Answer
$12 \sin^{-1} (\dfrac{2}{3})$
Work Step by Step
The surface area of the surface 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+(\dfrac{-y}{\sqrt {9-y^2}})^2} dA $
and, $\iint_{D} dA$ is the area of the region inside $D$.
Therefore, $A(S)=\int_{0}^{4} \int_{0}^{2} \sqrt{1+(0)^2+(\dfrac{-y}{\sqrt {9-y^2}})^2} dy dx \\ =\int_{0}^{4} \int_{0}^{2} \sqrt {\dfrac{9-y^2+y^2}{9-y^2}} dy dx \\ =\int_{0}^{4} \int_{0}^{2}[\dfrac{3}{9-y^2}] dy dx\\ = \int_0^4 3 \sin^{-1} \dfrac{2}{3} dx \\ =[ 3x \sin^{-1} (\dfrac{2}{3}]_0^4 \\ = 12 \sin^{-1} (\dfrac{2}{3})$
