Answer
$$\dfrac{15}{16} [6 \sqrt {14}+\ln [\dfrac{1}{5}(9+2 \sqrt {14})]]$$
Work Step by Step
The surface area of the part $z=f(x,y)$ can be computed as: $A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2} dx \ dy$
and, $\iint_{D} dA$ represents the projection of the surface on the $xy$-plane.
The area of the given surface is: $A(S)=\iint_{D} \sqrt{(2)^{2}+(3+8y)^{2}+1} d A \\ =\iint_{D} \sqrt{4+(9+64y^2+48 y)+1} d A \\ =\iint_{D} \sqrt{14+64y^2+48 y} d A $
Now, we will use a calculator to find the area of the given surface.
$$A(S)=\iint_{D} \sqrt{14+64y^2+48 y)} d A \\= \int_{0}^{1} \int_{1}^{4} \sqrt{14+64y^2+48 y)} \ dx \ dy \\=\dfrac{15}{16} [6 \sqrt {14}+\ln [\dfrac{1}{5}(9+2 \sqrt {14})]]$$
