Answer
$$\approx 3.3213$$
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{(2xy^2)^{2}+(2x^2y)^{2}+1} d A \\ =\iint_{D} \sqrt{4x^2y^4+4x^4y^2+1} d A \\=\iint_{D} \sqrt{4x^2y^2(x^2+y^2)+1} dA $
Next, we will apply the polar co-ordinates for the part $x^2+y^2$
Now, we will use a calculator to find the area of the given surface.
$$A(S)=\int_{0}^{2 \pi} \int_{0}^{1} \sqrt{1+4r^4 \cos^2 \theta \sin^2 \theta (r^2)} r dr d \theta \approx 3.3213$$
