Answer
$16$
Work Step by Step
$Surface \ Area =\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2} dA$
where $\iint_{D} dA$ is the area of the region $D$
The cylinders $y^2+z^2=1$ and $x^2+z^2=1$ intersect along the planes $x=y$ and $x=-y$ and the equation of the surface is $z=\sqrt{1-x^2}$
Therefore,
$Surface \ Area =\iint_{D} \sqrt {1+(0)^2+(\dfrac{-x}{z})^2} dA =\iint_{D} \sqrt {1+\dfrac{x^2}{z^2}} dA \\=\iint_{D} \dfrac{1}{\sqrt {1-x^2}} \ dA $
Since, $0 \leq x \leq 1$ and $ -x \leq y \leq x$
So,
$Surface \ Area =\int_{0}^{1} \int_{-x}^{x} \dfrac{1}{\sqrt {1-x^2}} \ dy \ dx \\=\int_0^1 \dfrac{2x}{\sqrt {1-x^2}} \\=[-2 \sqrt {1-x^2}]_0^1 \\=2$
So, the total area becomes:
$Surface \ Area =(2)(8)=16$
