Answer
16
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: the equation of the surface is: $z=\sqrt {1-x^2}$
Since the cylinders $y^2+z^2=1$ and $x^2+z^2=1$ intersect along the planes $x=y$ and $x=-y$, thus, we need to multiply the calculated area by 8 to get the total area.
The area of the given surface is:
$$A(S)=\iint_{D} \sqrt{1+(0)^2+(\dfrac{-x^2}{z^2})} dA \\= \iint_{D} \sqrt{\dfrac{x^2+z^2}{z^2}} dA \\ = \iint_{D} \dfrac{1}{\sqrt {1-x^2}} dA \\ =\int_{0}^1 \int_{-x}^x \dfrac{1}{\sqrt {1-x^2}} \ dy \ dx \\ =[-2 \sqrt {1-x^2}]_0^1 \\=2$$
Thus, the total area is equal to: $ A=(2) (8) =16$
