Answer
$\dfrac{13 \sqrt {2}}{12}$
Work Step by Step
We have $Surface \ Area=\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, $Surface \ Area=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA =\int_0^1 \int_{0}^{x} \iint_{D} \sqrt {2+16x^2} \ dy \ dx = \int_0^1 x \sqrt {2+16x^2} \ dx$
Set $2+16x^2 =a $ and $ 32 x dx = da $
Now, $Surface \ Area = \dfrac{1}{ 32} \times \int_{2}^{18} a^{1/2} da=\dfrac{1}{ 32} \times \dfrac{2 }{3} [a^{3/2}]_{2}^{18}=\dfrac{13 \sqrt {2}}{12}$
