Answer
$4.1073$
Work Step by Step
The surface area of the part $z=f(x,y)$ can be defined 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: $A(S)=\iint_{D} \sqrt {1+[-2x \sin (x^2+y^2]^2+[-2y \sin (x^2+y^2]^2} dA= \iint_{D} \sqrt {1+4(x^2+y^2)] \sin^2 (x^2+y^2)} dA$
Next, we need to use the polar co-ordinates because of 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 \times \int_0^{1} \sqrt {1+4r^2 \sin^2(r^2)} \ r dr $
Now, by using a calculator, we have:
$A(S)=2 \pi \times \int_0^{1} \sqrt {1+4r^2 \sin^2(r^2)} \ r dr \approx 4.1073$
