Answer
$9 \pi \sqrt {30}$
Work Step by Step
The surface area of the part $z=f(x,y)$ can be computed 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.
Area of the given surface is as follows: $A(S)=\iint_{D} \sqrt {1+(-2)^2+(-5)^2} dA=\sqrt {30} \iint_{D} dA$
and, $\iint_{D} dA$ is the area of the region inside $D$.
The area of the region inside $D$ $\pi r^2=\pi \times (3)^2=9 \pi \pi$
Then, we have: $A(S)=\sqrt {30} \iint_{D} dA \\ A(S) =9 \pi \sqrt {30}$
