Answer
$12 \sqrt {35}$
Work Step by Step
We write the surface area of the part $z=f(x,y)$ as follows: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ defines the projection of the surface on the xy-plane.
Our aim is to calculate the area of the given surface. $A(S)=\iint_{D} \sqrt {1+(5)^2+(3)^2} dA=\sqrt {35} \iint_{D} dA$
Now, Area of the rectangle $A_{R}=(4-1) \times (6-1) =12$
So, $A(S)=\sqrt {35} \times \iint_{D} dA=12 \sqrt {35}$
