Answer
$\approx 13.9783$
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: $A(S)=\iint_{D} \sqrt {1+(-2xe^{-x^2-y^2})^2+(-2ye^{-x^2-y^2})^2} dA= \iint_{D} \sqrt {4(x^2+y^2)e^{-2(x^2+y^2)}} dA$
Next, we need to use the polar co-ordinates because of the part $x^2+y^2$
$A(S)=\iint_{D} r \sqrt {1+4r^2 e^{-2r^2}} dA \\=\int_{0}^{2\pi} \int_{0}^{2} r \sqrt {1+4r^2 e^{-2r^2}} \ r dr d \theta \\= 2 \pi \times \int_0^{2} r \sqrt {1+4r^2 e^{-2r^2}} \ r dr $
Now, by using a calculator, we have:
$A(S)=2 \pi \int_0^{2} r \sqrt {1+4r^2 e^{-2r^2}} \ r dr \approx 13.9783$
