Answer
$16$
Work Step by Step
The cylinders $y^2+z^2=1$ and $x^2+z^2=1$ intersect along the planes $x=y$ and $x=-y$
We have $A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2} dA$; $\iint_{D} dA$ is the area of the region $D$
and the equation of the surface is $z=\sqrt{1-x^2}$
Therefore, $A(S)=\iint_{D} \sqrt {1+(0)^2+(\dfrac{-x}{z})^2} dA =\iint_{D} \sqrt {1+\dfrac{x^2}{z^2}} dA$
or, $=\iint_{D} \dfrac{1}{\sqrt {1-x^2}} dA $
Since, $0 \leq x \leq 1$ and $ -x \leq y \leq x$
So, $A(S)=\int_{0}^{x} \int_{-x}^{x} \dfrac{1}{\sqrt {1-x^2}} dy dx =\int_(0)^1 \dfrac{2x}{\sqrt {1-x^2}}=[-2 \sqrt {1-x^2}]_0^1=2$
By using a calculator, we get
Thus, the total area is: $S_{A}=(2)(8)=16$
