Answer
$A(S) = 2\pi \int_0^2 r \sqrt {4r^2 e^{-2r^2}+1} dr \approx 13.9783$
Work Step by Step
The parameterization for the given surface is: $r=\lt r \cos \theta, r \sin \theta, e^{-r^2} \gt$
We have $A(S)=\iint_{D} |r_r \times r_{\theta}| dA$
and $\iint_{D} dA$ is the area of the region $D$
and $|r_r \times r_{\theta} dA=r \sqrt {4r^4e^{-2r^2}+r^2}=r \sqrt {4r^2 e^{-2r^2}+1}$
Now, $A(S) \iint_{D} |r_r \times r_{\theta}| dA=\iint_{D} r \sqrt {4r^2 e^{-2r^2}+1} d \theta dr$
or, $=\int_0^2 \int_0^{2 \pi} r \sqrt {4r^2 e^{-2r^2}+1}d \theta dr$
or, $=2\pi \int_0^2 r \sqrt {4r^2 e^{-2r^2}+1} dr$
Therefore, by using a calculator, we get
$A(S) = 2\pi \int_0^2 r \sqrt {4r^2 e^{-2r^2}+1} dr \approx 13.9783$
