Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 18

The integrand is continuous on the compact rectangle, so Fubini's theorem allows integrating in $y$ first. For $x\ne 0$, ${\int }_0^{1}x{e}^{xy}dy=e^x-1$; the same formula holds by continuity at $x=0$. Hence

The domain of integration is the upper half of the ellipse $x^2+4y^2\le 1$. Reversing the order makes $-1\le x\le 1$ and $0\le y\le \frac{1}{2}\sqrt{1-x^2}$. ThusThe inner antiderivative is $y^2/2$, so the square of the upper limit removes the square root.

Use cylindrical coordinates $x=r\cos \theta$, $y=r\sin \theta$. The upper plane gives $z=4-r(\cos \theta +\sin \theta )>0$ throughout the quarter disk, and the Jacobian determinant contributes $r$. Thereforeso