Answer
$12 \sqrt {35}$
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: $A(S)=\iint_{D} \sqrt {1+(5)^2+(3)^2} dA=\sqrt {35} \iint_{D} dA$
and, $\iint_{D} dA$ is the area of the region inside $D$.
Area of the rectangle $=(4-1) \times (6-1) =12$
Thus, $A(S)=\sqrt {35} \iint_{D} dA=12 \sqrt {35}$
