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PROFESSOR: Now, today I need
to get started by finishing up
what I did last time.
Namely, talking about
numerical methods.
And I want to just
carry out one example.
And then I want to
fill in one loose end.
And then we'll talk
about the unit overall.
We were talking, last time,
about numerical integration.
I'm going to illustrate this
just with the simplest example
that I can.
We're going to look at the
integral from 1 to 2 of dx / x.
Which we know
perfectly well already
is the log of x evaluated
between 1 and 2, which is ln 2
- ln 1.
Which is just ln 2.
Now, if you punch that
into your calculator,
you're going to get
something like this.
I hope I saved it here.
Yeah.
It's about 0.693147.
That's more digits than we're
going to get in our discussion
here.
Anyway, that's about
how big this number is.
And the numerical
integration methods
will give you about as
much accuracy as you
can get on the function itself.
And, of course,
some functions we
may have more trouble
approximating.
But the function 1 / x, we
know pretty well how to do,
because we know how to divide.
So since the function that
we're integrating here is 1 / x,
it's going to be not
too difficult to get
some arithmetic.
Nevertheless, I'm
going to do this
in the simplest possible case.
Namely, just with two intervals.
Now, you really can't expect
things to work so well
with two intervals.
That's a pretty ridiculous
approximation to your function.
When you have two
intervals, that
means you're looking at the
graph of this hyperbola.
And you have 1 here, and you
have 2 here and you have 3/2.
And you're really only
keeping track of the values
at these three spots.
So the idea that you can
approximate the area just
by knowing the values
of three places
is a little bit of a
stretch of the imagination.
But we're going
to try it anyway.
Now, the trapezoidal rule
is the following formula.
It's delta x (1/2 the first
value + the second value
+ 1/2 the third value).
In this case, the pattern
is 1/2, 1, 1, 1, 1, 1, 1/2.
And in this case, delta x =
1/2 because this interval's
of length 1.
The b - a, right.
Let's just point that out here.
Here, b = 2. a = 1. b - a = 1.
And the number n is 2.
And so, delta x, which
is (b - a) / n, is 1/2.
So here's what we get.
And let's just see
what this number is.
It's 1/2 of the value at here.
Well, so let's just check
what these values are.
This value is 1, this
value over here is 2/3,
and the last value is 1/2.
Because the function, of
course, was y = 1 / x.
And those were the three
values that we have.
So y_0, this one is
y_0, this one is y_1,
and this one is y_2.
Now, here we have 1/2*
1 + 2/3 + 1/2 * 1/2.
Now, on an exam,
I don't expect you
to add up long messes
of numbers like this.
When you have two
numbers, I expect
you to add them up if they're
reasonable, or subtract them.
Just as we do when we
take antiderivatives.
Like, for example,
I don't want you
to leave the answer to
an integration like this
in this form.
I want you to simplify
it at least down to here.
And I of course don't
expect you to know
the numerical approximation.
But I certainly expect
you to be able to do that.
On the other hand,
when the arithmetic
gets a little bit long,
you can relax a little bit.
But I did carry this
out on my calculator.
Unless I'm mistaken,
it's about 0.96.
It's pretty far off.
So remember what it was.
It's what you get when you
get these straight lines.
And there are these little
extra pieces of junk there.
Now, don't trust that
too much, but the point
is that it's far off.
So now, let's take a
look at Simpson's Rule.
And I claim that Simpson's
Rule is surprisingly accurate.
In this case, really,
even a little more than it
deserves to be.
The formula is (delta x
/ 3) (y_0 + 4 y_1 + y_2).
So the pattern is
1, 4, 1, or 1, 4
and then it alternates 2's and
4's until 4, 1 at the very end.
And if I just plug in the
numbers now, what I get
is 1/6, because
delta x = 1/2 again.
And the value for y_0 was 1.
And the value for y_1 was 2/3.
And the value for y_2 was 1/2.
So here's the
estimate in this case.
And this one I did
carry out carefully.
And it came out to 0.69444.
Which is actually
pretty impressive,
if you think about it.
Given what the logarithm is.
Now, what's going on with
Simpson's Rule in general
is this.
If you-- Simpson's
minus the exact answer,
in absolute value, is
approximately of the size
of (delta x)^4.
That's really the
way it behaves.
Which means that if
delta x is about 1/10,
so if we had divided this
up into 10 intervals,
which we didn't, but if
we'd divided it up into 10
intervals, then you could
expect that delta x--
the error would
be about 10^(-4).
In other words, four digits of
accuracy here for this thing.
But the exact analysis of
this, a more careful analysis
of this, is in your textbook.
And I'm not going to do it.
But I just want to point out
that it is an effective method.
It really does give
you nice four-digit
with manageable-- you could
even really do it by hand.
It's so convenient,
the Simpson's Rule.
Whereas the other rules
aren't really that impressive
as far as giving fairly
accurate answers.
The last little
remark to make is
that the reason is
that Simpson's Rule is
matching a parabola.
And somehow the parabola
follows this curve better.
It's giving the exact answer.
So I'll mention this.
Simpson's Rule is derived
using the exact answer
for all degree 2 polynomials.
In other words, parabolas.
All parabolas.
But even all the
ones of lower degree.
So straight lines would
work, and constants
would work as well.
Whereas the other ones only
work for, say, straight lines.
The trapezoidal rule only
works for straight lines.
But there is a weird accident.
It turns out that it
also works for cubics.
Once you get the formulas,
it works for cubics.
So it's also exact for cubics.
And that's what explains
the fourth order validity.
The last thing that
I want to point out
is that this is extremely
vague, what I said there.
And you should be a little
bit cautious about it.
You need to watch out
for 1/x for x near 0.
All bets are off if the
function is singular.
And there's a lot
of area under there.
And it's also true that if
the derivative messes up,
you're in trouble too.
You really need for the function
to be nice and smooth in order
for Simpson's Rule to work.
This is wath out.
That's a real wath out, but
we'll try to-- Watch out.
Watch out for whenever x near 0.
Then this thing doesn't work.
This thing really depends
on bounds on derivatives.
But I'm going to be
relatively vague about that.
I'm not attempting to give
you an error analysis here.
OK, so if you were
doing this on an exam,
how do you remember this
strange pattern of numbers?
The one thing that I want to
recommend to you is, as a way
of remembering it, so
the one mnemonic device,
we'll call it a
mnemonic device here,
for remembering what it
is that you're doing,
is to remind yourself of
what happens for the simplest
possible case.
Which is f(x) = 1.
It seems very modest, but
if it doesn't give you
the exact answer
for f(x) = 1, you've
got the wrong weightings.
And here, if you check out what
happens in the first formula
here, y_0 / 2 + y_1 +..., well,
we'll go all the way to y_(n-1)
+ y_n / 2.
If you check that
formula out here,
this is the trapezoidal rule.
If you check it out for
this case, then what you get
is that this is equal
to delta x times what?
Well, all of these are 1's.
And how many are
there in the middle?
There are n - 1 of
them in the middle.
So it's 1/2 + n - 1 + 1/2.
At the tail end.
So all told it's delta x n.
And I remind you that
delta x = b - a / n.
So, delta x, this thing,
is equal to b - a.
And that's just as it should be.
What we just calculated is an
approximation to this integral
here.
Which is just the area of
the rectangle of base b -
a and height 1.
Which of course is b - a.
So this is the check that
you got your weighted average
correct here.
You've put the correct
weightings on everything.
And you can do this same
thing with Simpson's Rule.
And match up those quantities.
There was a question in
the room at some point.
No, OK.
So now, the next thing
I want to do for you
is the loose end
which I left hanging.
Namely, I want to compute that
mysterious constant square root
of pi / 2.
This is really one of the
most famous computations
in calculus.
And it's a very,
very clever trick.
I certainly don't expect you
to come up with this trick.
I certainly wouldn't
have myself.
But it's an important
thing to calculate.
And it's just very useful.
So I'm going to
tell you about it.
And it's just on the subject
that we're dealing with
in this unit; namely, slicing.
Or adding up.
So the first step, which is just
something that we already did,
was that we found the
volume under this curve.
This bell-shaped
curve, e^(-r^2).
But rotated around an axis.
Rotated around this axis.
Around this way.
So we figured that out.
And that was a relatively
short computation.
I'm just going to remind
you, it goes by shells.
We integrate the whole
range from 0 to infinity.
And we have 2 pi r
2 pi r e^(-r^2) dr.
So this again is the
circumference of the shell.
This is the height of
the shell, and this
is the thickness of the shell.
Circumference,
height, thickness.
So we're just taking a little
piece here and sweeping it
around.
And then adding up.
And then this antiderivative
is pi-- -pi e^(-r^2),
evaluated at 0 and infinity.
And we worked this
out last time.
This is pi.
It's pi (1 - 0).
Which is pi.
So the conclusion
is that V = pi.
We already know that.
Now, the problem that
we want to deal with now
is the problem not of
a volume, but an area.
And this looks quite different.
And of course the answer
is going to be different.
But let's do it.
So this is this
question mark here.
And I'm going to do the one
from minus infinity to infinity.
And I'll relate it
to what we talked
about earlier in this unit,
in just a couple of minutes
when I show you the procedure
that we're going to follow.
So here's the
quantity and now, what
this is interpreted as is the
area under this bell curve.
This time, Q is really an area.
Now, what's going to turn
out to happen, is this.
This is the trick.
We're going to compute
V in a different way.
And you'll see it laid
out in just a second.
We will compute V by slices.
We're going to slice it
like a piece of bread here.
We're going to solve for
that same thing here.
And then, amazingly, what's
going to happen is that we will
discover that V = Q^2.
That's going to be
what's going to come out.
And that's the end of the
computation that we want.
Because actually we
already know what V is.
We don't want to read
this equation forward.
We want to read
it the other way.
We want to say Q^2 = V,
which we already know is pi.
And so Q is equal to
the square root of pi.
I haven't shown this yet,
this is the weird part.
And I'm going to put
it in a little box
so that we know that this
is what we need to check.
We need to check this fact here.
We haven't done that yet.
Now, let me connect this with
what we did a few days ago.
With what I called one of
the important functions
of mathematics besides
the ones you already know.
And so the function
that we were faced with,
and that we discussed,
was this one.
And then, we were interested
in the value at infinity.
We were interested in this.
Which, if you draw
a picture of it,
and you draw the
same bell curve,
that's the area
under half. of it.
That's the area starting
from 0 and going to infinity.
That's the area under half.
So this chunk is F of infinity.
And now I hope that this
part of the connection
is not meant to be fancy.
The idea here is that
Q = 2 F(infinity).
This number here.
And so F F(infinity) is equal
to the square root of pi
over 2, if we believe what
we said on the last panel.
And that was the thing that I
drew a picture of on the board.
Namely, the graph of
F looked like this.
And there was this
asymptote, which
was the limit F(x) tends
to square root of pi
over 2, as x goes to infinity.
That was that limiting value.
Which is F of infinity.
So this is the asymptote.
And now I've explained
the connection
between what we claimed before,
which was quite mysterious,
and what we're actually going
to be able to check now.
Concretely, by making
this computation.
So how in the world can you
get something like this.
What's in that orange
box there, that V = Q^2.
Again, the technique
is to use slices here.
And I'm going to have to
draw you a 3-D picture
to visualize the slice.
Let's do that.
I'm going to draw
three axes now,
because we're now going to be
in three-dimensional space,
and I want you to imagine
the x-axis as coming out
of the blackboard, the
y-axis is horizontal,
and there's a new
axis, which I'll call
the z-axis, which is going up.
So what's happening
here is that I'm
thinking of this-- This is,
if you like, some kind of side
view.
And this is a view where I've
tilted things a little bit
up to the top.
Now, the distribution,
or you could
think of this target
in the plane, where
the most likely places
to hit were in the middle
and it died off.
As we went down.
Now, I want to draw a
picture of this graph.
I'm going to draw a
picture of e^(-r^2).
And it's basically a hump.
So I'm going to take the
first-- the slice along y = 0.
The y = 0 slice.
And I claim that that
goes up like this.
And then comes back down.
Let me shade this in,
so that you can see
what kind of a slice this is.
This is supposed to be along
this vertical plane here.
Which is coming out
of the blackboard
and coming towards you.
And that's a slice.
Now, I'm going to draw
one more slice so that you
can see what's happening.
I'm going to draw a
slice at another place.
Along here.
This will be y = b.
Some other level.
And now I'm going to
show you what happens.
What happens is that the
hump dies down a little bit.
So the bump is just
a little bit lower.
And it's going to look a
little bit the same way.
But it's just going
to be a bit smaller.
So there's another slice here.
Like that.
And I want to give a
name to these slices.
I'm going to call this A(b).
That is, the area
of the b slice.
Under the surface.
OK?
Yes, question.
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah, the solid.
Yeah.
We're trying to figure
out this volume here,
which is the one we started
out with, by slices.
So first I have to think
of-- I'm going to visualize--
So here I didn't even visualize.
I took a cross
section and I thought
about how to spin it around
without actually doing that
in three-dimensional space.
But now I'm going to take
a different kind of slice.
I'm going to take
that same bump, which
is a three-dimensional object.
I'm going to lay
it down on a plane.
Which looks like this.
And then it's a bump here.
It's a hump.
And now I'm going to try to
slice it by various planes.
STUDENT: [INAUDIBLE]
PROFESSOR: So one
way of defining
the bump, as you just suggested,
is you take this curve
and you rotate it
around this z-axis.
So in other words, you make
this the axis of rotation,
you spin it around.
That's correct.
So that shows you that the
peaks as you go down here
are going to descend
the same way.
But I don't want to
draw those lines.
I want to imagine what
the parallel slices are.
Because I don't want
to get cross slices.
I want all slices parallel
to the same thing.
STUDENT: [INAUDIBLE]
PROFESSOR: OK.
This is not particularly
easy to visualize.
Now, here's the formula
for volume by slices.
The formula for volume by
slices is that you add up
the areas of the slices.
That's how you do it.
You take each slice.
You add the
cross-sectional area,
and then you take a
little thickness, dy,
and then you add all of them up.
Because this is extending
over the whole plane,
we're going to have to go all
the way from minus infinity
to plus infinity.
And this is the formula
for volumes by slicing.
And now our goal, in order
to do this calculation,
we're going to just fix
y is equal to some b.
We're just going to fix
one of these slices.
And we're going
to calculate A(b).
That's what we
need to do in order
to make this procedure succeed.
This is the only place
where this method works.
But it's an important one.
In order to make
it work, I'm going
to have to again draw the plot
from a different point of view.
I'm going to do the top view.
So I want to look down
on this x-y plane here.
This is the x-direction,
and here's the y-direction.
And then again I want
to draw my slice.
My slice is here.
At y = b.
So we're just
right on top of it.
And it's coming up
at some kind of bump.
Here, with a little
higher in the middle
and going down on the sides.
Now, the formula for
the height is this.
If I take a distance r here,
the formula for the height
of the bump is e^(-r^2).
I'll store that over here.
e^(-r^2) is the
height at this place.
If this distance
to the origin is r.
That's true all the way around.
And in terms of b and x,
we can figure out that
by this right triangle.
This height is b, and
this distance is x.
So r^2 = b^2 + x^2.
Question.
STUDENT: [INAUDIBLE]
PROFESSOR: The question
is, is that the x-y plane.
So the answer is that over here
I cleverly used the letter r.
I avoided using y's
and z's or anything.
And over here, this
is the distance r.
And you like, this
is z, going up.
That's the way to think of it.
So that all of the
letters are consistent.
So I just avoided
giving it a name.
That's good, that's
exactly the point.
Alright.
So now, I claim I have
a formula for r^2.
And so I can write this down.
This e^(-b^2 + x^2).
But now I'm going to use
the rule of exponents.
Which is that this is the same
as e^(-b^2) times e^(-x^2).
And that's going to be
the main way in which we
use the particular function
that we're dealing with here.
That's really the
main step, amazingly.
So now I get to
compute what A(b) is.
A(b) is the area under a curve.
So it's going to be, let
me write it over here,
A(b) is the area
under this curve here.
Which is some constant
times-- so if you imagine,
call this thing the name c.
Under some curve, ce^(-x^2).
Where the c is
equal to e^(-b^2).
That's what our slice is.
In fact, it looks
like one of those.
It looks like one
of those bumps.
Here's its formula again.
It's the integral from
minus infinity to infinity
of e^(-b^2) e^(-x^2) dx.
We just recopied
what I had up there.
And this is the height at
each value of x, with b fixed.
And now, so we have
a lot of steps here.
But each of them
is very elementary.
The first one was just
that law of exponents.
That we could split
the two into products.
Now I'm going to make that
splitting even further.
This is a constant.
It's not varying with x.
So I'm going to factor
it out of the integral.
This is e^(-b^2) times the
integral from minus infinity
to infinity of e^(-x^2) dx.
So this might look frightening,
but actually it's just
the property of an integral.
All integrals have
this kind of property.
You can always factor
out a constant.
And now here comes
the remarkable thing.
This is e^(-b^2) times a number
which is now familiar to us.
What is this number?
This is what we're looking for.
This is our unknown, Q.
So I've computed A(b),
and now I'm ready to
finish the problem off.
A(b) = e^(-b^2) Q. Q is
that strange number which we
don't know yet.
What it is.
So now I'm going to
compute the whole volume.
The whole volume,
remember, it's over there,
it's minus infinity
to infinity, A(y) dy.
And now I'm just going to plug
in the formula that we've found
for A. Now I'm doing
this for each b,
so I'm doing it
varying over all b's.
So I have the integral from
minus infinity to infinity.
And here I have e^(-y^2).
I've replaced b by y.
And now I have Q. And I have dy.
I just recopied what
I had over there
into the formula for slicing.
And now, I'm going to do
this trick of factoring out
the constant a second time.
This is a constant.
It doesn't depend on y.
It's the same for all y,
it just will factor out.
So this is the same as Q
times the integral from minus
infinity to infinity,
e^(-y^2) dy.
And now, lo and behold,
this expression here.
Of course, notice how
I defined Q. Let's
go back carefully to
where Q is defined.
Here's Q. This t is
a dummy variable.
It doesn't matter
what I call it.
I can call it x,
I can call it u,
I can call it v.
In this case, I've
given it two different names.
At this stage, I called it x.
And at this stage
I'm calling it y.
But it's the same variable.
And so this little chunk is
Q and altogether I have two
of them, for Q^2
being the total.
And that's the end
of the argument.
It's a real miracle.
STUDENT: [INAUDIBLE]
PROFESSOR: Great question.
The question is, wait a minute.
As y changes, doesn't x change.
And so then this
wouldn't be a constant.
So that's the way in which
we've used the letters x and y
in this whole course.
When you get to 18.02,
you'll almost never do that.
Always y and x will be
different variables.
And they won't have to
depend on each other.
Now, let me show you where on
this picture the x and the y
are.
We've got a whole x-y plane,
and here I'm fixing y = b,
y isn't varying.
Whereas x is changing.
So, in other words, I don't have
a relationship between x and y,
unless I fix it.
In this case I've decided that
y is going to be constant.
For all x.
Over here, I made a computation.
And I have a Q, which
is just a single number.
No matter which b I took,
it didn't matter which.
No matter which y equals b.
Of course, I changed
the name to b so it
wouldn't be so jarring to you.
But in fact this
b was y all along.
It's just that the x varied
completely independently
of the y.
I could fix the y and vary
the x, I could fix the x
and vary the y.
So it's a different
use of the letters.
From what you're used to.
It happens that y is
not a function of x.
In this case.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: The question
is, because I'm
rotating around the z-axis,
doesn't x change exactly
as much as y does.
What happens is that x and
y are symmetric variables.
They can be treated equally.
But if I decide to take slices
with respect to y being fixed
and x varying, then of
course they're now separated,
and I have a separate role
for the x and a separate role
for the y.
Or if I'd sliced
it the other way,
I would have gotten
the same answer.
I just would have reversed
the roles of x and y.
So what's happening
is that x and y
are on equal footing with
each other in this picture,
and I could've
sliced the other way.
I would have gotten
the same answer.
That's more or less the
answer to your question.
OK.
Now I have given
you a review sheet,
and I want to run
through, briefly,
what's going to be on the exam.
And this list of
exam questions is
what's going to be on the exam.
There are, sorry this is
not displayed correctly.
So, exam questions,
but now I'm just
going to show you what they are.
There are five
questions on the exam.
They are completely parallel
to what you got last year.
So you should look at that test.
It's worth looking at.
And you'll see in the
descriptions on this sheet
that what I'm describing
is what's on that test.
So what's going to happen
is - and this is also
posted on the Web
- is that you'll
be expected to calculate
some definite integrals using
the fundamental
theorem of calculus.
Do a numerical approximation.
There'll be a Riemann,
a trapezoidal rule
and a Simpson's Rule.
Calculate areas and volumes.
And then some other
cumulative sum.
Either an average value or
probability or perhaps work.
And sketch a function
which is given
in this form as an integral.
So those are the
questions, and you'll
see by the example of last
year's exam exactly the style.
They're really going
to be very similar.
Yes, question.
STUDENT: [INAUDIBLE]
PROFESSOR: OK, good question.
So the question is,
for Riemann sums,
what's the difference
between upper and lower,
and right and left?
So here we have a Riemann sum.
And I'm going to give
you a picture which is,
maybe this function y = 1 /
x, which was the one that we
were discussing earlier.
If you take the
function y = 1 / x
and you break it up
into pieces here,
however it doesn't
matter how many pieces,
let's just say there
are four of them.
Then the lower Riemann
sum is the staircase
which fits underneath.
So this one is a picture
of the lower sum.
It's always less.
And in the case of a decreasing
function, it's going to be,
so since if you like,
since 1 / x decreases,
the lower sum equals
the right sum.
You can see that
visually on this picture.
The values you're
going to select
are going to be the right
ends of the rectangles.
The upper sum is the left one.
Now, if the function
wiggles up and down,
then you have to pick
whichever side is appropriate.
Or maybe it'll be a
point in the middle,
if the maximum is
achieved in the middle.
Yeah, another question.
STUDENT: [INAUDIBLE]
PROFESSOR: Correct.
If the function is
increasing, then the lower sum
is the left sum.
So it just exactly
reverses what's here.
So this is decreasing,
lower sum is right-hand sum.
Increasing, lower
sum is left-hand sum.
STUDENT: [INAUDIBLE]
PROFESSOR: Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: Good question.
Suppose you're faced
with a function like this
in this last problem.
Which, generally, these
are the trickiest problems.
And the question is,
how are you ever going
to be able to decide
on an asymptote,
even whether there
is an asymptote.
And the answer is, you're not.
It's going to be
pretty tricky to get
keep track of what's happening
as it goes to infinity.
We had an example
on the homework
where is was
oscillating and it's
very unclear what's going on.
You have to do a very
long analysis for that.
So in fact, just don't
worry about that now.
At the very end of
the class, we'll
talk a little bit
about these asymptotes.
And really, the first issue
is whether they exist or not.
And that's even something.
That's a serious question
which we'll address
at the very end of this course.
STUDENT: [INAUDIBLE]
PROFESSOR: That's right.
It's not going to be
anything that complicated.
Other questions?
We we still have a
five whole minutes,
and I have an example to give,
if nobody has a question.
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: The question,
uh, will I tell you
which one of what to use?
STUDENT: [INAUDIBLE]
PROFESSOR: When
I tell you the numeric
approximation is,
you'll see on the exam.
The practice exam that you have.
I will ask you for all three.
I will ask you for the Riemann
sum, the trapezoidal rule,
and the Simpson's rule.
I'm guaranteeing you they'll
all three be on the exam.
I'm guaranteeing that
every single thing which
is on that piece of
paper is on the exam.
And you'll see it on the
exam that you've got.
It's exactly parallel
to what's there.
STUDENT: [INAUDIBLE]
PROFESSOR: So with
areas and volume,
the question is will I tell
you which method to use.
So let's discuss that.
So with areas and volumes,
there's basically--
So this is always
true with areas.
And it's true with
volumes of revolution.
By the way you should
read this sheet.
Not everything that's
on here have I said.
But you should read it.
Because it's all relevant.
So with volumes of revolution,
you always work your way
back to some 2-D diagram.
So there's some 2-D
diagram which is always--
two-dimensional
diagram, which is always
connected with these problems.
I mean, something this hard
is really just too hard
to do on an exam, right?
I mean, I'm not going to ask
you something this complicated
on the exam.
Because this involves
a three-dimensional
visualization.
But once you're
down to 2-D, you're
supposed to be
able to handle it.
Now, what's the main issue after
you've got your 2-D diagram?
The main issue is, do you
want to integrate with respect
to dx or dy?
And the answer is
that it will depend.
And if there's one
that's going to cause
you incredible difficulty,
and I feel that you're not
able to dodge it, then
I might give you a hint
and say you'd better use
shells, or you'd better
use disks or washers
or something like that.
But if I feel that you're
grown up enough to figure out
which one it is, because
one of them is so ridiculous
you say forget it, immediately,
after thinking about it.
Then I won't tell you which one.
Because I figure,
in other words,
I don't want you
to waste your time.
But I'm willing to waste a
minute or two of your time
on a wild goose chase.
So let me give you
an example of this.
Suppose you're looking at the
curve y between 0 and x - x^3.
So this is some kind of lump.
Like this.
It goes from 0 to 1, because the
right-hand side is 0 at 0 and 1
here.
It's some kind of thing.
And there are these
two possibilities.
One of them is to do shells.
And then, so this is supposed
to be rotated around the y-axis.
In this case.
And the same would apply,
actually, to the area problem.
So I'm doing a slightly
more complicated problem.
But you could ask for the area
underneath this, and so forth.
OK.
So we can integrate this dx,
or we can integrate this dy.
This indicates that
I'm deciding that this
is going to be of thickness
dx, and I'm integrating dx.
So that's a choice
that I'm making.
Now, the minute I
made that choice
I know that these are shells.
Because they sweep around this
way and that makes them shells.
Cylindrical shells.
And if I do that,
the setup is this.
It's 2 2 pi x (x - x^3) dx.
Now, I claim that when
you get to this point,
you already know you've won.
Because this is an easy
integral to calculate.
So you're done here.
You're happy.
Now, if you happened to say,
oh gee, I hate to do this.
I want to do
something clever, you
could try to do it
with cutting this way.
Let's do this.
And this would be
the dy thickness.
And then when you
sweep this around,
you get what we call a washer.
Which is really just the
difference of two disks.
So the shape here is this
thing swung around this axis.
And it looks like this.
So it's going to be the
difference of radii.
So what's the formula for this?
It's some integral of
pi times the right end,
which I'll call x_2,
and here the left end,
which I'll call x_1.
So this is pi pi
(x_2^2 - x_1^2) dy.
Now, already at this stage,
you think to yourself
this is more complicated
than the other method.
So you've already abandoned it.
But I'm just going to go one
step further into this one
to see what it is
that's happening.
If you try to figure out what
these values x_1 and x_2 are,
that corresponds to solving
for x_1 and x_2 in terms of y.
So that's the
following equation.
x_1 and x_2 solve the
equation that-- the curve, x -
x^3 is equal y.
Now, look at this equation.
That's the equation x^3--
sorry, x^3 - x + y, I guess.
Let's see.
Yeah, that's right,
is equal to 0.
This is a cubic equation.
Although there is
a formula for this.
You've never been taught the
formula for this equation.
So therefore, you
will never, ever
be able to get a formula for
x_2 and x_1 as a function of y.
And you'll never be able
to compute this one.
This is more than just a dead
end, it's like crash, burn,
and, you know, self-destruct.
So there may be such a
thing, so do the other way.
Good luck, folks.
