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Simple Areas

We know the basic standard formulae for the area of basic shapes, but why are they true? From the point of view of calculus, area is the integral of , the area element.

In this chapter, we will use the following procedure to determine a quantity :
1. Determine the differential element .
2. Integrate to compute .

Length of an interval

Before getting to areas, first consider how this method works for computing the length of the interval from to . If the length is denoted , then the length element will be denoted , and . In this context, the appropriate length element would be if we're working along the -axis.

So, we want to integrate as goes from to .
The length,

Parallelogram

The formula for the area of a parallelogram is base height . Consider the following rearrangement into differential elements, where we carve the parallelogram into parallel horizontal strips of width and height , where is the -axis.

In this case, the area element, , is the area of this infinitesimal rectangle. The limits on should go from to the height, of the parallelogram.
The area,

We have our familiar answer . This means that we've done a rearrangement in terms of infinitesimal strips. Shearing that parallelogram preserves the area element and hence, the area. That is why a parallelogram has the same area as the corresponding rectangle.

Triangle

The formula for the area of a triangle is base height . Let's think in terms of a differential area element. Given the fact that we can shear and preserve the area element, and thus the area, let's present our triangle as having a hypotenuse modeled by the line .

To compute the area element, let's use a vertical strip.

where the height of that vertical strip is and the width is the length element .
The area,

Disc

We will use three ways to find the area of a circular disc of radius :
1. Using an angular area element.
2. Using a radial variable.
3. Using a lateral, or a vertical rectangular strip.

  1. Angular

In this case, we'll use an angular area element. We will take a wedge with angle . If we look at that close up, it's modeled fairly well as a triangle. It's not a perfect triangle, there's a bit of curvature at the end. This is a triangle with two sides of length whose included angle is . Such a triangle has area , since is very small. If we model that as a triangle with height , and width , we can ignore the higher order terms in the Taylor expansion of that area. We obtain an area element .

Integrating to get the area, has to spin all the way around the circle from to .
The area,

  1. Radial

Let's consider a radial variable. We can sweep out the area of the circular disk using annuli with a radial coordinate . Then, we're looking at an annular strip of width . The corresponding area element is the circumference thickness .

Integrating this from to the radius gives us the area.

  1. Lateral

We will use a vertical rectangular strip. Again, it is not a perfect rectangle and there's a little bit of curvature at the end. But, these are higher order terms, and we just care about the differential element. So, using a vertical strip with width , and knowing that the formula for the boundary circle is , we solve for along the upper and lower branches.

We then obtain an area element that is the area of this rectangular strip.

In the case of strips, assume the circle is centered at the origin, and let keep track of where the strip intersects the -axis. Thus, ranges from to . Integrating, and using a trigonometric substitution gives

The area between two curves

Let's say the is on top and the is below. Then as we sweep a vertical strip from left to right, we obtain the area. In this case, the area element is a vertical rectangle of width and of height , the length of the interval between the two.

The general formula for the area between two curves and ,

Example
Find the area of the region bounded above by and below by .

The logical choice for area element is a vertical strip:

The height of this strip is , and the width of the strip is . So the area element is . To find the intersection points, set the curves equal, which gives . This implies , which factors to . Thus, the intersections are and . It follows that

Gini Index (An application of area formula)

In economics, this ratio is used to quantify income inequality in a population.

Let = Fraction of total income earned by the lowest fraction of the populace, .
The Gini index quantifies how far is from a "flat" distribution. This means that f(0) = 0, f(1) = 1.
is probably going to be below the flat distribution where , the lowest fraction earns the lowest fraction.

The Gini index, is measuring the difference between these two distributions, in terms of area. It's the ratio of the area between the flat distribution and the given population's income distribution . One normalizes that by the area between the flat distribution and , namely the area of that triangle, or .

Example
Compute for a power law distribution .

The Gini Index doesn't tell you the income distribution, but we could approximate it in the assumption of a power law. For example, in the year 2010, in the state of New York in USA, the Gini Index was very close to . If we assume that it went by a power law distribution, that would imply a cubic distribution of income.


EXERCISES

  • What is the area between the curve and the -axis from to ?
  • Find the area of the bounded region enclosed by the curves and .
  • What is the area between the curve and the -axis for ?
  • Calculate the Gini index of a country where the fraction of total income earned by the lowest fraction of the populace is given by
.
  • Compute the area between the curves and for .
  • Consider a cone of height with base a circular disc of radius . Let's compute the "surface area" -- the area of the "outside" of the cone, not including the bottom. Following how we computed the area of a circular disc (which is, indeed, such a cone with ), we can decompose its area into infinitesimal triangles with base and height the slant length . The area element is then the area of this infinitesimal triangle. Integrating from to gives the "surface area" of the cone. What is its value?
  • Compute the area between the curves and from to .
  • Compute the area of a triangle with vertices at (0,0), (2,1), (3,6)