Answer
$4 \pi $
Work Step by Step
The surface area of the part $z=f(x,y)$ can be computed as: $A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2} dx \ dy$
and, $\iint_{D} dA$ represents the projection of the surface on the $xy$-plane.
Given: $f(x,y,z)=x^{2}+y^{2}+z^{2}=4z$
Thus when $z=1$, we have:
$f(x,y,z)=x^{2}+y^{2}+z^{2}=4$
Now, $x^{2}+y^{2}=3 \implies z=\sqrt {4-x^{2}-y^{2}}$
Now, $$ A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2} dx \ dy \\=\iint_{D} \sqrt{1+[-x/(4-x^{2}-y^{2})^{1 / 2}]^{2}+[(-y)(4-x^{2}-y^{2})^{-1 / 2}]^{2}} d A \\
=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} [\sqrt{\dfrac{r^{2}}{4-r^{2}}+1} ] r d r d \theta \\
=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \dfrac{2 r}{\sqrt{4-r^{2}}} \ d r \ d \theta \\ \left.=\int_{0}^{2 \pi}\left[-2\left(4-r^{2}\right)^{1 / 2}\right]_{0}^{\sqrt{3}} \ d \theta \\
=\int_{0}^{2 \pi}(-2+4) d \theta \\ =2 \times [\theta\right]_{0}^{2 \pi} \\=2( 2\pi-0) \\=4 \pi $$
