Circular uniform distribution using copula $\Theta$ is a uniform random variable on the interval $[0, 2\pi)$, and we have $X = \cos(\Theta)$ and $Y = \sin(\Theta)$. 
Suppose $(x,y)$ is a point inside the unit circle. Then, the arc length that is shown in white in Fig. 3.2.a on page 56 of An Introduction to Copulas (Nelson, 2006) is $2\pi H(x,y)$, where $H(x,y)$ is a joint distribution function of RVs $X$ and $Y$.
Could you please help me to understand why the arc length is $2\pi H(x,y)$?
[EDIT] 

 A: $H$ refers to a bivariate distribution function.  In general this means
$$H(x,y) = \Pr(X \le x,\ Y \le y).$$
In the $(x,y)$ plane, such a region is a rectangle bounded at the top by $y$, at the right by $x$, and infinite to the left and down, as shown by the gray regions in both figures.
These particular random variables are confined to the unit circle because $X^2 + Y^2 = \cos^2(\Theta) + \sin^2(\Theta) = 1$.  Moreover, they are uniformly distributed on that circle because $\Theta$ has a uniform distribution on the interval $[0, 2\pi)$.  Uniform means the chance that $(X,Y)$ lies along any particular (measurable) part of the circle is directly proportional to the length of that part.  Since the total length is the perimeter $2\pi$, the constant of proportionality must be $1/(2\pi)$.
Consequently, $H(x,y)$ must be proportional to the total length of any parts of the circle lying within the semi-infinite rectangle determined by $(x,y)$.  The figures show two examples, one where $(x,y)$ is in the disk bounded by the circle and another where $(x,y)$ is outside that disk.  In either case, though, $H(x,y)$ must equal $1/(2\pi)$ times this total arc length.  Multiplying this equality yields the statement in the question: $2\pi H(x,y)$ is the total length of arcs to the left of and below the point $(x,y)$.
