Answer
$16$
Work Step by Step
We are given that $z=\sqrt {1-x^2}$ is the equation of the surface.
Since, the cylinders $y^2+z^2=1$ and $x^2+z^2=1$ intersect along the planes $x=y$ and $x=-y$, so, we will take the total area as 8 times each small area.
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, the area of the given surface is: $A(S)=\iint_{D} \sqrt{1+(0)^2+(-x/z)^2} dA \\= \iint_{D} \sqrt{\dfrac{x^2+z^2}{z^2}} dA \\ = \iint_{D} \dfrac{1}{\sqrt {1-x^2}} dA \\ =\int_{0}^1 \int_{-x}^x \dfrac{1}{\sqrt {1-x^2}} \ dy \ dx \\ =[-2 \sqrt {1-x^2}]_0^1 \\=2$
Therefore, we have the total area:
$ A(S)= 2 \times 8 =16$
