0:00

Welcome to Calculus, I'm Professor Ghrist.

Â We're about to begin lecture 31, Bonus Material.

Â When you get to multivariable calculus, you're going to learn a simpler and more

Â fundamental approach to area elements. Recall that when we computed the area of

Â the region between two curves, the graphs of f and g, then, we used a vertical

Â strip as an area element and integrated with respect to dx.

Â In other problems it became more efficient to use a horizontal strip and

Â integrate with respect to dy. In multivariable calculus, you will take

Â the intersection of these two, obtaining an infinitesimal rectangle and filling up

Â the region of interest with those shapes. What is the area element in this case?

Â If we look at a rectangular element with dimensions dx and dy, infinitesimals,

Â then the area element dA is the product dx times dy.

Â Or if you like, dy times dx, since at this point, we're not so worried about

Â the order of things. In this case, to compute the area of this

Â domain, let's say the domain d between these two graphs.

Â Then, how would you integrate the area element?

Â In multivariable calculus, you will write this as a double integral using two

Â integral signs. This means that you need to integrate

Â both with respect dx and with respect to dy.

Â Now the double integral of this area element is the double integral of, let's

Â say dy dx and the reason I write it in that order is that we're going to insert

Â some parenthesis and first integrate with respect to dy.

Â Then integrate with respect to dx. Now all of the complexity in this problem

Â is built into how you set up the limits of integration.

Â Let's work from the inside moving out. If we consider integrating with respect

Â to dy for a fixed value of x, then what are the limits on y?

Â Well, clearly y goes from g of x at the bottom to f of x at the top.

Â That is really what's giving you your veritcal strip.

Â And then, our limits on x are, x is going from a to b, since once we've integrated

Â out with respect to y, then we just need to perform that sweep from left to right.

Â And now, although the notation is a bit unusual, the computation is completely

Â elementary. Let's do that inner integral first.

Â What is the integral of dy? Well of course, that's just y.

Â But, we need to evaluate this. What are the limits?

Â Y is going from g of x to f of x, then we need to take that output.

Â And integrate with respect to d x. That leads to the formula, the integral x

Â goes from a to b of f of x, the upper limit, minus g of x, the lower limit, d

Â x. And you've seen that formula before.

Â That is one principled matter in which to derive it.

Â You're going to see a lot more about this.

Â Later in multivariable calculus and a little bit later in this course.

Â Well, let's take an extra moment or two and see something remarkable that we can

Â do with this approach. Let's consider a domain D that is a round

Â disk, radius 1 Now we could split this up into vertical or horizontal strips and

Â compute the area. But let's use our infinitesmal square da

Â and fill up the region in that manner. Then, what is the area, A, of this

Â domain, D? It is of course the double integral of dx

Â times dy. Now, I'm not going to fill in the limits

Â and do that work, you've already done that in this world.

Â You and I both know that the area of this region is pi.

Â Now, consider what happens when we look at an eliptical region.

Â Let's say, the minor and major axes are a and b respectively aligned along the x

Â and the y axis let's say. Then one way that I could think about

Â computing the area of e is how well I could take my domain d and stretch it out

Â and squeeze it into this elliptical shape.

Â When you know a little bit about matrices and vectors then you'll be able to

Â specify this exactly. But for now let's assume that this disk

Â were made out of some stretchy material and we've.

Â Simply stretched out the round disk into this elliptical region, stretching or

Â compressing as need be. In so doing, we have not only deformed

Â the region and changed it's area, we have deformed the area element dA, how have we

Â done so? Well if you look at what is happening in

Â the x direction and in the y directions. The radius of one gets changed to a in

Â the horizontal axis and b in the vertical axis.

Â That means at the infinitesimal level you can think of small rectangles whose

Â dimensions are a times dx and b times dy, and these squares will match up perfectly

Â with the rectangles on the right. So, in this case we could say that the

Â area element for this region e is a times b times dx times dy.

Â Then integrating that area element not over e but over d.

Â And then pushing it over to the elliptical region allows us to do this

Â integral very simply. We can pull out the constants a times b,

Â and we're left with the double integral over D of dxdy that we know to be pi.

Â That gives final area of pi, times a times b, and that is the area enclosed by

Â this ellipse. Maybe you should think about what happens

Â in that formula when a equals b equals some constant r.

Â Then we obtain the familiar formula for the area of a circular region radius r.

Â More generally what we've done in that computation which seems like a bit of a

Â trick is actually a very deep idea. It is a type of change of variables

Â formula and that's one of the most important principles.

Â You're going to learn about in multivariable calculus.

Â