Answer
$$\dfrac{13 \sqrt {2}}{12}$$
Work Step by Step
We have $A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA= \iint_{D} \sqrt {1+(4x)^2+(1)^2 } dA= \iint_{D} \sqrt {2+16x^2} dA $
and $\iint_{D} dA$ is the area of the region $D$
Therefore, $A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA=\int_0^1 \int_{0}^{x} \iint_{D} \sqrt {2+16x^2} dA dy dx= \int_0^1 x \sqrt {2+16x^2} \ dx$
Substitute $a=2+16x^2 $ and $da= 32 x dx$
Now, $A(S)= \dfrac{1}{ 32} \int_{2}^{18} a^{1/2} da=\dfrac{1}{ 32} [\dfrac{2 a^{3/2}}{3}]_{2}^{18}$
This implies that, $A(S)=\dfrac{13 \sqrt {2}}{12}$
