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PROFESSOR: Welcome
back to recitation.
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In this video I want us to
look at the following problem.
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We're going to let f
of x equal 1 over x.
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And I want to consider
the solid generated
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by rotating f of x about
the x-axis between x
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equal 1 and x equal infinity.
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I want you to find the area
of a cross-sectional slice,
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and then I want you to find
the volume of the solid.
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In the cross-sectional slice,
the one I'm interested in
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is the one that you
see in the xy-plane.
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So my suggestion to you is
that right away draw yourself
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a picture, just at least a
rough sketch of the curve y
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equals f of x, and then what
that cross-sectional slice
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will look like and
start off from there.
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And take a while to work
on that and then I'll
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be back and show you what I did.
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OK welcome back.
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Well, hopefully you were
able to make some headway
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on this problem, and maybe you
found some interesting things.
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Hopefully you did.
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And we will see what I find
and if they are interesting,
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which I think they are.
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So let's start off.
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I told you the first
thing you should do
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is get a rough picture.
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So I'm going to
draw a rough picture
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and make sure that
those match up well,
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that my picture
matches up with yours.
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So let me give you the xy-plane.
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And then y equals 1 over x is
a curve that at x equal 1--
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I'm going to make
y equal 1 up here.
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y equal 1 up there. x equal 1.
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So when x is 1, y is 1.
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And then it's going
to decay down.
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As x goes to infinity it decays
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So it looks something like that.
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And then I actually
should have moved this 1
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because I'm going to
need that side to get
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my cross-sectional slice.
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So my cross sectional
slice is going
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to come down here and
look something like this.
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Oops, that's supposed
to be symmetric.
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That's still supposed to
be something like this.
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It's supposed to be
symmetric about the x-axis.
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And the cross-sectional
slice is the area of that.
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Now, if you remember
what you saw in lecture.
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You saw, I believe,
that the integral from 1
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to infinity of 1
over x dx diverges,
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because you end up with
log evaluated at infinity
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and that is infinite and
so you wind up with this.
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This actually diverges.
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That was the area of the
top part from 0 to f of x.
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So of course if I
double that, I'm still
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going to get that it diverges.
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So in fact the cross sectional
slice area, CSS area,
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is infinite.
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OK, so that's one part.
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The other part is
to find the volume
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of the solid of revolution.
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So the second part was take
this piece from 0 up to f of x,
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rotate it around the x-axis
and compute the volume.
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Now we know how to
compute the volume.
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This is the disc method
that I'm going to use here.
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And I will write out
what we need to do
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and then we'll look at
what the integral gives us.
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So for the disc method, so let's
say this is the volume part.
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So the volume, it's going
to be the disc method.
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So I need to do pi r squared.
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And in this case I'm
integrating from 1 to infinity,
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so I need pi r squared dx.
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I'm going to be
integrating something
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that's pi r squared dx.
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And I know the bounds
are 1 to infinity,
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and so I need r in terms
of, as a function of x.
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That's fairly easy. r is
just the measure from 0 up
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to f of x.
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So that's just 1 over x.
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So this is actually
the integral--
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I'm going to pull
the pi out-- of 1
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over x squared from
1 to infinity dx.
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OK so how do I do this integral?
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Well you were shown in
class that in actuality this
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is the limit, as some value
up here goes to infinity,
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of this integral,
but we're just going
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to do it the shorthand way
that he also mentioned in class
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and keep the infinity around.
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So we can keep
that as our bounds,
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knowing we're taking limits as
this thing goes to infinity.
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So the integral of 1 over x
squared is negative 1 over x.
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So I get minus pi times 1 over
x, evaluated at 1 in infinity.
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I can just check
that if I need to,
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but this is x to the minus
1 and its derivative is
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negative x to the minus 2.
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So I needed that negative there.
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And then I evaluate this.
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Well at infinity, 1 over
x-- as x goes to infinity,
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this first part goes to 0.
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So this is 0 minus
negative pi times 1 over 1.
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So negative pi times 1.
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So this is just pi.
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OK, so hopefully that kind of
blows your mind a little bit,
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that you could have
something where
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this cross-sectional slice
is infinite, but in fact
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if you look at it the way
we computed the volume,
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we have in fact a finite volume.
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So I don't know
what else I'm going
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to say about that except
I think it's really cool.
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And you can think
about why that is.
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And in fact you
might want to notice
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that we had to--
we computed things
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in terms of
cross-sectional slices
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coming from the other direction.
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So we looked at these
cross-sectional slices
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and we got our-- we showed
our volume was finite there.
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So we had a sum of a
bunch of finite things
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and so it made sense that
you were going to get--
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and the finite things were
getting small fast enough.
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That's the point.
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That when you add up
these finite things that
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are getting small fast
enough, you can still
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end up with a finite number.
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But I guess, yeah,
that's where I'll stop.
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So just to say that we were
looking at this sort of solid
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of revolution
problem again, but we
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were dealing with
improper integrals,
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and we were showing how you
can do these kinds of problems
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with improper integrals.
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So now I really will stop there.