Answer
$\approx 22.1652$
Work Step by Step
$Surface \ Area ; A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2}\ dA$
and $\iint_{D} dA$ is the area of the region $D$
Therefore, $A(S)=\iint_{D} \sqrt {1+(2)^2+(3+8y)^2} dA $
or, $=\iint_{D} \sqrt {1+4+9+64y^2+48 y} \ dA $
Since, $1 \leq x \leq 4$ and $ 0 \leq y \leq 1$
So, $Surface \ Area ; A(S)=\int_{0}^1 \int_{1}^4 \sqrt {14+64y^2+48 y} dx dy $
Now, we will use calculator to get as follows:
$Surface \ Area ; A(S) = \int_{0}^1 \int_{1}^4 \sqrt {14+64y^2+48 y} \ dx \ dy \\=\dfrac{15}{16} (6 \sqrt {14}+\ln (9+2 \sqrt {14})-\ln 5) \\ \approx 22.1652$
