Answer
$3 \sqrt {14}$
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.
$A(S)=\iint_{D} \sqrt {1+(-3)^2+(-2)^2} dA=\sqrt {14} \iint_{D} dA$
Now, the area of the triangle is: $=\dfrac{1}{2} b h =\dfrac{2 \times 3}{2}=3$
So, $A(S)=\sqrt {14} \iint_{D} dA=3 \sqrt {14}$
