Answer
$$\approx 3.3213$$
Work Step by Step
We write the surface area of the part $z=f(x,y)$ as follows: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ defines the projection of the surface on the xy-plane.
Our aim is to calculate the area of the given surface.
$$A(S)=\iint_{D} \sqrt{1+(2xy^2)^{2}+(2x^2y)^{2}} \space d A \\ =\iint_{D} \sqrt{1+4x^2y^4+4x^4y^2} \space d A \\=\iint_{D} \sqrt{1+4x^2y^2(x^2+y^2)} \space dA $$
Next, use the polar co-ordinates for of the part $x^2+y^2$
Now, we will use calculator to compute the result as follows:
$$A(S)=\int_{0}^{2 \pi} \int_{0}^{1} \sqrt{1+4r^4 \cos^2 \theta \sin^2 \theta \times (r^2)} \times (r) dr d \theta \approx 3.3213$$
