= Solution
The <gradient> is $\nabla f=(2x-4x^3,-2y-4y^3)$, so the real <stationary points> are $(0,0)$ and $(\pm1/\sqrt2,0)$. The <Hessian matrix> is diagonal, with entries $2-12x^2$ and $-2-12y^2$. At the origin it has opposite-sign <eigenvalues>, giving a <saddle point>. At each of the other two <stationary points> it is negative definite, giving \b[strict maxima of value $1/4$]. These are also the global maxima, since
$$
f(x,y)=\frac14-\left(x^2-\frac12\right)^2-y^2-y^4\leq\frac14.
$$
There are \b[no local minima], and the <function> is unbounded below. The zero <contour> has two lobes touching at the saddle; positive <contours> are closed curves around the maxima.
Under the stated <change of variables>,
$$
f=\frac12-X^2-Y^2,\qquad x>0\Longleftrightarrow X>-\frac12,\qquad y>0\Longleftrightarrow Y>\frac12.
$$
Inside the resulting circular cap the constraint $X>-1/2$ is automatic. Comparing its <circle> and horizontal boundary gives
$$
\boxed{r=\frac1{\sqrt2},\qquad a=\frac12,\qquad b=\frac12.}
$$
Here $a$ is the positive half-width used in the source figure, independent of the pulse-height notation in the preceding question. The <Jacobian determinant> is $\partial(X,Y)/\partial(x,y)=4xy>0$ in the chosen quadrant, so the integral is one-quarter of the <circular segment> area:
$$
\int_Rxy\,dx\,dy=\frac14\left[r^2\arccos\frac br-b\sqrt{r^2-b^2}\right]
=\frac14\left(\frac\pi8-\frac14\right)
=\boxed{\frac{\pi-2}{32}.}
$$
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2018-quartic-contours.png]
{title=The <contours>, maxima and saddle, and the circular cap produced by the <change of variables>}
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