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 wriiten 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+(2)^{2}+(3+8y)^{2}} d A \\ =\iint_{D} \sqrt{1+4+(9+64y^2+48 y)} d A \\ =\iint_{D} \sqrt{14+64y^2+48 y)} d A $
Therefore, by using a calculator, we have:
$A(S)=\iint_{D} \sqrt{14+64y^2+48 y)} d A \\= \int_{0}^{1} \int_{1}^{4} \sqrt{14+64y^2+48 y)} \ dx \ dy \\=\dfrac{15}{16} [6 \sqrt {14}+\ln [\dfrac{1}{5}(9+2 \sqrt {14})]]$
