Answer
$3 \sqrt {11}+2 \sinh^{-1} (\dfrac{3}{\sqrt 2})$
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.
Since, $-2 \leq x \leq 1$ and $-1 \leq y \leq 1$
Our aim is to calculate the area of the given surface.
$$A(S)=\iint_{D} \sqrt{1+(2x+1)^{2}+(1)^{2}} d A \\ =\iint_{D} \sqrt{1+4x^2+4x+1+1} d A \\ =\iint_{D} \sqrt{4x^2+4x+3} \space d A $$
Next we will use calculator to compute the result as follows:
$$A(S)=\iint_{D} \sqrt{4x^2+4x+3} \space d A \\= \int_{-1}^{1} \int_{-2}^{1} \sqrt{4x^2+4x+3} \space dx \space dy \\=3 \sqrt {11}+2 \sinh^{-1} (\dfrac{3}{\sqrt 2})$$
