Answer
$4 \pi $
Work Step by Step
given: $f(x,y,z)=x^{2}+y^{2}+z^{2}=4z$; and Now, when $z=1$ , so $f(x,y,z)=x^{2}+y^{2}+z^{2}=4$
and $x^{2}+y^{2}=3 \implies z=\sqrt {4-x^{2}-y^{2}}$
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.
Now, $ f_{x}=\dfrac{-x}{\sqrt {4-x^{2}-y^{2}}} , \quad f_{y}=\dfrac{-y}{\sqrt {4-x^{2}-y^{2}}}$
Our aim is to calculate the area of the given surface.
Thus, $ 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 $
or, $=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} [\sqrt{\dfrac{r^{2}}{4-r^{2}}+1} ] r d r d \theta $
or, $=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \dfrac{2 r}{\sqrt{4-r^{2}}} \ d r \ d \theta$
or, $=-2 \times \int_{0}^{2 \pi}[(4-r^{2})^{1 / 2}]_{0}^{\sqrt{3}} \ d \theta $
or, $=\int_{0}^{2 \pi}(-2+4) d \theta$
or, $=2\times [\theta]_{0}^{2 \pi} $
or, $=4 \pi $
