Answer
$\dfrac{\pi}{6} (101 \sqrt {101}-1) $
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+(2x)^2+(2z)^2} dA \\=\sqrt{1+4(x^2+z^2)} \iint_{D} dA $
We can use the polar co-ordinates because of the part $x^2+z^2$
Therefore, we have:
$A(S)=\int_{0}^{2 \pi} \int_{0}^5 \sqrt {1+4r^2} r dr d \theta \\= \int_{0}^{2 \pi} d \theta \times \int_{0}^5 \sqrt {1+4r^2} r dr $
Substitute $1+4r^2 = a \implies da= 8r dr$
$A(S)=[ \theta]_0^{2 \pi} \int_{0}^{5} \sqrt a \dfrac{da}{8} \\= \dfrac{\pi}{4} [\dfrac{2}{3} (4r^2+1)^{3/2}]_0^5 \\= \dfrac{\pi}{6} (101 \sqrt {101}-1) $
