Answer
$\approx 4.1073$
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.
Now, $A(S)=\iint_{D} \sqrt {1+[-2x \sin (x^2+y^2]^2+[-2y \sin (x^2+y^2]^2} dA $ and $ A(S)= \iint_{D} \sqrt {1+4(x^2+y^2)] \sin^2 (x^2+y^2)} dA$
Apply the polar co-ordinates for the part $x^2+y^2$.
$$A(S)=\iint_{D} \sqrt {1+4(x^2+y^2)] \sin^2 (x^2+y^2)} dA \\=\int_{0}^{2\pi} \int_{0}^{1} \sqrt {1+4r^2 \sin^2 r^2 (\cos^2 \theta +\sin^2 \theta ) } \ r dr d \theta \\= 2 \pi \int_0^{1} \sqrt {1+4r^2 \sin^2(r^2)} \ r \space dr $$
Now, we will use a calculator to find the area of the given surface.
$$A(S)=2 \pi \int_0^{1} \sqrt {1+4r^2 \sin^2(r^2)} \ r dr \\ \approx 4.1073$$
