) | x^{2}+y^{2} \leq 3\right\}.
$$
and
$$
z=\sqrt {4-x^{2}-y^{2}},
$$
$$
f_{x}=\frac{-x}{\sqrt {4-x^{2}-y^{2}}} , \quad f_{y}=\frac{-y}{\sqrt {4-x^{2}-y^{2}}}
$$
Thus
$$
\begin{aligned} A(S) &=\\
&\iint_{D} \sqrt{\left[(-x)\left(4-x^{2}-y^{2}\right)^{-1 / 2}\right]^{2}+\left[(-y)\left(4-x^{2}-y^{2}\right)^{-1 / 2}\right]^{2}+1} d A \\
&=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \sqrt{\frac{r^{2}}{4-r^{2}}+1} r d r d \theta \\
&=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \sqrt{\frac{r^{2}+4-r^{2}}{4-r^{2}}} r d r d \theta \\
&=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \frac{2 r}{\sqrt{4-r^{2}}} d r d \theta \\ &\left.=\int_{0}^{2 \pi}\left[-2\left(4-r^{2}\right)^{1 / 2}\right]_{r=0}^{r=\sqrt{3}} d \theta \\
=\int_{0}^{2 \pi}(-2+4) d \theta=2 \theta\right]_{0}^{2 \pi}=4 \pi \end{aligned}
$$
Thus, the area of the surface $z$ is
$$
A(S) =4 \pi .
$$
