Answer
$25 \pi \sqrt {14}$
Work Step by Step
The surface area of the part $z=f(x,y)$ 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: $A(S)=\iint_{D} \sqrt {1+(-3)^2+(-2)^2} dA=\sqrt {14} \iint_{D} dA$
and, $\iint_{D} dA$ is the area of the region inside $D$.
The area of the region inside $D$ $\pi r^2=25 \pi$
Thus, $A(S)=\sqrt {14} \iint_{D} dA=25 \pi \sqrt {14}$
