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PROFESSOR: Today, I'm going to
continue the idea of setting up
integrals.
And what we'll deal with
is volumes by slices.
By slicing.
And it's lucky that
this is after lunch.
Maybe it's after
breakfast for some of you,
because there's the typical way
of introducing this subject is
with a food analogy.
There's a lot of ways
of slicing up food.
And we'll give a few more
examples than just this one.
But, suppose you
have, well, suppose
you have a loaf of bread here.
So here's our loaf
of bread, and I
hope that looks a little
bit like a loaf of bread.
It's supposed to be
sitting on the kitchen
counter ready to be eaten.
And in order to figure out
how much bread there is there,
one way of doing it is
to cut it into slices.
Now, you probably know
that bread is often
sliced like this.
There are even
machines to do it.
And with this setup
here, I'll draw the slice
with a little bit of a
more colorful decoration.
So here's our red
slice of bread.
It's coming around like this.
And it comes back down behind.
So here's our bread slice.
And what I'd like to
figure out is its volume.
So first of all, there's
the thickness of the bread.
Which is this
dimension, the thickness
is this dimension dx here.
And the only other dimension
that I'm going to give,
because this is a very
qualitative analysis for now,
is what I'll call the area.
And that's the area on
the face of the slice.
And so the area of
one slice, which
I'll denote by delta V,
that's a chunk of volume,
is approximately the area
times the change in x.
And in the limit, that's going
to be something like this.
And maybe the areas
of the slices vary.
There might be a little hole
in the middle of the bread
somewhere.
Maybe it gets a little
small on one side.
So it might change as x changes.
And the whole volume
you get by adding up.
So if you like,
this is one slice.
And this is the sum.
And you should think of it
in a sort of intuitive way
as being analogous to the
Riemann sum, where you would
take each slice individually.
And that would look like this.
Alright, so that's just a
superficial and intuitive way
of looking at it.
Now, we're only going
to talk about one
kind of systematic slice.
It's already on
your problem set,
you had an example of
a slice of some region.
But we're only going to talk
systematically about something
called solids of revolution.
The idea here is this.
Suppose you have some
shape, some graph,
which maybe looks like this.
And now I'm going to revolve it.
This is the x-axis and
this is the y-axis.
In this case, I'm going to
revolve it around the x-axis.
If you do that, then
the shape that you get
is maybe like this.
If I can draw it a little bit.
It's maybe a football.
So that's the shape that you get
if you take this piece of disk
and you revolve it around.
If you had made this
copy underneath,
it still would have
been the same region.
So we only pay attention to
what's above the axis here.
So that's the basic idea.
Now, I'm going to apply
the method of slices
to figure out the
volume of such regions
and give you a general formula.
And then apply it
in a specific case.
I want to take one little
slice of this football,
maybe a football
wouldn't work too well,
maybe we should go back
to a loaf of bread.
Anyway, the key point
is that you never really
have to draw a 3-D picture.
And 3-D pictures are awful.
They're very hard to deal with.
And it's hard to
visualize with them.
And one of the reasons why
we're dealing with solids
of revolution is that we don't
have as many visualization
problems.
So we're only going
to deal with this.
And then you have
to imagine from
the two-dimensional
cross-section what
the three-dimensional
picture looks like.
So we'll do a little
bit of an exercise
with the
three-dimensional picture.
But ultimately, you
should be used to,
getting used to, drawing
2-D diagrams always.
To depict the
three-dimensional situation.
Since it's much harder to draw.
The first step is to consider
what this slice is over here.
And again it's going
to have width dx.
And we're going to consider
what it looks like over
on the 3-D picture.
So it starts out being
more or less like this.
But then we're going
to sweep it around.
We're revolving
around the x-axis,
so it's spinning
around this way.
And if you take
this and think of it
as being on a hinge, which
is down on the x-axis,
it's going to swing down and
swoop around and come back.
Swing around.
And that traces out
something over here.
Which I'm going
to draw this way.
It traces out a disk.
So it's hard for me to draw, and
I'm not going to try too hard.
Maybe I drew it more like a
wheel, looking like a wheel.
But anyway, it's this
little flat disk here.
And so the method
that I'm describing
for figuring out the volume
is called the method of disks.
This is going to be
our first method.
Now I'm going to
apply the reasoning
that I have up on the
previous blackboard here.
Namely, I need to get
the volume of this chunk.
And the way I'm going to
get the volume of this chunk
is by figuring out its
thickness and its area,
its cross-sectional area.
And that's not too
difficult to do.
If this height, so this height
is what we usually call y.
And y is usually a function
of x, it's varying.
And this particular
distance is y.
Then the area of the face
is easy to calculate.
Because it's a circle, or a disk
if you like, with this radius.
So its area is pi y^2.
So that's, if you like,
one of the dimensions.
And then the thickness is dx.
So the incremental
volume is this.
So this is the method of disks.
And this is the integrand.
Now, there's one peculiar
thing about this formula.
And there are more peculiar
things about this formula.
But there's one
peculiar thing that you
should notice immediately.
Which is that I'm integrating
with respect to x.
And I haven't yet
told you what y is.
Well, that will depend on
what function y = f(x) I use.
So we have to plug
that in eventually.
If we're actually going
to calculate something,
we're going to have
to figure that out.
There's another
very important point
which is that in order to get
a definite integral, something
I haven't mentioned,
we're going to have
to figure out where
we're starting
and we're ending the picture.
Which is something we dealt
with last time in 2-D pictures.
So let's deal with an example.
And we'll switch over from
a football to a soccer ball.
I'm going to take a circle,
and we'll say it has radius a.
So this is 0 and this is a.
And I'm putting it in this
particular spot for a reason.
You can do this in
lots of different ways,
but I'm picking this one
to make a certain exercise
on your homework easier for you.
Because I'm doing half
of it for you right now.
Appreciate it, yeah.
I'm sure especially
today it's appreciated.
So again, the formula
has to do with keeping
track of these slices here.
And we're sweeping
things around.
So the full region that
we're talking about
is the volume of the
ball of radius a.
That's what our goal
is, to figure out
what the volume of the
ball of radius a is.
Alright, again as I say,
the thing sweeps around.
Coming out of the blackboard,
spinning around on this x-axis.
So the setup is the following.
It's always the same.
Here's our formula.
And we need to figure out what's
going on with that formula.
And so we need to solve
for y as a function of x.
And in order to do that,
what we're going to do
is just write down the
equation for the circle.
This is the circle.
It's centered at (a, 0),
so that's its formula.
And now there's one
nice thing, which
is that we really didn't need
it to find the formula for y,
we only needed to find
the formula for y squared.
So let's just solve for y^2
and we won't have to solve
a quadratic or anything.
Take a square root, that's nice.
This is a^2 - (x^2 - 2ax + a^2).
The a^2's cancel.
These two terms cancel.
And so the formula
here is 2ax - x^2.
Alright now, that is what's
known as the integrand.
Well, except for this
factor of pi here.
And so the answer for the volume
is going to be the integral
of pi times this
integrand, (2ax - x^2) dx.
And that's just the
same thing as this.
But now there's also
the issue of the limits.
Which is a completely
separate problem,
which we also have to solve.
The range of x is, from
this leftmost point
to the rightmost point.
So x varies, starts at 0,
and it goes all the way up
to what, what's
the top value here.
2a.
And so now I have a
completely specified integral.
Again, and this was
the theme last time,
the whole goal is
to get ourselves
to a complete
formula for something
with an integrand and limits.
And then we'll be
able to calculate.
Now, we have clear sailing
to the end of the problem.
So let's just finish it off.
We have the volume is, if I
take the antiderivative of that,
that's pi ax^2, whose derivative
is 2ax, minus x^3 / 3.
That's the thing whose
derivative is -x^2.
Evaluated at 0 and 2a.
And that is equal to
pi times, let's see.
2^2 = 4.
So this is 4a^3 -
8a^3 / 3, right?
And so all told, that is,
let's see, (12/3 - 8/3) pi a^3.
Which is maybe a familiar
formula, 4/3 pi a^3.
So it worked, we got it right.
Let me just point out a
couple of other things
about this formula.
The first one is that
from this point of view,
we've actually accomplished
more then just finding
the volume of the ball.
We've also found the
volume of a bunch
of intermediate regions,
which I can draw schematically
this way.
If I chop this thing, and
this portion is x here,
then the antiderivative
here, this region here,
which maybe I'll fill
in with this region
here, which I'm
going to call V(x),
is the volume of the
portion of the sphere.
Volume of portion of
width x of the ball.
And, well, the formula for it is
that it's the volume equals pi
(ax ^2 - x^3 / 3).
That's it.
So we've got something
which is actually
a lot more information.
For instance, if
you plug in x = a,
not surprisingly, and
this is a good idea
to do because it checks
that we've actually
got a correct formula here.
So if you like you can
call this a double-check.
If you check V(a), this should
be the volume of a half-ball.
That's halfway.
If I go over here and
I only go up to a,
that's exactly half of the ball.
That had better be
half, so let's just see.
V(a) in this case is pi, and
then I have (a^3 - a^3 / 3).
And that turns out to be pi
times a total of 2/3 a^3,
which is indeed half.
Now, on your problem set, what
you're going to want to look at
is this full formula here.
Of this chunk.
And what it's going to be good
for is a real-life problem.
That is, a problem
that really came up
over the summer, and this fall,
at a couple of universities
near here, where
people were trying
to figure out a phenomenon
which is well-known.
Namely, if you have a bunch
of particles in a fluid,
and maybe the size of
these things is 1 micron.
That is, the radius is 1 micron.
And then you have a bunch
of other little particles,
which are a lot smaller.
Maybe 10 nanometers.
Then what happens is that the
particles, the big particles,
like to hug each other.
They like to clump
together, they're very nice.
Friendly characters.
So what's the
explanation for this?
The explanation is
that actually they
are not quite as friendly
as they might seem.
What's really happening
is that the little guys
are shoving them around.
And pushing them together.
And they have sharp
elbows, the little ones
and they're pushing them.
Don't like them to be around and
they're pushing them together.
But there's actually
another possibility.
Which is that they
also will stick
to the sides of the container.
So there are two things
that actually happen here.
And if you want to get
a quantitative handle
on how much of this
happens, it has
to do with how much space
these things take up.
And so the issue is
some kind of overlap
between a band around
one sphere and a band
around the other sphere.
And this overlap region is
what you have to calculate.
You have to calculate
what's in here.
And you can do that
using this formula here.
It's not even difficult. So
this is, if you cut it in half,
turns out to be
two of these guys.
And then you're on your way
to figuring out this problem.
And the question is,
which do they prefer.
Do they prefer to
touch each other,
or do they prefer
to touch the wall.
Do they all cluster to the wall.
So you can actually see this
in solutions, what they do.
And the question
is, to what extent
do they prefer one
configuration to the other.
So that's a real, live
problem, which really comes up.
Came up just this year.
And frequently comes up.
Which is solved by
our first calculation.
So now I have, I
called this Method 1.
For solids of revolution, which
is called the method of disks.
And now I need to tell you
about the other standard method.
Which is called the
method of shells.
So this is our second method.
I'm going to illustrate this one
with a holiday-themed example
here.
This is supposed to be
a witches' cauldron.
Whoops, witch, witches, well.
Maybe more than one witch
will have this cauldron here.
So here's this shape.
And we're going to figure out
how much liquid is in here.
I'm going to plot this.
Maybe I'll put this down
just a bit lower here.
And I'm going to
make it a parabola.
This is y = x^2.
And I'm going to make
the top height be y = a.
So here's my situation.
And I want to figure out
how much liquid is in here.
Now, the reason why I presented
the problem in this form,
of course, is really to get
you used to these things.
And the first new thing that
I want you to get used to
is the idea that now we can also
revolve around the y-axis, not
just the x-axis.
So this one is going to be
revolved around the y-axis.
And that's what happens.
If you spin the
parabola around, you
get this kind of shape, this
kind of solid shape here.
Now, I'm going to
use the same kind
of slicing that I did before.
But it's going to look
totally different.
Namely, I'll draw
it in red again.
I'm going to take a
little slice over here.
And now I want to imagine
what happens if it gets
revolved around the y-axis.
This time it's not a disk.
Actually, if I revolve
this around the other way,
it would have had a hole in it.
Which is also possible to do.
That's practically the same
as the method of disks.
You'll maybe discuss
that in recitation.
Anyway, we're going to
revolve it around this way.
So again, I need
to sweep it around,
swing it around like this.
And I'll draw the shape.
It's going to sweep
around in a circle
and maybe it'll have a little
bit of thickness to it.
And this is the thing
that people call a shell.
This is the so-called
shell of the method.
I would maybe call
it a cylinder,
and another way of
thinking about it
is that you can maybe
wrap up a piece of paper.
Like this.
There it is, there's a
cylinder you can see.
Very thin, right?
Very thin.
Now, the reason why I
used the piece of paper
as an example of
this is that I'm
going to have to figure out
the volume of this thing.
Its thickness, as
usual, is equal to dx.
Its height is what?
Well, actually I have to
use the diagram to see that.
The top value is a, and the
bottom value is what we call y.
So the height is equal
to, I'm sorry, a - y.
Now, again this is an
incredibly risky thing here.
And I've done this
before I pointed this out
on the very first
day of lecture.
The letter y represents a
lot of different things.
And in disguise, when I
call this y, I mean y = x^2.
In other words, the
interesting curve.
I don't mean y = a,
which is the other part.
In general, you might think of
it as being equal to y_(top) -
y_(bottom).
So there are, of course,
two y's involved.
In disguise.
But we have a
shorthand, and the sort
of uninteresting
one we call by its--
we just evaluate immediately,
and the interesting one
we leave as the symbol y.
Now, the last bit that
I have to do, I claim,
in order to figure this
out, is the circumference.
And the reason for that is
that if I think of this thing
as like this tube,
or piece of paper
here, in order to figure out
how much stuff there is here,
all I have to do is unfold it.
Its size, the whole
quantity of paper
here is the same whether
it's rolled up like this
or whether it's stretched out.
So if I unwrap it,
it looks like what?
Well, it looks like
kind of a slab?
Right, it looks like
just a slab like this.
And again, the thickness is dx.
The height is a - y.
And now we can see
that the length here
is all the way around.
It's the circumference.
So this is going to be the
circumference when I unwrap it.
And in order to figure
out the circumference,
I need to figure out the radius.
So the radius is, on this
diagram, is right over there.
This is the radius.
And that distance is x.
So this length here is x.
And so this
circumference is 2 pi x.
And this height is
still ay. a - y, sorry.
And the thickness is still dx.
So in total, we're just going to
multiply these numbers together
to get the total volume.
We have, in other words,
dV is equal to the product
of the (2 pi x) dimension
dimension, the (8 - y)
dimension, and the dx dimension.
Incidentally,
dimensional analysis
is very useful and
important in these problems.
You can see that there
are three lengths being
multiplied together.
So we'll get a volume in end.
Something cubic.
And we will be coming
back to that, because it's
a quite subtle issue sometimes.
So here's the formula, and
let's simplify it a little bit.
We have 2 pi x times, remember,
first I have to substitute,
otherwise I'm never going
to be able to integrate.
And then I rewrite that as 2
pi (ax - x^2, whoops, x^3) dx.
Better not get that wrong.
And now the last little
bit here, that I had better
be careful about in
order to figure out
what the total volume
is, is the limits.
So the volume is going to be the
integral of this quantity 2 pi
(ax - x^3) dx.
And now I have to pay attention
to what the limits are.
Now, here you have
to be careful.
x is possibly the-- You
have to always go back
to the 2-D diagram.
I went to it immediately,
but that's the whole point.
Is that everything gets read
off from this diagram here.
When you take this guy
and you sweep it around,
you take care of everything
that's to the left.
So we only have to count
what's to the right.
We don't have to count
anything over here.
Because it's taken care
of when we sweep around.
So the starting place
is going to be x = 0.
That's where we start.
And where we end is the
farthest, rightmost spot for x.
Which is down here.
You've got to watch out
about where that is.
In the y variable,
it's up at y = a.
But in the x variable, we
can see that it's what?
It's the square root of a.
So these limits,
this is where you're
going to focus all your
attention on the integrand
and getting it just right.
And then you're going
to lose your steam
and not pay attention
to the limits.
They're equally important.
You've no hope of getting the
right answer without getting
the limits right.
So this is the integral
from 0 to square root of a.
Again, that's just because y =
a and y = x^2 implies x = square
root a.
That's that upper limit there.
And now, we're ready
to carry out this,
to evaluate this integral.
So we get 2-- sorry, we get
2 pi ax, that's pi ax^2.
Maybe I'll leave
the 2's in there.
2 pi ax^2 is the antiderivative
of this ax^2 / 2,
and then here x^4 / 4, evaluated
at 0 and square root of a.
And finally, let's
see, what is that.
That's 2 pi (a^2 / 2 - a^2 / 4).
Which is a total of 1/4,
right, 2 pi a^2 / 4.
Which is pi/2 a^2.
Yes, question.
STUDENT: [INAUDIBLE]
PROFESSOR: Right.
So the question is,
why did I integrate
only from the
middle to this end,
instead of all the
way from over here,
minus square root of a, all
the way to the plus end.
And the reason is that
you have to look at what's
happening with the rotation.
This red guy, when
I swept it around,
I counted the stuff to
the right and the left.
So in other words, if I just
rotate the right half of this,
I'm covering the left half.
So if I counted the stuff
from minus square root of a,
I would be doing it twice.
I would be doubling what I need.
So it's too much.
Another way of saying
it is if I wanted
to take the whole
region, if I rotate it
around only 180
degrees, only pi, that
would fill up the whole
region, if I did both halves.
And then instead of a
circumference, instead of a 2
pi x, I could use a pi x.
But then I would have
double what I had.
So there are two ways
of looking at it.
The same goes, actually,
for the football case.
When I have that football, I
didn't count the bottom part.
Because when I swung
it around the x-axis,
the top part sufficed and I
could ignore the bottom half.
Yeah, another question.
STUDENT: [INAUDIBLE]
PROFESSOR: Ooh, good question.
The question is, when
do you know, how do you
know when to take the rectangle
to be vertical or horizontal.
So far we've only done
vertical rectangles.
And I'm going to do a
horizontal example in a second.
And the answer to the question
of when you do it is this.
You can always set
it up both ways.
One way may be a
difficult calculation
and one way may be an
easier calculation.
Yesterday, we did it -
or, sorry, the last time.
Yeah, I guess it was yesterday.
We did it with-- and the
horizontal and the vertical
were quite different
in character.
One of them was really
a mess, and one of them
was a little easier.
So very often, one will
be easier than the other.
Every once in a while,
one of them is impossible
and the other one is possible.
In other words, the difference
in difficulty can be extreme.
So you don't know
that in advance.
Yeah, another question.
STUDENT: [INAUDIBLE]
PROFESSOR: The
question is, did we
just find the volume
when you rotate
this green region around.
Or, did we find the volume when
we rotate this whole region.
In other words, just the right
half or the right and the
left half.
The answer is both.
The region that you
get is the same.
You always get this
cauldron, whether you
take this right half
when you rotate it around
or you take both and
you rotate it around.
So the answer to both of
those questions is the same
and it's this.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: So that if you
rotated them both around
and you only wanted
to cover things once,
you would rotate halfway around.
Only by 180 degrees.
That's true.
But you can rotate around
as many times as you want.
You're still covering
the same thing.
Over and over and
over and over again.
So let's go on.
I have a very subtle point
that I need to discuss with you
in order to go on to
the next application.
So here's my, and
I do want to get
to the point of horizontal
cross-sections as well.
So let's continue here.
So the first thing that I
want to point out to you now
is, I want you to
beware of units.
There's something a little
fishy in this problem.
And it can be summarized in the
character of the answer, which
is just a little bit not clear.
Namely, it looks
like it's a^2, right?
And we know that a
is in units and we
should've gotten cubic units.
So there's something a little
bit tricky about this question.
And so I want to illustrate
the paradox right now.
So, suppose that a
= 100 centimeters.
And suppose the units
are centimeters.
Then the formula for the
volume is pi/2 100^2.
and the units we must take
are centimeters cubed.
Despite the fact
that you kind of
want to square the centimeters.
But that's not what
this problem says.
OK, so this is the
situation that we've got.
Now, if you work
out what this is,
to figure out what the
volume of this cauldron is,
what you find is that
it's pi / 2 times, well,
10^4 is 10 * 1,000
centimeters cubed.
And those are otherwise
known as liters.
So this is approximately
10 pi / 2 liters,
which is about 16 liters.
And so that's how much
was in the cauldron
under this choice of units.
Now, I'm going to make
another choice of units now.
And we're going to
make a comparison.
Suppose that the
units are 1 meter.
Looks like it
should be the same,
but if I calculate the volume,
it's going to be pi / 2,
and 1^2 times meters cubed.
That's what the
formula tells us to do.
And if you calculate that
out, that's pi / 2 (100 cm)^3.
And with this unit
notation we really do
want to cube the centimeters
and cube the 100.
So we get pi / 2.
And here we get 10 ^ 6 cm^3.
And that comes out to
pi / 2 * 1000 liters.
Or, in other words,
about 1600 liters.
So I'd like to ask you
first to contemplate this.
And this is a paradox.
And this is a serious paradox.
If you really want
to apply problems,
you actually have to understand
what your answers mean.
So what do you think
is going on here?
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: Right.
So the question is, how could
either of these make sense.
How am I dealing with
the units in either case.
So now I'm going to
explain to you the answer.
Because this is
really quite puzzling.
But it has a resolution.
The answer to this question is
that both answers are correct.
This is correct
reasoning in both cases.
What's the matter is that you
have to interpret the equation
y = x^2 in two different ways.
Two ways.
So let me explain what they are.
For instance, you can take
y = x^2 in centimeters.
So y = x ^2 in centimeters.
In which case, the picture
looks like the following.
a = 100 centimeters.
And this distance here, which
is the x value, this is 10.
This is 10 centimeters.
And that's what the
relationship means.
So the top of the
cauldron, if you like,
this distance here
is 20 centimeters.
This is actually very
badly drawn to scale.
It's actually very,
very, deep, this thing.
It's a rather skinny, deep one.
So this is very much not
to scale, this picture.
The other picture, the other
picture is interpreting y = x^2
in meters.
And that's more like what
I had in mind, actually.
I had in mind this big vat here.
And this distance
here is 1 meter,
and then the square
root of 1 is 1.
So this distance
here is also 1 meter.
And the top is 2 meters.
Now, it's not that crazy.
And in fact it's easy to
check that, it's pretty
reasonable in terms of scale.
That this thing has
16 liters in it.
And this guy has
1600 liters in it.
So you actually have to
know what your symbols mean
when you're dealing with these
kinds of applied problems.
And if you're ever really going
to do some real consequences,
you have to know
what the units mean.
And the problem with the
equation y = x^2 is that
it's the one that
violated scaling rules.
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: No.
STUDENT: [INAUDIBLE]
PROFESSOR: OK, so the question
is whether the formula V = pi /
2 a^2.
This is the correct
answer to the problem.
But it is not
consistent in units.
If you plug in a equals
some number of centimeters,
some number of millimeters,
some number of inches and so on,
every single time you'll
get a different answer.
And they're all inconsistent.
In other words, this
formula violates scaling.
STUDENT: [INAUDIBLE]
PROFESSOR: If you study
each step correctly,
you will discover that these
are the consistent and correct
statements, what I'm
writing on the blackboard.
And this makes sense
in a unit-less sense.
But then if you actually stick
units on them, one of them,
they both are correct.
And one of them describes
this situation and one of them
describes this situation.
And it's a mistake to think
of this as being 1 meter
and cubing the meters.
That will be an error that
will cause you problems.
This 1 is just unit-less,
and then the meters cubed
got converted.
So I encourage you to
study this on your own.
So now I'm going to
introduce the next problem.
We'll have to
solve it next time.
But the reason why I spent
all the time on units
is that otherwise it
would be impossible
for you to believe me when
I do this next calculation.
Because we're
trying to get a real
answer out of a real question.
And I'm going to make
conversions between centimeters
and meters back and forth.
And we have to get it
consistent in order
to have the right answer.
So there was a reason for
illustrating this pitfall.
So this second, the next
thing that I'd like to do,
is I'd like to boil the water
in the witches' cauldron.
This is definitely
seasonally appropriate,
since we're
approaching Halloween.
And we'll work it
out fully next time.
Now, I'm going to
introduce another feature
into the problem.
And this is the one that I
want you to understand now.
We'll set it all up
tomorrow, but right now I
need you to understand what
the new main idea that we're
going to get.
There is the new
physical feature
that I'm going to add to this
problem is that if when you're
boiling, when the witches
are boiling this water,
the temperature of
the water is not
the same at each
level in the kettle.
At the bottom of the kettle,
where you're heating it up,
it's at its highest temperature.
So at the bottom it's going
to be, say, 100 degrees.
That is, it's going
to be totally boiling.
100 degrees Celsius.
And at the top, it's going
to be, say, 70 degrees.
Right, it's very cold outside.
In fact, it's 0 degrees outside.
Which is the temperature at
which all witches operate,
I think.
Anyway, so they're
boiling their stuff.
And the question that
we're going to ask
is how much heat, how much
heat, do they need to do it.
Now, the thing starts
out at 0 degrees Celsius.
And we're going to heat it up to
this temperature configuration
here.
But it's rising from 100
down here to 70 up here.
So the temperature
is varying in height.
And for simplicity
I'm going to make
the formula for the temperature
be 70 at the top and 100
at the bottom.
So it's going to be 100 - 30y.
We'll let the level be 1, here.
Sorry, 30.
I said 3-- I wrote
3, but I meant 30.
So this is the
situation that we have.
Now, the point
about this problem
is we're going to figure
out the total temperature,
the total amount of heat
that you need to add in order
to heat this thing up.
That's going to be
temperature times volume.
But some places will
count more than others.
These will be hotter, but
there's less water down here.
This is wider up here, so
there's more water up here.
So there are various things that
are varying in this problem.
Now, the only way to
set up so that it works
is to chop things
up horizontally
instead of vertically.
Because it's on the
horizontal levels
that the temperature
is constant.
So we'll have an
easy calculation
for how much it takes to heat
up a layer, a horizontal layer.
When we rotate this
guy around the y axis,
that is which kind of shape.
It's a disk.
So actually this one's going
to be an easier problem.
It's going to be a disk
problem, not a shell problem.
But we're going to have to
work things out with respect
to the dy variation.
In other words, the integral
will be with respect to dy.
So we will do that next time.
We'll figure out how much heat
it takes to boil the kettle.
