Answer
$4 \pi$
Work Step by Step
$f(x,y,z)=x^{2}+y^{2}+z^{2}=4z \implies f(x,y,z)=x^{2}+y^{2}+z^{2}=4$ ; when $z=1$
and $x^{2}+y^{2}=3 \implies z=\sqrt {4-x^{2}-y^{2}}$
So $ f_{x}=\dfrac{-x}{\sqrt {4-x^{2}-y^{2}}} , \quad f_{y}=\dfrac{-y}{\sqrt {4-x^{2}-y^{2}}}$
We write the surface area of the part $z=f(x,y)$ as follows: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ defines the projection of the surface on the xy-plane.
Our aim is to calculate the area of the given surface.
$ A(S)=\iint_{D} \sqrt{[\dfrac{(-x)}{(4-x^{2}-y^{2})^{1 / 2}}]^{2}+[(-y)\times (4-x^{2}-y^{2})^{-1 / 2}]^{2}+1} d A \\
=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} [\sqrt{\dfrac{r^{2}}{4-r^{2}}+1} ] r \space d r \space d \theta \\
=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \dfrac{2 r}{\sqrt{4-r^{2}}} \space d r \ d \theta \\ \left.=\int_{0}^{2 \pi}\left[-2\left(4-r^{2}\right)^{1 / 2}\right]_{0}^{\sqrt{3}} \space d \theta \\
=\int_{0}^{2 \pi}(-2+4) d \theta \\=2 \times [\theta\right]_{0}^{2 \pi} \\=4 \pi $
