Answer
$$\dfrac{\pi}{6} (101 \sqrt {101}-1) $$
Work Step by Step
The surface area of the part $z=f(x,y)$ can be computed as: $A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2} dx \ dy$
and, $\iint_{D} dA$ represents the projection of the surface on the $xy$-plane.
The area of the given surface is:
$A(S)=\iint_{D} \sqrt{1+(2x)^2+(2z)^2} dA $
or, $=\sqrt{1+4(x^2+z^2)} \iint_{D} dA $
Now, apply the polar co-ordinates because of the part $x^2+z^2$
Thus, $$A(S)=\int_{0}^{2 \pi} \int_{0}^5 \sqrt {1+4r^2} r dr d \theta $$
or, $$= \int_{0}^{2 \pi} d \theta \times \int_{0}^5 \sqrt {1+4r^2} r dr $$
Let us substitute that $a=1+4r^2 $ and $ 8r dr= da$
$$=[ \theta]_0^{2 \pi} \int_{0}^{5} \sqrt a \dfrac{da}{8} \\= \dfrac{\pi}{4} \times [\dfrac{2}{3} (4r^2+1)^{3/2}]_0^5 $$
and $$A(S)= \dfrac{\pi}{6} (101 \sqrt {101}-1) $$
