Uploaded by MIT on 07.01.2011

Transcript:

PROFESSOR: Welcome back to recitation.

In this video I want us to look at the following problem.

We're going to let f of x equal 1 over x.

And I want to consider the solid generated by rotating f

of x about the x-axis between x equal

1 and x equal infinity.

I want you to find the area of a cross sectional slice, and

then I want you to find the volume of the solid.

In the cross sectional slice, the one I'm interested in is

the one that you see in the xy plane.

So my suggestion to you is that right away draw yourself

a picture, just at least a rough sketch of the curve y

equals f of x and then what that cross sectional slice

will look like and start off from there.

And take a while to work on that and then I'll be back and

show you what I did.

OK welcome back.

Well hopefully you were able to make some headway on this

problem, and maybe you found some interesting things.

Hopefully you did.

And we will see what I find and if they are interesting,

which I think they are.

So let's start off.

I told you the first thing you should do

is get a rough picture.

So I'm going to draw a rough picture and make sure that

those match up well, if my picture matches up with yours.

So let me give you the xy plane.

And then y equals 1 over x is a curve that at x equal 1 I'm

going to make y equal 1 up here.

y equal 1 up there. x equal 1.

So when x is 1, y is 1.

And then it's going to decay down.

As x goes to infinity it decays

So it looks something like that.

And then I actually should have moved this 1 because I'm

going to need that side to get my cross sectional slice.

So my cross sectional slice, is going to come down here and

look something like this.

Oops that's supposed to be symmetric.

That's still supposed to be something like this.

It's supposed to be symmetric about the x-axis.

And the cross sectional slice is the area of that.

Now, if you remember what you saw in lecture.

You saw I believe that the integral from 1 to infinity of

1 over x dx diverges, because you end up with log evaluated

at infinity and that is infinite and so you

wind up with this.

This actually diverges.

That was the area of the top part from 0 to f of x.

So of course if I double that, I'm still going

to get that it diverges.

So in fact the cross sectional slice area,

CSS area, is infinite.

OK, so that's one part.

The other part is to find the volume of the solid of

revolution.

So the second part was take this piece from 0 up to f of

x, rotate it around the x-axis and compute the volume.

Now we know how to compute the volume.

This is the disc method that I'm going to use here.

And I will write out what we need to do and then we'll look

at what the integral gives us.

So for the disc method, so let's say this

is the volume part.

So the volume, it's going to be the disc method.

So I need to do pi r squared.

And in this case I'm integrating from 1 to

infinity, so I need pi r squared dx.

I'm going to be integrating something that's

pi r squared dx.

And I know the bounds are 1 to infinity, and so I need r in

terms of, as a function of x.

That's fairly easy. r is just the measure from

0 up to f of x.

So that's just 1 over x.

So this is actually the integral--

I'm going to pull the pi out --of 1 over x squared from 1

to infinity dx.

OK so how do I do this integral?

Well you were shown in class that in actuality this is the

limit-- as some value up here goes to infinity --of this

integral, but we're just going to do it the shorthand way

that he also mentioned in class and keep

the infinity around.

So we can keep that as our bounds, knowing we're taking

limits as this thing goes to infinity.

So the integral of 1 over x squared is negative 1 over x.

So I get minus pi times 1 over x, evaluated at 1 in infinity.

I can just check that if I need to, but this is x to the

minus 1 and it's derivative is negative x to the minus 2.

So I needed that negative there.

And then I evaluate this.

Well at infinity 1 over x as x goes to infinity, this first

part goes to 0.

So this is 0 minus negative pi times 1 over 1.

So negative pi times 1.

So this is just pi.

OK, so hopefully that kind of blows your mind a little bit,

that you could have something where this cross sectional

slice is infinite, but in fact if you look at it the way we

computed the volume, we have in fact a finite volume.

So I don't know what else I'm going to say about that except

I think it's really cool.

And you can think about why that is.

And in fact you might want to notice that we had to, we

computed things in terms of cross sectional slices coming

from the other direction.

So we looked at these cross sectional slices and we got

our, we showed our volume was finite there.

So we had a sum of a bunch of finite things and so it made

sense that you were going to get, and the finite things

were getting small fast enough.

That's the point.

That when you add up these finite things that are getting

small fast enough, you can still end up

with a finite number.

But I guess, yeah, that's where I'll stop.

So just to say that we were looking at this sort of solid

of revolution problem again, but we were dealing with

improper integrals, and we were showing how you can do

these kinds of problems with improper integrals.

So now I really will stop there.