Answer
$\dfrac{13 \sqrt{26} - 5 \sqrt {10}}{12}$
Work Step by Step
The surface area of the part $z=f(x,y)$ can be written as: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ is the projection of the surface on the xy-plane.
Now, the area of the given surface is: $A(S)=\iint_{D} \sqrt{1+(3)^2+(4y)^2} dA $ \\
and, $\iint_{D} dA$ is the area of the region inside $D$.
We have: $
D=\left\{ (x, y) | 0 \leq x\leq 2y, \ 0 \leq y \leq1\right\}
$
Therefore, $A(S)=\int_{0}^{1} \int_{0}^{2y} \sqrt {10+16y^2} d x \ dy \\= \int_{0}^1 2y \sqrt {10+16y^2} dx dy $
Let us suppose that $10+16y^2 =a \implies da = 32 y dyr$
Thus, $A(S)=(1/16) \int_{10}^{26} \sqrt a da\\= [\dfrac{a \sqrt a }{24}]_{10}^{26} \\=\dfrac{26 \sqrt{26}}{24}-\dfrac{10 \sqrt{10}}{24} \\= \dfrac{13 \sqrt{26} - 5 \sqrt {10}}{12}$
