Answer
$4 \pi $
Work Step by Step
The surface Area of the part $z=f(x,y)$ can be computed 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.
Given : $f(x,y,z)=x^{2}+y^{2}+z^{2}=4z$; and when $z=1$, then, we have: $f(x,y,z)=x^{2}+y^{2}+z^{2}=4$ ;
Now, $x^{2}+y^{2}=3$ and $z=\sqrt {4-x^{2}-y^{2}}$
We have: $ 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}+[\dfrac{(-y)}{(4-x^{2}-y^{2})^{1 / 2}}]^{2}+1} d A $
We need to apply the polar co-ordinates for the part
$x^2+y^2+z^2$
$A(S)=\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 \\ =\int_{0}^{2 \pi}\left[-2(4-r^{2}\right)^{1 / 2}]_{0}^{\sqrt{3}} \ d \theta \\
=\int_{0}^{2 \pi}(-2+4) d \theta\\=2 [\theta]_{0}^{2 \pi} \\=4 \pi $
