Answer
$3 \sqrt {11}+2 \sinh^{-1} (\dfrac{3}{\sqrt 2})$
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+(2x+1)^{2}+(1)^{2}} d A \\ =\iint_{D} \sqrt{1+4x^2+4x+1+1} d A \\ =\iint_{D} \sqrt{4x^2+4x+3} d A $
Since, $-2 \leq x \leq 1$ and $-1 \leq y \leq 1$
Therefore, by using a calculator, we have:
$A(S)=\iint_{D} \sqrt{4x^2+4x+3} d A \\= \int_{-1}^{1} \int_{-2}^{1} \sqrt{4x^2+4x+3} \ dx \ dy \\=3 \sqrt {11}+2 \sinh^{-1} (\dfrac{3}{\sqrt 2})$
