## Calculus: Early Transcendentals 8th Edition

The area of the surface $z$ is $$A(S) =4 \pi .$$
$$f(x,y,z)=x^{2}+y^{2}+z^{2}=4 ,$$ when $z=1$ we get $$x^{2}+y^{2}=3.$$ so, $$D=\left\{ (x, y) | x^{2}+y^{2} \leq 3\right\}.$$ and $$z=\sqrt {4-x^{2}-y^{2}},$$ $$f_{x}=\frac{-x}{\sqrt {4-x^{2}-y^{2}}} , \quad f_{y}=\frac{-y}{\sqrt {4-x^{2}-y^{2}}}$$ Thus \begin{aligned} A(S) &=\\ &\iint_{D} \sqrt{\left[(-x)\left(4-x^{2}-y^{2}\right)^{-1 / 2}\right]^{2}+\left[(-y)\left(4-x^{2}-y^{2}\right)^{-1 / 2}\right]^{2}+1} d A \\ &=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \sqrt{\frac{r^{2}}{4-r^{2}}+1} r d r d \theta \\ &=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \sqrt{\frac{r^{2}+4-r^{2}}{4-r^{2}}} r d r d \theta \\ &=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \frac{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 \end{aligned} Thus, the area of the surface $z$ is $$A(S) =4 \pi .$$