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Home Applications of the Integral Length of a Curve Theorem and Proof
 
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Theorem and Proof

We next prove a theorem which shows the connection between the length of an arc and the area of a sector of a circle. Given two points P and Q on a circle with center 0, the arc PQ is the portion of the circle traced out by a point moving from P to Q in a counterclockwise direction. The sector POQ is the region bounded by the arc PQ and the radii OP and OQ as shown in Figure 6.3.7.

06_applications_of_the_integral-144.gif

Figure 6.3.7

THEOREM

Let P and Q be two points on a circle with center 0. The area A of the sector POQ is equal to one half the radius r times the length s of the arc PQ,

A = ½rs.

DISCUSSION

The theorem is intuitively plausible because if we consider an infinitely small arc As of the circle as in Figure 6.3.8, then the corresponding sector is almost a triangle of height r and base As, so it has area ΔA ≈ ½r Δs (compared to Δs). Summing up, we expect that A = ½rs.

06_applications_of_the_integral-145.gif

Figure 6.3.8

We can derive the formula C = 2πr for the circumference of a circle using the theorem. By definition, π is the area of a circle of radius one,

06_applications_of_the_integral-146.gif

Then a circle of radius r has area

06_applications_of_the_integral-147.gif

Therefore the circumference C is given by

A = ½rC,       πr2 = ½rC,       C = 2πr.

06_applications_of_the_integral-148.gif

Figure 6.3.9

PROOF OF THEOREM

To simplify notation assume that the center 0 is at the origin, P is the point (0, r) on the x-axis, and Q is a point (x, y) which varies along the circle (Figure 6.3.9). We may take y as the independent variable and

 

use the equation 06_applications_of_the_integral-149.gif for the right half of the circle. Then A and s

depend on y. Our plan is to show that

06_applications_of_the_integral-150.gif

First, we find dx/dy:

06_applications_of_the_integral-151.gif

Using the definition of arc length,

06_applications_of_the_integral-152.gif

The triangle OQR in the figure has area ½xy, so the sector has area

06_applications_of_the_integral-153.gif

Then 06_applications_of_the_integral-154.gif=06_applications_of_the_integral-155.gif=06_applications_of_the_integral-156.gif=06_applications_of_the_integral-157.gif=

06_applications_of_the_integral-158.gif

Thus 06_applications_of_the_integral-159.gif, 06_applications_of_the_integral-160.gif,06_applications_of_the_integral-161.gif

So A and ½rs differ and only by a constant. But when y = 0, A = ½rs = 0. Therefore A = ½rs.

To prove the formula A = ½rs for arcs which are not within a single quadrant we simply cut the arc into four pieces each of which is within a single quadrant.
Home Applications of the Integral Length of a Curve Theorem and Proof

Last Update: 2010-11-25