 Solution

The gradient is $\nabla f=(2x-4x^3,-2y-4y^3)$, so the real stationary points are $(0,0)$ and $(\pm 1/\sqrt{2},0)$. 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 strict maxima of value $1/4$. These are also the global maxima, sinceThere are 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,Inside the resulting circular cap the constraint $X>-1/2$ is automatic. Comparing its circle and horizontal boundary givesHere $a$ 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: