Answer
$$\approx 22.1652$$
Work Step by Step
We have $$A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2} \ dA \\=\iint_{D} \sqrt {1+(2)^2+(3+8y)^2} dA \\=\iint_{D} \sqrt {1+4+9+64y^2+48 y} \ dA $$
Since, $1 \leq x \leq 4$ and $ 0 \leq y \leq 1$
and$$A(S)=\int_{0}^1 \int_{1}^4 \sqrt {14+64y^2+48 y} dx dy $$
Now, by using a calculator, we get
$$A(S) = \int_{0}^1 \int_{1}^4 \sqrt {14+64y^2+48 y} dx dy \\=\dfrac{15}{16} (6 \sqrt {14}+\ln (9+2 \sqrt {14})-\ln 5) \\ \approx 22.1652$$
