Yes, this is essentially what the author derived (by means of calculus rather than geometric argument but the result is unsurprisingly the same). The 2D analog is easy to grok: to compute the area of a polygon, find the sum of the signed areas of each of the trapezoids formed by an edge and its projection on the x-axis. Turns out the negative areas of the right-to-left trapezoids cancel precisely out any excess area of the left-to-right trapezoids (or in the case of edges below the x-axis, add precisely the "missing" area).