Problem 12: The Isoperimetric Inequality and other Geometric Applications

Suppose that C is a closed curve in the plane such that it is of class C²; that is, both the first and second derivatives of any parameter functions for C exist and are continuous.

Let's review curvature. Just as tangent lines can approximate a curve, so can circles. Similarly, at a given point p there is exactly one circle that best approximates the curve at that point: the circle intersects the curve at p, and has the same first and second derivates as the curve at p. The radius of curvature is the radius of this circle, called the osculating circle. In the picture below, the red circle best approximates the curve. As you can see, the larger the radius the less "curvy" C is. The curvature is then .

Let s be the arc length of C and let denote angle between the tangent to C and, say, the positive x-axis (any fixed direction will do). The curvature of C can also be shown to equal , and the total absolute curvature is defined by the following line integral:


		⁽¹⁾

Let

denote the number of tangents to the curve C that are parallel to a given direction

. In the figure above, notice that

= 6. It follows that

appears

times for all

; in other words, every point of C can be defined by a

. As a result, the integral in equation (1) can be rewritten as:

		(2)

Here is a clever observation. Suppose that G is a line that intersects our closed planar curve C in the above picture. At most, G intersects C in six points, as in the figure below.

Observe that between any two of these interesction points, the Mean Value Theorem asserts that there is a point such that the tangent line at that point is parallel to the line G. In our example, there are six points of intersection, namely P1, P2, ..., P6, and there are six differentiable arcs, namely P1P2, P2P3, P3P5, P4P5, P4P6, and P6P1. Notice that the arc lengths do not necessarily move consecutively from each point. In general, if a line interects a closed planar curve at n points, then there are at least n points on C whose tangent lines are parallel to G.

Suppose K is a bounded convex set that has area F and perimter L. Let G, G' be two lines that intersect

, and let the arc lengths between the two points of each line's intersection with the boundary of K be s and s', respectively. Also, let

and

denote the angle between G, G' and its repective support line, and finally let

and

denote the length of the chord made by the intersection of K and the two lines G and G' respectively. We will consider the following integral with the restrictions

. This may seem out of the blue, but the justification for this equation is quite complex and unecessary for this course.

		(3)

From this diagram, we can see that that is a function of s and . This fact will come in handy later. Then, from equation (5) from problem 11, we have and .

In order to solve the integral let's suppose the following:

		(4)

Now let's prove that equation (4) is true. Let P be a point on the boundary of K and let G be the tangent line at that point. Then is the length of the chord formed by the intersection of G with K. Imagine that G "sweeps over" K, as shown in the picture below.

Two chords formed by G create a triangle within the circle. If we add up the area of these infintely small triangles we will arive at the area of K.

As gets smaller and smaller, the length of the two legs of the triangle approach equal length. So the area of this isosceles triangle is . The area of K is this function integrated from = 0 and = . We have: or



		or

Calculating area by slicing up your figure and adding up the area of the slices can also be done for volume. This is the idea behind Cavalieri's Principle, which states that

This is one method used to estimate the volume of human organs and tumors.

Problem 12:

Using Cavalieri's Principle, find the volume of a sphere centered at the origin of radius R.

Solution

Now back to the integral at hand. Clearly:

		(5)

Evaluate.


	Separate the integrals
	Recall that is a function of s and

	Using Equation (16) from Problem 10 and Equation (5) from this problem:

	Now use Equation (4)

	Using the fact that each line is counted twice,
	that is once for each intersection point, and equation (5) from problem 11 we get

							(6)

Note that since both area and perimeter must be greater than or equal to zero, it is implied that

		(7)

which is well-known as the isoperimetric inequality. One implication of this famous inequality is that the geometric shape that encompasses the greatest area with respect to its perimeter is the circle, in which case . This is often referred to as Dido's problem and is mentioned in Virgil's Aneid. Read about her tale here.

References: Santalo; De-lin; Eric Weissten's Mathworld