Answer
$\dfrac{\sqrt {21}}{2}+\dfrac{17}{4} \ln (\dfrac{2 +\sqrt {21}}{\sqrt {17}})$
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+(4)^2+(2z)^2} dA =\iint_{D} \sqrt {4z^2+17} dA $
and $\iint_{D} dA$ is the area of the region $D$
Therefore, $A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA=\iint_{D} \sqrt {4z^2+17} dA =\int_0^{1} \int_{0}^{1} \sqrt {4z^2+17} dx dz=\sqrt {17} \int_0^1 \sqrt {\dfrac{4z^2}{17}+1} dz$
Substitute $a=\dfrac{2z }{\sqrt {17}} $ and $da=\dfrac{\sqrt {17} da}{ 2}$
Now, $A(S)= \dfrac{17}{2} \int_0^{2/\sqrt {17}} \sqrt {a^2+1} da$
This implies that,
$A(S)=\dfrac{\sqrt {21}}{2}+\dfrac{17}{4} \ln (\dfrac{2 +\sqrt {21}}{\sqrt {17}})$
