= Solution
The integration limits describe the first-quadrant quarter of the unit disk: $x\ge0$, $y\ge0$, $x^2+y^2\le1$. Its circular boundary and the axes are shown in the right-hand panel below; the left panel gives the bounded region used in the preceding calculation.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015-planar-integration-regions.png]
{title=The parabolic strip and quarter-disk integration regions}
Use <polar coordinates> $x=r\cos\theta$, $y=r\sin\theta$ with $0\le r\le1$ and $0\le\theta\le\pi/2$. The <Jacobian determinant> gives $dx\,dy=r\,dr\,d\theta$, so
$$
I=\int_0^{\pi/2}\int_0^1 e^{-r^2}r\,dr\,d\theta=\frac\pi2\left[-\frac12e^{-r^2}\right]_0^1.
$$
Thus $\boxed{I=\tfrac\pi4(1-e^{-1})}$. Including the radial Jacobian is essential; the exponential becomes radial, but the area element does not remain $dr\,d\theta$.
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