Area of a sector is the integral of arc length?

[24forChromium]

Area_sector = 0.5 (radius)^2 * angle
Arc length= radius * angle

Can it be said and proven that the area of a sector is the integral of the arc length? What would that even mean?

 

[haruspex]

Area_sector = 0.5 (radius)^2 * angle
Arc length= radius * angle

Can it be said and proven that the area of a sector is the integral of the arc length? What would that even mean?

I assume you mean the integral with respect to the radius.
That being so, draw a diagram which shows a sector divided into arcs of width dr. Can you see the interpretation?

 

[slider142]

Well, there is something there. To see it, we have to think in terms of the effects of small changes in r on the area of a circle.
First, note that both the area of a sector and arc length of a sector are just a linear fraction of the area of a circle and the circumference of a circle, respectively. So let us look at those.
First, consider the circumference of a circle, which is [itex]2\pi r[/itex]. If we increase r by an extremely small amount [itex]\Delta r[/itex], consider how much area is added to the circle: a thin strip that can be unrolled, which, if we ignore the slanted ends, is approximately of area [itex](2\pi r)\cdot \Delta r[/itex].
So the ratio of change in area with respect to change in radius is approximately [itex]\frac{\Delta A}{\Delta r} = 2\pi r = C[/itex], the circumference of the original circle. If we start with a circumference, then, and have a series of extremely small changes in r, we just multiply each circumference by each small change in r and sum them up to get the change in area. This is essentially the physicist's quick and dirty Riemann integral of the circumference with respect to r.
To make this formal, we have to use more rigorous language than "approximately", so we would add in a formalism such as Cauchy's limits and series program, or Robinson's infinitesimal program.