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Take a circle of radius $R$, a vertex $A$ on its circumference, and two chords from $A$ symmetric about the diameter through $A$. If their angle is $0\le \theta \le \pi$, choose the arc between their other endpoints that does not contain $A$. The area bounded by the two chords and that arc is $R^2(\theta +\sin \theta )$. To calculate it, set $A=(0,R)$ and the other endpoints to $(\pm R\sin \theta ,-R\cos \theta )$. Their triangle has area $R^2\sin \theta (1+\cos \theta )$, and the adjoining circular segment, of central angle $2\theta$, has area $R^2(\theta -\sin \theta \cos \theta )$. The sum gives the formula, including the limiting zero-area and full-disk cases.

A circular segment is the part of a disk bounded by a chord and one of the two circular arcs joining its endpoints. For radius $R$ and the chosen arc's central angle $0\le \varphi \le 2\pi$ in radians, its area is $R^2(\varphi -\sin \varphi )/2$. Subtract the signed central triangle of area $R^2\sin \varphi /2$ from the circular sector of area $R^2\varphi /2$. Angles below $\pi$ give the smaller segment and angles above $\pi$ the larger one.

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:

Let $A$ be the top vertex, $O$ the center, and $B,C$ the two other points on the circle. The diameter bisects the angle at $A$, so the two right triangles obtained by drawing the diameter to the opposite point show that $B,C$ have coordinates $(\mp \sin \theta ,-\cos \theta )$ when $O=(0,0)$ and $A=(0,1)$. The chord $BC$ cuts off a lower circular segment of central angle $2\theta$. The triangle $ABC$ has area $\sin \theta (1+\cos \theta )$, while the segment has area $\theta -\sin \theta \cos \theta$, by subtracting the central triangle from its circular sector. Adding giveswith $\theta$ measured in radians. This is the unit-radius case of the area enclosed by two chords and a circular arc.