= Solution
In first-quadrant <polar coordinates>, $x=r\cos\theta$, $y=r\sin\theta$, and the <Jacobian determinant> contributes $r$. The integrand and area element together become $r^2e^{-r^2}\sin\theta\cos^2\theta\,dr\,d\theta$. Hence
$$
I=\left(\int_0^{\pi/2}\sin\theta\cos^2\theta\,d\theta\right)\left(\int_0^\infty r^2e^{-r^2}dr\right)=\frac13\frac{\sqrt\pi}{4}=\boxed{\frac{\sqrt\pi}{12}}.
$$
The radial integral follows by <integration by parts> from the <Gaussian integral>. The isolated undefined ratio at the origin does not affect this convergent <double integral>.
Back to article page
