Answer
$\dfrac{13 \sqrt {26}-5 \sqrt {10}}{12}$
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+(3)^2+(4y)^2 } dA= \iint_{D} \sqrt {10+16y^2} dA $
and $\iint_{D} dA$ is the area of the region $D$
Therefore, $A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA=\int_0^1 \int_{0}^{ 2y} \sqrt {10+16y^2} dy dx=(1/16) \int_0^1 32y \sqrt {10+16y^2} dy$
Substitute $a=10+6y^2 $ and $da= 32 y dy$
Now, $A(S)= \dfrac{1}{16} \int_{10}^{26} a^{1/2} da=\dfrac{1}{16} [\dfrac{2 a^{3/2}}{3}]_{10}^{26}$
This implies that,
$A(S)=\dfrac{13 \sqrt {26}-5 \sqrt {10}}{12}$
