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You might think that when computing areas there are a lot of formula that you need

Â to memorize. Well, fortunately, that's not the case.

Â There's one formula that you need to know for area.

Â And that is that area is the integral of the area element dA.

Â What do I mean by that? Let's recall some of the classical area

Â formulae that you learned when you were in school.

Â The area of a rectangle, or a parallelogram, or a triangle, or a

Â circular disk. These you know, but do you know why

Â they're true? They're true because they are The

Â integral of the corresponding area element.

Â Let's see how this works in a few examples.

Â Recall a parallelogram. Now you would probably argue by

Â rearrangement that if you cut into pieces and reshuffle.

Â Then you can build a rectangle, and we all know the area of a rectangle.

Â Well, that's good but there's another way to think about this.

Â Consider your parallelogram in the x y plane with the base on the x axis and

Â then slice it into horizontal strips. Of, if you will, an infinitismal

Â thickness, dy. Then what is the width of each strip?

Â It is, let's call it b, the length of the base of the parallelogram.

Â Now, how do we get the area out of this? The area element?

Â The area of this thin strip is well, b times dy, thinking of it as a rectangle

Â with one edge of infinitesmal length. The area is the integral of the area

Â element that is the integral of bdy as y goes from 0 to h the height in the

Â parallelogram and now lets see how it falls out the integral of a constant dy

Â is that constant b times y. Evaluate from 0 to h, and we obtain

Â simply the times h which we all know is the area.

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This also helps us make sense of the area of a triangle.

Â You're probably used to thinking of it as half the area of the bounding rectangle

Â by using geometry and similar triangles. But we could do things a little bit

Â differently. If we also arrange this triangle so that

Â the base is along the X axis and shear it horizontally.

Â So that it becomes a right triangle. We see, we have not changed the area

Â since we've preserved the horizontal area elements.

Â Now the hypotenuse has the equation of y equals h over b times x.

Â Now let's consider a slightly different area element.

Â This one, veritical The height of that vertical strip is, from our equation for

Â y, h/b*x, the thickness is dx, and thus, computing the area is the intergral of

Â the area element, and what do we get. We get the integral of h over b x dx as x

Â goes from 0 to b. Now, what do we see?

Â Well, h over b is a constant x integrated step x squared over 2.

Â Evaluating from 0 to b gives h b squared over 2b.

Â Minus 0. Simplifying we get one half base times

Â height. Very simple.

Â The example of a circular disc finaly takes us to a nonlinear case.

Â The classical means for determining his area was to use an Angular variable.

Â And to approximate as a limit of triangles.

Â Let us do likewise. This is perhaps not exactly how

Â Archimedes did it, but it's kind of close.

Â Consider an angular variable, theta. And consider a wedge.

Â Defined by an infinitesmal angle d theta. Then, thinking of this wedge as

Â approximately a triangle. Not exactly, but approximately a triangle

Â to first order, what would the dimensions be?

Â Well, we would say that the height of the triangle is roughly r, the radius of the

Â circle. The base is going to have length r times

Â d theta. With that in mind, the area element is

Â what? It is one half the base times the height,

Â that is r times r d theta. The area, being the integral of the area

Â element, is the integral of 1/2 r squared d theta, as theta sweeps all the way

Â around from zero to 2 pi. Now, we know what, well, wait a minute,

Â there's no theta in the integrand. There's just r squared, which is a

Â constant, as is 1/2, so we can pull out the 1/2 r squared.

Â The integral of d theta is, of course, theta.

Â Evaluating that from zero to 2 pi gives 2 pi times 1/2 r squared.

Â That is, pi R squared, the classical answer done very simply.

Â But this is not the only way to do this problem.

Â Consider if we used a radial variable, and we sweep out thin annuli and fill up

Â the disc in that manner. Let's call that radial variable t.

Â The annular strip is going to have thickness dt, and now we need to argue

Â that the area of this infinitesimal annular strip is The circumference, 2 pi

Â t times the thickness, d t. I'm not going to argue that that's really

Â correct right now. Let me just state it, and let's see what

Â happens when we integrate it. We get the integral of 2 pi t d t Aa t

Â goes from zero to r, the radius. That is, we get pi t squared as t goes

Â from zero to r. That again yields our answer, which, if

Â nothing else, tells us that our area element was indeed correct.

Â That is a very simple integral. But notice that it's not the only way

Â that we can do it. Still, we can cut the disk into slices.

Â Let's say vertical slices. And we sweep from left to right.

Â Placing the circular disk at the center of the x,y plane.

Â Allows us to use a strip of width dx, and height given by what?

Â Well, the equation of the circle is x squared plus y squared equals r squared.

Â Solving for y gives plus or minus square root of r squared minus x squared.

Â The area element, in this case, is going to be the height, that is, twice root r

Â squared minus x squared, times the width dx.

Â Then we see that integrating to get the area gives the integral of twice root r

Â squared minus x squared dx. Evaluating the limits from minus r to r,

Â what do we get? Well, this integral is not as easy as the

Â other two that we've done. It is an even integrand or a symmetric

Â domain, so could pull out the two, double it, and integrate from zero to r.

Â But we're still not done. We would need to use trigonometric

Â substitution, something of the form x equals r sine u and do a little bit of

Â work. I'm going to leave it to you to verify

Â that what one gets for that integral is pi r squared over 4.

Â Leading to the final answer that we all know and love.

Â Notice that depending on which way you split the domain up into area elements.

Â The resulting integral may be trivial, or may be quite involved.

Â This then leads to the example that we all know and love, the area between two

Â curves. Let's say f on the top, and g on the

Â bottom. Now, it's very clear given what we've

Â done, how to determine the area element. In this case, using a vertical strip,

Â it's height is going to be f(x) minus g(x).

Â its width is d(x). And so, integrating this, we see the

Â classical formula that you've no doubt seen before.

Â The area is the integral f or x minus g of x dx, as x goes from a to b.

Â One interesting application of this formula is to economics.

Â In the definition of something called the genie index.

Â This is a ratio that is used to quantify income and equality.

Â In a fixed population, it relies on one slightly technical definition and the

Â rest is simple calculus. >> Let us denote by f of x, the

Â fraction of the total income earned by the lowest x fraction of the populous,

Â the domain is x goes from 0 to 1 and you should think of x as a percentage of the

Â population And f(x) as the amount of income earned by that lowest x percent.

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The genie index, denoted g, quantifies how far from a flat, or even

Â distribution. The income is.

Â And what do I mean by that? Well, consider what this function, f,

Â looks like. It tells you how much of the net income

Â is being earned by what fraction of the population.

Â A flat income would be the diagonal line where x equals y.

Â Why? Well the diagonal line would mean that if

Â you look at the lowest 50% of the population, they earn 50% of all income.

Â Of course, in practice that's not true. They are in Less.

Â Folks at a higher income earn more and a larger percentage of the net income

Â earned. The Gini index is a measure of how far

Â from the perfectly flat, or equitable, distribution you are.

Â And it's defined as follows. It's the ratio of the area between the

Â flat distribution, y equals x, and the true income distribution, y equals f.

Â It's the ratio of that. And, the area between y equals x and y

Â equals 0. Well, that denominator is just telling

Â you the area of the triangle, and since the base and the height are 1, that's

Â 1/2. So, another way to write the Gini index

Â is as the numerator, this area between F and x.

Â Namely x minus f of x, d x value of it from zero to one divided by one-half,

Â that is doubled. Let's look at this in a specific example.

Â Let's compute the gini index for a power law distribution.

Â Assuming that income is distributed like a polynomial, x to the n, then what would

Â we get? Well, we integrate x minus f(x) from 0 to

Â 1 and double it. Well, that integral is almost trivial.

Â It's going to be x squared over two minus x to the end plus one over and plus one.

Â Evaluating from zero to one and then doubling.

Â What do we get? We get one minus two over and plus one.

Â Putting that over a common denominiator yields the result of N minus one over N

Â plus one. If we look at a little bit of economic

Â data, we find that in the year 2010, the state of New York had a GINI index of one

Â half approximately. If the income in New York state were

Â distributed according to a monomial, then what would that monomial be?

Â Well we see that it would be a cubic distribution.

Â We can't conclude that, but if it were a monomial it would be a cubic

Â distribution. That's a nice application.

Â