Answer
$$4 \pi $$
Work Step by Step
We are given that $f(x,y,z)=x^{2}+y^{2}+z^{2}=4z$; and $f(x,y,z)=x^{2}+y^{2}+z^{2}=4$
When $z=1$, we have:
$x^{2}+y^{2}=3$
$z=\sqrt {4-x^{2}-y^{2}}$
The surface area of the part $z=f(x,y)$ can be written as: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ is the projection of the surface on the xy-plane.
Now, $ f_{x}=\dfrac{-x}{\sqrt {4-x^{2}-y^{2}}}, \quad f_{y}=\dfrac{-y}{\sqrt {4-x^{2}-y^{2}}}$
Thus, $ A(S)=\iint_{D} \sqrt{[\dfrac{(-x)}{(4-x^{2}-y^{2})^{1 / 2}}]^{2}+[(-y)(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 d r d \theta \\
=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \sqrt{\dfrac{r^{2}+4-r^{2}}{4-r^{2}}} \ r \ dr \ d\theta \\
=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \dfrac{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 $
