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.