Answer
$3.3213$
Work Step by Step
The surface area of the part $z=f(x,y)$ can be wriiten 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+(2xy^2)^{2}+(2x^2y)^{2}} d A \\ =\iint_{D} \sqrt{1+4x^2y^4+4x^4y^2} d A \\=\iint_{D} \sqrt{1+4x^2y^2(x^2+y^2)} dA $
We can use the polar co-ordinates because of the part $x^2+y^2$
Therefore, by using a calculator, we have:
$A(S)=\int_{0}^{2 \pi} \int_{0}^{1} \sqrt{1+4r^4 \cos^2 \theta \sin^2 \theta (r^2)} r dr d \theta \approx 3.3213$
