Welcome to calculus, I'm Professor 
Ghrist. We're about to begin lecture 30 on simple 
areas. As we begin chapter four of the course, 
we're going to turn from our classical tools of Taylor series, limits, 
derivatives, and integrals and begin putting them to work. 
We'll focus on applications of the definite integral, beginning with some 
simple area formula. We'll start small, but work our way up to 
bigger things. You might think that when computing areas 
there are a lot of formula that you need to memorize. 
Well, fortunately, that's not the case. There's one formula that you need to know 
for area. And that is that area is the integral of 
the area element dA. What do I mean by that? 
Let's recall some of the classical area formulae that you learned when you were 
in school. The area of a rectangle, or a 
parallelogram, or a triangle, or a circular disk. 
These you know, but do you know why they're true? 
They're true because they are The integral of the corresponding area 
element. Let's see how this works in a few 
examples. Recall a parallelogram. 
Now you would probably argue by rearrangement that if you cut into pieces 
and reshuffle. Then you can build a rectangle, and we 
all know the area of a rectangle. Well, that's good but there's another way 
to think about this. Consider your parallelogram in the x y 
plane with the base on the x axis and then slice it into horizontal strips. 
Of, if you will, an infinitismal thickness, dy. 
Then what is the width of each strip? It is, let's call it b, the length of the 
base of the parallelogram. Now, how do we get the area out of this? 
The area element? The area of this thin strip is well, b 
times dy, thinking of it as a rectangle with one edge of infinitesmal length. 
The area is the integral of the area element that is the integral of bdy as y 
goes from 0 to h the height in the parallelogram and now lets see how it 
falls out the integral of a constant dy is that constant b times y. 
Evaluate from 0 to h, and we obtain simply the times h which we all know is 
the area. Now, there's a slightly different 
perspective we can put on this, namely that when we shear that parallelogram and 
slide the horizontal slices. So that it becomes a rectangle then we're 
not changing the area element, this shearing preserves the area element. 
Thus the area, that's a different sort of rearrangement argument one at the 
infinite as well level. This also helps us make sense of the area 
of a triangle. You're probably used to thinking of it as 
half the area of the bounding rectangle by using geometry and similar triangles. 
But we could do things a little bit differently. 
If we also arrange this triangle so that the base is along the X axis and shear it 
horizontally. So that it becomes a right triangle. 
We see, we have not changed the area since we've preserved the horizontal area 
elements. Now the hypotenuse has the equation of y 
equals h over b times x. Now let's consider a slightly different 
area element. This one, veritical The height of that 
vertical strip is, from our equation for y, h/b*x, the thickness is dx, and thus, 
computing the area is the intergral of the area element, and what do we get. 
We get the integral of h over b x dx as x goes from 0 to b. 
Now, what do we see? Well, h over b is a constant x integrated 
step x squared over 2. Evaluating from 0 to b gives h b squared 
over 2b. Minus 0. 
Simplifying we get one half base times height. 
Very simple. The example of a circular disc finaly 
takes us to a nonlinear case. The classical means for determining his 
area was to use an Angular variable. And to approximate as a limit of 
triangles. Let us do likewise. 
This is perhaps not exactly how Archimedes did it, but it's kind of 
close. Consider an angular variable, theta. 
And consider a wedge. Defined by an infinitesmal angle d theta. 
Then, thinking of this wedge as approximately a triangle. 
Not exactly, but approximately a triangle to first order, what would the dimensions 
be? Well, we would say that the height of the 
triangle is roughly r, the radius of the circle. 
The base is going to have length r times d theta. 
With that in mind, the area element is what? 
It is one half the base times the height, that is r times r d theta. 
The area, being the integral of the area element, is the integral of 1/2 r squared 
d theta, as theta sweeps all the way around from zero to 2 pi. 
Now, we know what, well, wait a minute, there's no theta in the integrand. 
There's just r squared, which is a constant, as is 1/2, so we can pull out 
the 1/2 r squared. The integral of d theta is, of course, 
theta. Evaluating that from zero to 2 pi gives 2 
pi times 1/2 r squared. That is, pi R squared, the classical 
answer done very simply. But this is not the only way to do this 
problem. Consider if we used a radial variable, 
and we sweep out thin annuli and fill up the disc in that manner. 
Let's call that radial variable t. The annular strip is going to have 
thickness dt, and now we need to argue that the area of this infinitesimal 
annular strip is The circumference, 2 pi t times the thickness, d t. 
I'm not going to argue that that's really correct right now. 
Let me just state it, and let's see what happens when we integrate it. 
We get the integral of 2 pi t d t Aa t goes from zero to r, the radius. 
That is, we get pi t squared as t goes from zero to r. 
That again yields our answer, which, if nothing else, tells us that our area 
element was indeed correct. That is a very simple integral. 
But notice that it's not the only way that we can do it. 
Still, we can cut the disk into slices. Let's say vertical slices. 
And we sweep from left to right. Placing the circular disk at the center 
of the x,y plane. Allows us to use a strip of width dx, and 
height given by what? Well, the equation of the circle is x 
squared plus y squared equals r squared. Solving for y gives plus or minus square 
root of r squared minus x squared. The area element, in this case, is going 
to be the height, that is, twice root r squared minus x squared, times the width 
dx. Then we see that integrating to get the 
area gives the integral of twice root r squared minus x squared dx. 
Evaluating the limits from minus r to r, what do we get? 
Well, this integral is not as easy as the other two that we've done. 
It is an even integrand or a symmetric domain, so could pull out the two, double 
it, and integrate from zero to r. But we're still not done. 
We would need to use trigonometric substitution, something of the form x 
equals r sine u and do a little bit of work. 
I'm going to leave it to you to verify that what one gets for that integral is 
pi r squared over 4. Leading to the final answer that we all 
know and love. Notice that depending on which way you 
split the domain up into area elements. The resulting integral may be trivial, or 
may be quite involved. This then leads to the example that we 
all know and love, the area between two curves. 
Let's say f on the top, and g on the bottom. 
Now, it's very clear given what we've done, how to determine the area element. 
In this case, using a vertical strip, it's height is going to be f(x) minus 
g(x). its width is d(x). 
And so, integrating this, we see the classical formula that you've no doubt 
seen before. The area is the integral f or x minus g 
of x dx, as x goes from a to b. One interesting application of this 
formula is to economics. In the definition of something called the 
genie index. This is a ratio that is used to quantify 
income and equality. In a fixed population, it relies on one 
slightly technical definition and the rest is simple calculus. 
 >> Let us denote by f of x, the fraction of the total income earned by 
the lowest x fraction of the populous, the domain is x goes from 0 to 1 and you 
should think of x as a percentage of the population And f(x) as the amount of 
income earned by that lowest x percent. The genie index, denoted g, quantifies 
how far from a flat, or even distribution. 
The income is. And what do I mean by that? 
Well, consider what this function, f, looks like. 
It tells you how much of the net income is being earned by what fraction of the 
population. A flat income would be the diagonal line 
where x equals y. Why? 
Well the diagonal line would mean that if you look at the lowest 50% of the 
population, they earn 50% of all income. Of course, in practice that's not true. 
They are in Less. Folks at a higher income earn more and a 
larger percentage of the net income earned. 
The Gini index is a measure of how far from the perfectly flat, or equitable, 
distribution you are. And it's defined as follows. 
It's the ratio of the area between the flat distribution, y equals x, and the 
true income distribution, y equals f. It's the ratio of that. 
And, the area between y equals x and y equals 0. 
Well, that denominator is just telling you the area of the triangle, and since 
the base and the height are 1, that's 1/2. 
So, another way to write the Gini index is as the numerator, this area between F 
and x. Namely x minus f of x, d x value of it 
from zero to one divided by one-half, that is doubled. 
Let's look at this in a specific example. Let's compute the gini index for a power 
law distribution. Assuming that income is distributed like 
a polynomial, x to the n, then what would we get? 
Well, we integrate x minus f(x) from 0 to 1 and double it. 
Well, that integral is almost trivial. It's going to be x squared over two minus 
x to the end plus one over and plus one. Evaluating from zero to one and then 
doubling. What do we get? 
We get one minus two over and plus one. Putting that over a common denominiator 
yields the result of N minus one over N plus one. 
If we look at a little bit of economic data, we find that in the year 2010, the 
state of New York had a GINI index of one half approximately. 
If the income in New York state were distributed according to a monomial, then 
what would that monomial be? Well we see that it would be a cubic 
distribution. We can't conclude that, but if it were a 
monomial it would be a cubic distribution. 
That's a nice application. This lesson may have seemed a bit 
redundant. After all, you all know the formula for 
the area of a triangle or a circular disk. 
These are the classics. But, by revisiting the classics, we 
deepen our understanding and make it possible to accomplish more difficult 
works later. That will happen in our next lesson when 
we consider more complex plainer shapes.