Welcome to Calculus, I'm Professor 
Ghrist. We're about to begin lecture 31, Bonus 
Material. When you get to multivariable calculus, 
you're going to learn a simpler and more fundamental approach to area elements. 
Recall that when we computed the area of the region between two curves, the graphs 
of f and g, then, we used a vertical strip as an area element and integrated 
with respect to dx. In other problems it became more 
efficient to use a horizontal strip and integrate with respect to dy. 
In multivariable calculus, you will take the intersection of these two, obtaining 
an infinitesimal rectangle and filling up the region of interest with those shapes. 
What is the area element in this case? If we look at a rectangular element with 
dimensions dx and dy, infinitesimals, then the area element dA is the product 
dx times dy. Or if you like, dy times dx, since at 
this point, we're not so worried about the order of things. 
In this case, to compute the area of this domain, let's say the domain d between 
these two graphs. Then, how would you integrate the area 
element? In multivariable calculus, you will write 
this as a double integral using two integral signs. 
This means that you need to integrate both with respect dx and with respect to 
dy. Now the double integral of this area 
element is the double integral of, let's say dy dx and the reason I write it in 
that order is that we're going to insert some parenthesis and first integrate with 
respect to dy. Then integrate with respect to dx. 
Now all of the complexity in this problem is built into how you set up the limits 
of integration. Let's work from the inside moving out. 
If we consider integrating with respect to dy for a fixed value of x, then what 
are the limits on y? Well, clearly y goes from g of x at the 
bottom to f of x at the top. That is really what's giving you your 
veritcal strip. And then, our limits on x are, x is going 
from a to b, since once we've integrated out with respect to y, then we just need 
to perform that sweep from left to right. And now, although the notation is a bit 
unusual, the computation is completely elementary. 
Let's do that inner integral first. What is the integral of dy? 
Well of course, that's just y. But, we need to evaluate this. 
What are the limits? Y is going from g of x to f of x, then we 
need to take that output. And integrate with respect to d x. 
That leads to the formula, the integral x goes from a to b of f of x, the upper 
limit, minus g of x, the lower limit, d x. 
And you've seen that formula before. That is one principled matter in which to 
derive it. You're going to see a lot more about 
this. Later in multivariable calculus and a 
little bit later in this course. Well, let's take an extra moment or two 
and see something remarkable that we can do with this approach. 
Let's consider a domain D that is a round disk, radius 1 Now we could split this up 
into vertical or horizontal strips and compute the area. 
But let's use our infinitesmal square da and fill up the region in that manner. 
Then, what is the area, A, of this domain, D? 
It is of course the double integral of dx times dy. 
Now, I'm not going to fill in the limits and do that work, you've already done 
that in this world. You and I both know that the area of this 
region is pi. Now, consider what happens when we look 
at an eliptical region. Let's say, the minor and major axes are a 
and b respectively aligned along the x and the y axis let's say. 
Then one way that I could think about computing the area of e is how well I 
could take my domain d and stretch it out and squeeze it into this elliptical 
shape. When you know a little bit about matrices 
and vectors then you'll be able to specify this exactly. 
But for now let's assume that this disk were made out of some stretchy material 
and we've. Simply stretched out the round disk into 
this elliptical region, stretching or compressing as need be. 
In so doing, we have not only deformed the region and changed it's area, we have 
deformed the area element dA, how have we done so? 
Well if you look at what is happening in the x direction and in the y directions. 
The radius of one gets changed to a in the horizontal axis and b in the vertical 
axis. That means at the infinitesimal level you 
can think of small rectangles whose dimensions are a times dx and b times dy, 
and these squares will match up perfectly with the rectangles on the right. 
So, in this case we could say that the area element for this region e is a times 
b times dx times dy. Then integrating that area element not 
over e but over d. And then pushing it over to the 
elliptical region allows us to do this integral very simply. 
We can pull out the constants a times b, and we're left with the double integral 
over D of dxdy that we know to be pi. That gives final area of pi, times a 
times b, and that is the area enclosed by this ellipse. 
Maybe you should think about what happens in that formula when a equals b equals 
some constant r. Then we obtain the familiar formula for 
the area of a circular region radius r. More generally what we've done in that 
computation which seems like a bit of a trick is actually a very deep idea. 
It is a type of change of variables formula and that's one of the most 
important principles. You're going to learn about in 
multivariable calculus.