MAGDALENA TODA: According to my watch,
we are right on time to start.
I may be one minute early, or something.
Do you have questions out of the material we covered last time?
What I'm planning on doing-- let me tell you
what I'm planning to do.
I will cover triple integrals today.
And this way, you would have accumulated
enough to deal with most of the problems in homework four.
You have mastered the double integration by now,
in all sorts of coordinates, which is a good thing.
Triple integrals are your friend.
If you have understood the double integration,
you will have no problem understanding
the triple integrals.
The idea is the same.
You look at different domains, and then you
realize that there are Fubini-Tunelli type of results.
I'm going to present one right now.
And there are also regions of a certain type, that
can be treated differentially, and then you
have cases in which reversing the order of integration
for those triple integrals is going to help you a lot.
OK, 12.5 is the name of the section, triple integrals.
So what should you imagine?
You should imagine that somebody gives you
a function of three variables.
Let's call that-- it doesn't often have a name as a letter,
but let's call it w.
Being a function of three coordinates, x, y, and z,
where x, y, z is in R3.
And we have some assumptions about the domain D
that you are working over, and you
have D as a closed-bounded domain in R3.
Examples that you're going to do use frequently.
Frequently used.
A sphere, a ball, actually, because in here,
if a sphere is together with a shell, it is the ball.
Then you have some types of polyhedra in r3 of all types.
And by that, I mean the classical polyhedra
whose sides are just polygons.
But you will also have some curvilinear polyhedra, as well.
What do I mean?
I mean, we've seen that already.
For example, somebody give you a graph
of a function, g of x and y, a continuous function, and says,
OK, can you estimate the volume under the graph?
Right?
And until now, we treated this volume under the graph
as double integral of g of x, y, continuous function over d, a,
where d, a was dx dy.
And we said double integral over the projected domain
in the plane.
That's what I have.
But can I treat it, this volume, can I treat it
as a triple integral?
This is the question, and answer is--
so can I make three snakes?
The answer is yes.
And the way I'm going to define that
would be a triple integral over a 3D domain.
Let's call it curvilinear d in r3, which is the volume-- which
is the body under the graph of this positive function,
and above the projected area in plane.
So it's going to be a cylinder in this case.
And I'll put here 1 dv.
And dv is a mysterious element.
That's the volume element.
And I will talk a little bit about it right now.
So what can you imagine?
They give you a way to look at it in the book.
I mean, we give you a way to look at it in the book.
It's not very thorough in explanations,
but it certainly gives you the general idea of what you want,
what you need.
So somebody gives you a potato.
It doesn't have to be this cylinder.
It's something beautiful, a body inside a compact surface.
Let's say there are no self-intersections.
You have some compact surface, like a sphere or a polyhedron,
assume it's simply connected, and it doesn't
have any self-intersections.
So a beautiful potato that's smooth.
If you imagine a potato that has singularities,
like most potatoes have singularities, boo-boos,
and cuts, so that's bad.
So think about some regular surface
that's closed, no self-intersections,
and that is a potato that is [INAUDIBLE].
Oh let's call it-- p for potato, no,
because I have got to use p for the partition.
So let me call it D from 3D domain,
because it's a three-dimensional domain, enclosed
by a curve, enclosed by a compact surface.
So think potato.
What do we do in terms of partitions?
Those pixels were pixels for the 2D world in flat line.
But now, we don't have pixels anymore.
Yes we do.
I was watching lots of sci-fi, and the holograms
have the three-dimensional pixels.
I'm going to try and make a partition.
It's going to be a hard way to partition this potato.
But you have to imagine you have a rectangular partition,
so every little pixel will be a-- is not cube.
It has to be a little tiny parallelepiped.
So a 3D pixel, let me put pixel in quotes,
because this is kind of the idea.
Well have what kind of dimensions?
We'll have three dimensions, right?
Three dimensions, a delta xk, a delta yk, and a delta zk
for the pixel number k.
That's pixel number k.
We have to number them, see how many they are.
Where k is from 1 to n, and is the total number of 3D
pixels that I'm covering the whole thing.
So don't think graphing paper, anymore,
because that's outdated.
Don't even think of 2D image.
Think of some hologram, where you cover everything
with tiny, tiny, tiny 3D pixels so
that in the limit, when you pass to the limit, with respect to n
and going to infinity, the discrete image,
you're going to have something like a diamond shaped thingy,
will convert to the smooth potato.
So as n goes to infinity, that surface
made of tiny, tiny squares will convert to the data.
So how do you actually find a triple snake integral f
of x, y, z over the domain D, dx dy dz.
OK, this theoretically should be what?
Think pixels.
Limit as n goes to infinity of the sum of the--
what do I need to do?
Think the whole partition into pixels.
How many pixels? n pixels total is called a p.
Script p.
And the normal p will be the highest diameter of-- highest
diameter among all pixels.
And you're going to say, Oh, what the heck?
I don't understand it.
I have these three-dimensional cubes, or three-dimensional
barely by p that get tinier, and tinier, and tinier.
What in the world is going to be a diameter of such a pixel?
Well, you have to take this pixel
and magnify it so we can look at it a little bit better.
What do we mean by diameter of this pixel?
Let's call this pixel k.
The maximum of all the distances you can compute between two
arbitrary points inside.
Between two arbitrary points inside,
you have many [INAUDIBLE].
So the maximum of the distance between points,
let's call them r and q inside the pixel.
So if it were for me to ask you to find
that diameter in this case, what would that be?
STUDENT: It's the diagonal.
MAGDALENA TODA: It's the diagonal between this corner
and the opposite corner.
So this would be the highest distance inside this pixel.
If it's a cube-- you see that I wanted a cube.
If it's a parallelepiped, it's the same idea.
So I have that opposite corner distance kind of thing.
OK.
So I know what I want.
I want n to go to infinity.
That means I'm going to have the p going to 0.
The length of the highest diameter
will go shrinking to 0.
And then I'm going to say here, what do I have inside?
F of some intermediate point.
In every pixel, I take a point.
Another pixel, another point, and so on.
So how many such points do I have?
n, because it's the number of pixels.
So inside, let's call this as pixel p k.
What is the little point that I took out of the [INAUDIBLE]
inside the cube, or inside the pixel?
Let's call that mister x k star, y k star, x k star.
Why do we put a star?
Because he is a star.
He wants to be number one in these little domains,
and says I'm a star.
So we take that intermediate point, x k star,
y k star, x k star.
Then we have this function multiplied by the delta v k.
somebody tell me what this delta v k will mean,
because ir really looks weird.
And then k will be from 1 to n.
So what do you do?
You sum up all these weighted volumes.
This is a weight.
So this is a volume.
All these weighted volumes.
We sum them up for all the pixels k from 1 to n.
We are going to get something like this cover
with tiny parallelepipeds in the limit,
as the partitions' [? norm ?] go to 0, or the number of pixels
goes to infinity.
This discrete surface will converge
to the beautiful smooth potato, and give you
a perfect linear image.
Actually, if we saw a hologram, this is what it is.
Our eyes actually see a bunch of I
tiny-- many, many, many, many, millions of pixels
that are cubes in 3D.
But it's an optical illusion.
We see, OK, it's a curvilinear, it's a smooth body of a person.
It's not smooth at all.
If you get closer and closer to that diagram
and put your eye glasses on, you are
going to see, oh, this is not a real person.
It's made of pixels that are all cubes.
just the same, you see your digital image of your picture
on Facebook, whatever it is.
If you would be able to be enlarge it,
you would see the pixels, being little tiny squares there.
The graphical imaging has improved.
The quality of our digital imaging has improved a lot.
But of course, 20 years ago, when you weren't even born,
we could still see the pixels in the photographic images
in a digital camera.
Those tiny first cameras, what was that, '98?
STUDENT: Kodak.
MAGDALENA TODA: Like AOL cameras that were so cheap.
The cheaper the camera, the worse the resolution.
I remember some resolutions like 400 by 600.
STUDENT: Black and white.
MAGDALENA TODA: Not black and white.
Black and white would have been neat.
But really nasty in the sense that you had the feeling
that the colors were not even-- they
were blending into each other, because the resolution
was so small.
So it was not at all pleasing to the eye.
What was good is that any kind of defects you
would have, something like a pimple
could not be seen in that, because the resolution was
so slow that you couldn't see the boo-boos,
the pimples, the defects of a face or something.
Now, you can see.
With the digital cameras we have now, we can do,
of course, Adobe Photoshop, and all of us
will look great if we photoshop our pictures.
OK.
So this is what it is in the limit.
But in reality, in the everyday reality,
you cannot take Riemann sums like that--
this is a Riemann approximating sum--
and then cast to the limit, and get ideal curvilinear domains.
No.
You don't do that.
You have to deal with the equivalent
of the fundamental theorem of calculus
from Calc I, which is called the Fubini-Tonelli type of theorem
in Calc III.
So say it again.
So the Fubini-Tunelli theorem that you
learned for double integrals over a rectangle
can be generalized to the Fubini-Tonelli theorem
over a parallelepiped.
And it's the same thing, practically,
as applying the fundamental theorem of calculus in Calc I.
So somebody says, well, let me start with a simple example.
I give you a-- you will say, Magdalena,
you are offending us.
This is way too easy.
What do you think, that we cannot understand the concept?
I'll just try to start with the simplest possible example
that I think of.
So x is between a and b.
In my case, they will be positive numbers,
because I want everything to be in the first octant.
First octant means x positive, y positive, and z positive
all together.
To make my life easier, I take that example,
and I say, I know the numbers for x, y, z.
I would like you to compute two integrals.
One would be the volume of this object.
Let's call it body.
It's not a dead body, it's just body in 3D.
The volume of the body, we say it
like a mathematician, V of B. What is that by definition?
Who's going to tell me?
Triple snake.
Don't say triple snake to other people,
because other professors are more orthodox than me.
They will laugh-- they will not joke about it.
So triple integral over the body of-- to get the volume,
the weight must be 1.
f integral must be 1, and then you have exactly dV.
How can I convince you what we have here,
in terms of Fubini-Tonelli?
It's really beautiful.
B is going to be a, b segment cross product, c, d segment,
cross product.
What is the altitude?
E, f, e to f.
Interval e to f means the height.
So length, width, height.
This is the box, or carry-on, or USPS parcel, or whatever
box you want to measure.
So how am I going to set up the Fubini-Tonelli integral?
a to b, c to d, e to f, 1.
And now, who counts first?
dz, dy, dx.
So it is like the equivalent of the vertical strip
thingy in double corners, double integrals.
Yes, sir?
STUDENT: Professor, why did you use 1 dV?
Why 1?
MAGDALENA TODA: OK.
You'll see in a second.
This is the same thing we do for areas.
So when you compute an area-- very good question.
If you use 1 here, and you put delta [? ak ?]
that is the graphing paper area.
It's going to be all the tiny areas,
summed up, sum of all the delta [? ak ?] which
means this little pixel, plus this little pixel,
plus this little pixel,
plus 1,000 pixels all together will cover up the area.
If you have the volume of a potato,
a body that is alive, but shouldn't move.
OK, it should stay in one place.
Then, to compute the volume of the potato,
you have to say, the potato, the smooth potato,
is the limit of the sum of all the tiny cubes of potato,
if you cut the potato in many cubes, like you cut cheese.
They got a bunch of cheddar cheese into small cubes,
and they feed us with crackers and wine-- OK, no comments.
So you have delta vk, you have 1,000 little cubes,
tiny, tiny, tiny, like that Lego.
OK, forget about the cheese.
The cheese cubes are way too big.
So imagine Legos that are really performing
with millions of little pieces.
Have you seen the exhibit, Lego exhibit with almost invisible
Legos at the Civic Center?
They have that art festival.
How many of you go to the art festival?
Is it every April?
Something like that.
So imagine those little tiny Legos, but being cubes
and put together.
This is what it is.
So f Vy.
Now, can we verify the volume of a box?
It's very easy.
What do we do?
Well, first of all, I would do it in a slow way,
and you are going to shout at me, I know.
But I'll tell you why you need to bear with me.
So integral of 1 dz goes first.
That's z between f and e.
So it's f minus e, am I right?
You say, duh, that's to easy for me.
I'm know it's too easy for me, but I'm
going somewhere with it.
dy dx.
The one inside, f minus e is a constant, pulls out,
completely out of the product.
And then I have integral from a to b of-- what is that left?
1 dy, y between d and c.
So d minus c, right?
d minus c dx.
And so on and so forth, until I get it.
If minus c times d minus c times b minus a,
and goodbye, because this is the volume of the box.
It's the height.
This is the height.
No, excuse me, guys.
The height is-- this one is the height.
This is the width, and this is the length, whatever you want.
All right.
How could I have done it if I were a little bit smarter?
STUDENT: You could have just put it in three integrals.
MAGDALENA TODA: Right Hey, I have
a theorem, just like before, which says
three integrals in a product.
This is what Matt immediately remembered.
We had two integrals in a product last time.
So what have we proved in double integrals
remains valid in triple integrals
if we have something like that.
So I'm going the same theorem.
It's in the book.
We have a proof.
So you have integral from a to b, c to d, e to f.
And then, some guys that you like, f of x, times g of y,
times h of z.
Functions of x, y, z, separated variables.
So f, a function of x only, g a function
of y only, h a function of z only.
This is the complicated case.
And then I have
STUDENT: dz, dy, dx.
MAGDALENA TODA: dz, dy, dx.
Excellent.
Thanks for whispering, because I was a little bit
confused for a second.
So, just as Matt said, go ahead and observe
that you can treat them one at a time like you did here,
and integrate one at a time, and integrate again, and pull out
a constant, integrate again, pull out a constant.
But practically this is exactly the same as integral
from a to b of f of x alone, dx, times integral
from c to d, g of y alone, dy, and times integral from e
to f of h of z, dz, and close.
So you've seen the version of the double integral,
and this is the same result for triple integrals.
And it's practically-- what is the proof?
You just pull out one at a time, so the proof
is that you start working and say, mister z counts here,
and he's the only one that counts.
These guys get out for a walk one at a time outside
of the first integral inside.
And then, integral of h of z, dz,
over the corresponding domain, will be just a constant, c1,
that pulls out.
And that is that-- c1 that pulls out.
Ans since you pull them out in this product one at a time,
that's what you get.
I'm not going to give you this as an exercise in the midterm
with a proof, but this is one of the first exercises
I had as a freshman in my multi--
I took it as a freshman, as multivariable calculus.
And it was a pop quiz.
My professor just came one day, and said, guys,
you have to try to do this [? before ?] by yourself.
And some of us did, some of us didn't.
To me, it really looked very easy.
I was very happy to prove it, in an elementary way, of course.
OK.
So how hard is it to generalize, to go
to non-rectangular domains?
Of course it's a pain.
It's really a pain, like it was before.
But you will be able to figure out what's going on.
In most cases, you're going to have
a domain that's really not bad, a domain that
has x between fixed values.
For example y between your favorite guys,
something like f of x and g of x, top and bottom.
That's what you had for double integral.
Well, in addition, in this case, you
will have z between-- let's make this guy
big F and big G, other functions.
This is going to be a function of x, y.
This is going to be a function of x, y,
and that's the upper and the lower.
And find the triple integral of, let's say 1 over d dV
will be a volume of the potato.
Now, I'm sick of potatoes, because they're not
my favorite food.
Let me imagine I'm making a tetrahedron, a lot of cheese.
I'm going to draw this same tetrahedron from last time.
So what did we do last time?
We took a plane that was beautiful,
and we said let's cut with that plane.
This is the plane we are cutting the cheese with.
It's a knife.
x plus y plus z equals 1.
Imagine that there's an infinite knife that
comes into the frame.
Everything is cheese.
The space, the universe is covered in solid cheese.
So the whole thing, the Euclidean space
is covered in cheddar cheese.
That's all there.
From everywhere, you come with this knife,
and you cut along this plane-- hi
let's call this [? high plane. ?]
And then you cut the x plane along the x, y plane,
y, z plane and x, z plane.
What are these called?
Planes of coordinates.
And what do you obtain?
Then, you throw everything away, and you
maintain only the tetrahedron made of cheese.
Now, you remember what the corners were.
This is 0, 0, 0.
It's a piece of cake.
But I want to know the vertices.
And you know them, and I don't want to spend time
discussing why you know them.
So
STUDENT: 0--
MAGDALENA TODA: 1, 0, 0.
Thank you.
Huh?
STUDENT: 0, 1, 0.
MAGDALENA TODA: Yes.
And 0, 0, 1.
All right.
Great.
The only thing is, if we see the cheese being a solid,
we don't see this part, the three axes of corners behind.
so I'm going to make them dotted, and you see the slice,
here, it has to be really planar.
And you ask yourself, how do you set up
the triple integral that represents
the volume of this object?
Is it hard?
It shouldn't be hard.
You just have to think what the domain will be like,
and you say the domain is inside the tetrahedron.
Do you want d or t?
T from tetrahedron.
It doesn't matter.
We have a new name.
We get bored of all sorts of names and notations.
We change them.
Mathematicians have imagination, so we change our notations.
Like we cannot change our identities,
and we suffer because of that.
So you can be a nerd mathematician imagining
you're Spiderman, and you can take,
give any name you want, and you can adopt a new name,
and this is behind our motivation
why we like to change names and change notation so much.
OK?
So we have triple integral of this T. All right, of what?
1 dV.
Good.
Now we understand what we need to do,
just like [? Miteish ?] asked me why.
OK, now we know this is going to be a limit of little cubes.
If were to cover this piece of cheese
in tiny, tiny, infinitesimally small cubes.
But now we know a method to do it.
So according to-- Fubini-Tonelli type of result.
We would have a between-- no, x-- is first, dz.
z is first, y is moving next, x is moving last.
z is constrained to move between a and b.
But in this case, a and b should be prescribed by you guys,
because you should think where everybody lives.
Not you, I mean the coordinates in their imaginary world.
The coordinates represent somebodies.
STUDENT: 0.
MAGDALENA TODA: x, 0 to 1.
How should I give you a feeling for that?
Just draw this line.
This red segment between 0 to 1.
That expresses everything instead of words
into pictures, because every picture is worth 1,000 words.
y is married to x, unfortunately.
y cannot say, oh, I am y, I'm going wherever I want.
He hits his head against this purple line.
He cannot go beyond that purple line.
He's constrained, poor y.
So he says, I'm moving.
I'm mister y.
I'm moving in this direction, but I cannot go past the purple
line in plane here.
I need you, because if you go, I'm lost.
y is between 0 and--
STUDENT: 1 minus x.
MAGDALENA TODA: 1 minus x.
Excellent Roberto.
How did we think about this?
The purple line has equation-- how do you
get to the equation of the purple line, first of all?
In your imagination, your plug in z equals 0.
So the purple line would be x plus y equals 1.
And so mister y will be 1 minus x here.
That's how you got it.
And finally, z is that-- mister z foes from the floor
all the way-- imagine somebody who
is like-- z is a helium balloon, and he
is left-- you let him go from the floor,
and he goes all the way to the ceiling.
And the ceiling is not flat like our ceiling.
The ceiling is this oblique plane.
So z is going to hit his head against the roof at some point,
and he doesn't know where he is going
to hit his head, unless you tell him where that happens.
So he knows he leaves at 0, and he's going to end up where?
STUDENT: 1 minus y minus x.
MAGDALENA TODA: Excellent.
1 minus x minus y.
How do we do that?
We pull z out of that, and say, 1 minus x minus y.
So that is the equation of the shaded purple plane,
and this is as far as you can go.
You cannot go past the roof of your house,
which is the purple plane, the purple shaded plane.
So here you are.
Is this hard?
No.
In many problems on the final and on the midterm,
we tell you, don't even think about solving that,
because we believe you.
Just set up the integral.
I might give you something like that again, just
set up the integral and you have to do that.
But now, I would like to actually work it out,
see how hard it is.
So is this hard to work this out?
I have to do it one at a time, because you see,
I don't have fixed endpoints.
I cannot say, I'm applying the problem with the integral if f
times the integral of g, times-- so I have to integrate one
at a time, because I don't have fixed endpoints.
And the integral of 1dz is z between that and that.
So z, 1 minus x minus y will be what's left over,
and then I have dy, and then I have dx.
And at this point it looks horrible enough,
but we have to pray that in the end
it's not going to be so hard, and I'm going to keep going.
So we have integral from 0 to 1.
We have integral from 0 to 1 minus x..
I'll just copy and paste it.
Which is integral from 0 to 1.
Now I have to think, and that's dangerous.
I have 1 minux x with respect to y.
This is going to be ugly.
That's a constant with respect to y, and times y,
minus-- integrate with respect to y, y is [? what? ?]
y squared over 2.
Between y equals 0 down.
That's going to save my life, because for y equals 0,
0 is going to be a great simplification.
And for y equals 1 minus x on top,
hopefully it's not going to be the end of the world.
It looks ugly now, but I'm an optimistic person,
so I hope that this is going to get better.
And I can see it's going to get better.
So I have integral from here to 1.
And now I say, OK, let me think.
Life is not so bad.
Why?
1 minus x, 1 minus x is 1 minus x squared.
I could think faster, you could think faster than me,
but I don't want to rush.
1 minus x squared over 2.
So it's not bad at all.
Look, I'm getting this guy who is beautiful in the end, when
I'm going to integrate, and you have
to keep your fingers crossed for me,
because I don't know what I'm going to get.
So I get integral from 0 to 1, 1/2 out, 1 minus x squared dx.
Is this bad?
Can you do this by yourself without my help?
What are you going to do?
x squared minus 2x plus 1.
That's the square.
STUDENT: Why not just change it?
MAGDALENA TODA: Huh?
STUDENT: Why not just change it?
MAGDALENA TODA: You can do it in many ways.
You can do whatever you want.
I don't care.
I want you to the right answer one way or another.
So I'm going to clean a little bit around here.
It's dirty.
You do it.
You have one minute and a half to finish.
And tell me what you get.
STUDENT: 1 minus x cubed over six negative.
MAGDALENA TODA: No, no.
In the end is the number.
What number?
But you have to go slow.
I need three people to give me the same answer.
Because then it's like in that proverb, if two people tell
you drunk, you go to bed.
I need three people to tell me what the answer is in order
to believe them.
Three witnesses.
STUDENT: 1 [? by ?] 6.
MAGDALENA TODA: Who got 1 over 6, raise hand?
Wow, guys, you're fast.
Can you raise hands again?
OK, being fast doesn't mean you're the best,
but I agree you do a very good job, all of you in general.
So I believe there were eight people or nine people.
1 over 6.
Now, how could I have cheated on this problem on the final?
STUDENT: It's a [? junction ?] from this--
MAGDALENA TODA: Right.
In this case, being a volume, I would have been lucky enough,
and say, it is the volume of a tetrahedron.
I go, the tetrahedron has area of the base 1/2,
the height is 1.
1/2 times 1 divided by 3 is 1/6.
And just pretend on the final that I actually
computed everything.
I could have done that, from here jump to here, or from here
jump straight to here.
And ask you, how did you get from here to here?
And you say, I'm a genius.
Could I not believe you?
I have to give you full credit.
However, what would you have done if I said compute,
I don't know, something worse, something
like triple integral of x, y, z over the tetrahedron 2.
In that case, you cannot cheat.
You're not lucky enough to cheat.
You're lucky enough to cheat when
you have a volume of a prism, you
have a volume of-- and volume means this should be the number
1 here, number 1.
So if you have number 1, here, or I ask you for the volume,
and it's a prism, or tetrahedron, or sphere,
or something, go ahead and cheat,
and pretend that you're actually solving the integral.
Yes, sir.
STUDENT: What would that represent, geometrically,
the triple integral of x, y, z?
MAGDALENA TODA: It's a weighted triple integral.
I'm going to give you examples later.
When you have mass and momentum, when you compute the center
map, or you compute the mass, and somebody give you
densities.
Let me get -- If you have a triple integral over row at x,
y, z, this could be it, but I [? recall ?] it row
for a reason, not just for fun.
And here, dx, dy, dz.
Very good question, and it's very insightful.
For a physicist or engineer, the guy
needs to know why we take this weighted [? integral. ?]
If row is the density of an object,
if it's everywhere the same, if row is a homogeneous density,
for that piece of cheddar cheese-- Oh my God
I'm so hungry-- row would be constant.
If it's a quality cheddar made in Vermont in the best
factory, whatever, row would be considered to be a constant,
right?
And in that case, what happens?
If it's a constant, it's a gets out,
and then you have row times triple integral 1
dV, which is what?
The volume.
And then the volume times the density of the piece of cheese
will be?
STUDENT: [INAUDIBLE]
MAGDALENA TODA: The mass of the piece of cheese,
in kilograms, because I think in kilograms because I
can eat more.
OK?
Actually, no, I'm just kidding.
You guys have really-- I mean, 2 pounds and 1 kilogram
is not the same thin.
Can somebody tell me why?
I mean, you know it's not the same thing because,
the approximation.
But I'm claiming you cannot compare pounds with kilograms
at all.
STUDENT: Pounds is a measure of weight,
whereas kilograms is a measure of mass.
MAGDALENA TODA: Excellent.
Kilogram is a measure of mass, pound
is a measure of the gravitational force.
It's a force measure.
So OK.
Which reminds me, there was-- I don't
know if you saw this short movie for 15 minutes that
got an award the previous Oscar last year,
and there was an old lady telling another old lady
in Great Britain, get 2 pounds of sausage.
And the other one says, I thought we got metric,
because we are in the European Union.
And she said, then get me just the one meter of sausage,
or something.
So it was funny.
So it can be mass.
But what if this density is not the same?
This is exactly why we need to do the integral.
Imagine that the density is-- we have
a piece of cake with layers.
And again, you see how hungry I am.
So you have a layer, and then cream, or whipped cream,
or mousse, and another layer, and another mousse.
The density will vary.
But then there are bodies in physics where the density is
even a smooth function.
It doesn't matter that you have such a discontinuous function.
What would you do?
You just split.
You have triple row 1 for the first layer, then triple row 2
for the second later, the layer of mousse,
and then let's say it's tiramisu,
you have another layer, row three, dV3 for the top layer
of the tiramisu.
STUDENT: Can any row be kept constant?
MAGDALENA TODA: So these are discontinuous.
They are all constant, though.
That would be the great advantage,
because presumably mousse would have the constant density,
the dough has a constant, homogeneous density, and so on.
But what if the density varies in that body from point
to point?
Then nobody can do it by approximation.
You'd say volume, mass 1 plus mass 2 plus mass 3 plus mass 1.
You have to have a triple integral where this row varies,
constantly varies.
And for an engineer, that would be a puzzle.
Poor engineers says, oh my God, the density
is different from one point to another.
I have to find an approximated function
for that density moving from one point to another on that body.
And then the only way to do it would be to solve an integral.
Imagine that somebody-- now it just occurred,
I never thought about it-- we would
be measured in terms of this type of integral.
Of course, people would be able to measure mass right away.
But then, if you were to know the density--
you cannot even know the density at every point of the body.
It varies a lot, so every point of our bodies
has a different material and a density.
OK.
STUDENT: Tiramisu. [INAUDIBLE]
MAGDALENA TODA: Huh?
STUDENT: So you use the tiramasu,
you're making me hungry.
MAGDALENA TODA: Yeah, because now,
OK take your mind off the tiramisu.
Think about an exam.
Then you don't--
STUDENT: Now I'm sick.
MAGDALENA TODA: Exactly.
Now you need something against nausea.
Let's see what else is interesting to do.
I'll give you ten minutes.
How much did I steal from you?
I stole constantly about five minutes of your breaks
for the last few Tuesdays.
STUDENT: So the integral--
MAGDALENA TODA: The integral of that.
I think I would be fair to give you 10 minutes
as a gift today to compensate.
OK, so remind me to let you go 10 minutes early.
Especially since spring break is coming.
We have a 3D application.
We have several 3D applications.
Let me see which one I want to mimic first.
Yeah.
I'm going to pick my favorite, because I just want to.
So imagine you have a disc that is
x squared plus y squared equals 1 would be the circle.
That's the unit disc on the floor.
And then I have the plane x plus y plus z equals 8.
Then I'm going to draw that plane.
I'll try my best.
It's similar to two examples from the book,
but I did not want to repeat the ones in the book
because I want you to actually read them.
That's kind of the idea.
So you have this picture, and you
realize that we had that in the first octant before.
So I say, I don't want the volume
of the body over the whole disc, only over the part of the disc
which is in the first octant.
So I say, I want this domain D, which is going to be what?
x squared plus y squared less than or equal to 1
in plane, with x positive, y positive.
Do you know what we call that in trigonometry?
Does anybody know what we call this in trigonometry?
Let me put the points while you think.
Hopefully, you are thinking about this.
This is 1 in x-axis.
1, 0, 0, and this is 0, 1, 0, and this is y.
If I were to go up until I meet the plane,
what point would this--
STUDENT: [INAUDIBLE]
MAGDALENA TODA: What point would this-- on the thing.
STUDENT: 1, 0, 7 and then 0, 1, 7.
MAGDALENA TODA: 1, 0, 7.
This would be you said 0, 1, 6.
And this would be 1, 0, 7.
How did you think about this?
How do you know?
STUDENT: Y plus z--
MAGDALENA TODA: Because z, because it's on the y-axis,
and since you are on the x-axis here, y has to be 0.
So you're right.
Very good.
Excellent.
Now I'm going to say, I'd like to know
the-- compute the volume of the body that is bounded above
from above by x plus y plus z equals 8, who's projection
on the floor is the domain D. And I'll say
volume of the cylindrical body.
So how could you obtain such a, again-- No,
this is Murphy's Law.
OK, how could you obtain such an object, such a cylinder?
STUDENT: Take a pencil, and cut it into fourths.
MAGDALENA TODA: Huh?
STUDENT: Take like a cylindrical pencil and cut it into fourths.
MAGDALENA TODA: Take a salami, a piece of salami.
Cut that piece of salami into four, into four quarters.
And then we take, we slice, and we slice like that.
So we have something like--
STUDENT: I tried to think of a non- food example.
MAGDALENA TODA: --a quarter.
How can I draw this?
OK, this is what it means.
You don't see this one.
You don't see this part.
This is curved.
And here, instead of cutting with another perpendicular
plane, along the salami-- so this
is the axis of the salami-- instead of taking the knife
and cutting like that, I'm cutting an oblique plane,
and this is what this oblique plane will do.
STUDENT: If you cut that way, then you
would have only squares.
MAGDALENA TODA: Hmm?
So I'm going to have some oblique-- I cannot draw better.
I don't know how to draw better.
So it's going to be an oblique cut in the salami.
Let's think how we do this problem.
Elementary, it will be a piece of cake--
it would be a piece of--
STUDENT: A piece of salami.
MAGDALENA TODA: No.
It wouldn't be apiece of salami.
STUDENT: It could be done.
MAGDALENA TODA: How could we do that quickly with the Calculus
III we know?
STUDENT: Find the triple integral.
Oh, you want us to do the double integral?
MAGDALENA TODA: Double, triple, I don't know what to do.
What do you think is best?
Let's do that triple integral first,
and you'll see that it's the same thing as double integral.
Triple integral over B, the body of the salami, 1 dV.
How can we set it up?
Well, this is a little bit tricky.
It's going to be like that.
We can say, I have a double integral over my domain,
D. When it comes to the z, mister z has to be first.
So mister z says, I'm first.
I know where I'm going.
You guys, x and y are bound together,
mired in the element of area of the circles.
This is like dx dy.
But I am independent from you.
I am z.
So I'm going all the way from the floor to what?
You taught me that.
8 minus x minus y, and 1.
This is the way to do it as a triple integral,
but then Alex will say, I could have
done this as a double integral.
Let me show you how.
I could have done it over the domain D in plane.
Put the function, 8 minus x minus y
is [? B and ?] z from the very beginning,
because that's my altitude function, f of x and y.
So then I say dx dy, dx dy, it doesn't matter.
That's the only theory element.
Fine.
It's the same thing.
This is what I wanted you to observe.
Whether you view it like the triple integral like that,
or you view it as the double integral like that,
it's the same thing.
This is not a headache.
The headache is coming next.
This is not a headache.
So you can do it in two ways.
And I'd like to look at the-- check
the two methods of doing this.
And set up the integrals without solving them.
Can you read my mind?
Do you realize what I'm asking?
Imagine that would be on the midterm.
What do you think I'm asking, the two methods?
This can be interpreted in many ways.
There are two methods.
I mean, one method by doing it with Cartesian
coordinates x and y.
The other method is switching to polar coordinates
and set up the integral without solving.
And you say, why not solving?
Because I'm going to cheat.
I'm going to use a TI-92 to solve it,
or I'm going to use a Matlab or Maple.
If it looks a little bit complicated,
then I don't want to spend my time.
Actually, engineers, after taking Calc III,
they know a lot.
They understand a lot about volumes, areas.
But do you think if you work on a real-life problem like that,
that your boss will let you waste your time
and do the integral by hand?
STUDENT: No.
MAGDALENA TODA: Most integrals are really complicated
in everyday life.
So what you're going to do is going
to be a scientific software, like Matlab, which is primarily
for engineers, Mathematica, which is similar to Matlab,
but is mainly for mathematicians.
It was invented at the University
of Illinois Urbana-Champaign, and they're still
very proud of it.
I prefer Matlab because I feel Matlab
is stronger, has higher capabilities than Mathematica.
You can use Maple.
Maple lets you set up the endpoints even as functions.
And then it's user friendly, you type in this,
you type in the endpoints.
It has little windows, here.
You don't need to know any programming.
It's made for people who have no programming skills.
So it's going to show a little window on top,
here, here, here and here.
You [? have ?] those, and you press Enter,
and it's going to spit the answer back at you.
So this is how engineers actually
solve the everyday integrals.
Not by hand.
I want to be able to set it up in both ways
before I go home or eat something, right?
So we don't have to spend a lot of time on it.
But if you want to tell me how I am going to set it up,
I would be very grateful.
So this is Cartesian, and this is polar.
All right.
Who helps me?
In Cartesian-- which one do you prefer?
I mean, it doesn't matter.
You guys are good and smart, and you'll
figure out what I need to do.
If I want to do it in terms of vertical strip-- so
for vertical strip method-- first
I integrate with respect to y, and then with respect to x.
And maybe, to test your understanding,
let me change the order of integrals
and see how much you understood from that last time.
STUDENT: [INAUDIBLE]
MAGDALENA TODA: So x is between what and what?
STUDENT: 0 and 1.
MAGDALENA TODA: Look at this picture.
I have to reproduce this picture like that.
0 to 1, says Alex, and he's right.
And why will he decide against--
STUDENT: 1 minus x squared.
MAGDALENA TODA: --square root 1 minus x squared.
So we know very well what we are going to do,
what Maple is going to do for us.
1 square root 1 minus x squared.
And then what do I put here?
8 minus x minus y.
Can I do it by hand?
Yes, I guarantee to you I can do it by hand.
Let me tell you why.
Because when we integrate with respect to y, I get xy.
So I get xy, and y will be plugged in 1 minus x squared.
How am I going to solve an integral like this?
I can the first one with a table, the second one with a u
substitution.
On the last one is a little bit painful.
I'm going to have y squared over 2--
STUDENT: That's the easiest part.
MAGDALENA TODA: According to Alex, yes, you're right.
Maybe that is the easiest.
STUDENT: That's the [INAUDIBLE] part you can integrate--
MAGDALENA TODA: And I can integrate one at a time,
and I'm going to waste all my time.
So if I want to be an efficient engineer,
and my boss is waiting for the end-of-the-day project,
of course I'm not going to do this by hand.
How about the other integral?
Same integral.
Same idea, y between 0 and 1.
And x between 0 and
STUDENT: Square root of 1 minus y squared.
MAGDALENA TODA: Square root of 1 minus y squared.
Because I'll do this guy with horizontal strips,
and forget about the vertical strips.
And here's the y-- I rotate my head and it cracks,
so that means that I need some yoga.
y is between 0 and 1.
Or gymnastics.
So x is between 0 and square root
1 minus y squared. [INAUDIBLE].
And I'll leave it here on the meter.
And I'm going to make a sample like I promised.
OK, good.
How would you do this to set up the polar coordinate integral?
And that is why Alex said maybe that's a pain because
of a reason.
And he's right, it's a little bit painful to solve by hand.
But again, once you switch to polar,
you can solve it with a calculator or a computer
software, scientific software in no time.
In Maple, you just have to plug in the numbers.
You cannot plug in theta, I think, as a symbol.
I'm not sure.
But you can put theta as t and r will be r,
or you can use whatever letters you
want that are roman letters.
So you have to integrate smartly, here,
switching to r and theta, and think
about the meaning of that.
So first of all, if I put dr d theta,
I'm not worried that you won't be able to get r and theta,
because I know you can do it.
You can prove it to me right now. r between 0 and
STUDENT: 1.
MAGDALENA TODA: Excellent.
And theta, pay attention, between 0 and
STUDENT: pi over 2
MAGDALENA TODA: Excellent.
I'm proud.
Yes, sir?
STUDENT: Is it supposed to be r dr 2 theta,
or are you going to add that later?
MAGDALENA TODA: I will add it here.
So the integrand will contain the r.
Now what do I put in terms of this?
I left enough room.
STUDENT: Is it pi over 2, or is it negative pi over 2?
MAGDALENA TODA: It doesn't matter,
because I'll have to take that-- we assume always theta to go
counterclockwise, and go between 0 and pi over 2,
so that when you start-- let me make this motion.
You are here at theta equals 0.
STUDENT: Oh, OK.
Sorry.
I got my coordinates mixed around--
MAGDALENA TODA: --and counterclockwise to pi over 2.
[INTERPOSING VOICES]
MAGDALENA TODA: Yeah.
So you go in the trigonometric-- Here, you have 8 minus,
and who tells me what I'm supposed to type?
STUDENT: r over x.
MAGDALENA TODA: r cosine theta minus
STUDENT: Sine.
MAGDALENA TODA: r sine theta.
And let mister whatever his name is,
the computer, find the answer.
Can I do it by hand?
Actually, I can.
I can, but again, it's not worth it, because it drives me crazy.
How would I do it by hand?
I would split the integral into three,
and I would easily compute 8 times r,
integrand is going to be easy.
Right?
Agree with me?
Then what am I going to do?
I'm going to say, an r out times an r, out comes r squared.
And I have integral of r squared times a function
of theta only, which is going to be
sine theta plus cosine theta.
We are going to say, yes, with a minus, with a minus.
Now, when I compute r and theta thingy,
theta will be between 0 and pi over 2.
r will be between 0 and 1.
But I don't care, because Matthew reminded me,
if you have a product of separate variables,
life becomes all of the sudden easier for you.
STUDENT: You've also got to add your integral of [? 8r ?]
[? dr. ?]
MAGDALENA TODA: Yeah.
At the end, I'm going to add the integral of 8r.
So I take them separately.
I just look at one chunk.
And this chunk will be what?
Can you even see how easy it's going to be with the naked eye?
Firs of all, integral from 0 to 1,
r squared dr is a piece of cake.
How much is that-- piece of salami.
STUDENT: 1/3.
MAGDALENA TODA: 1/3.
Right?
Because it's r cubed over 3.
Then you have 1/3.
That's easy.
With a minus in front, but I don't care about it in the end.
What is the integral of sine theta cosine theta?
STUDENT: Negative [INAUDIBLE].
MAGDALENA TODA: Minus cosine theta plus sine theta
taken between 0 and pi over 2.
Will this be hard?
Who's going to tell me what, or how I'm going to get what--
we don't compute it now, but I just give you.
Cosine of pi over 3 is?
STUDENT: 0.
MAGDALENA TODA: 0.
Sine of pi over 2 is?
STUDENT: Oh yeah.
1.
MAGDALENA TODA: 1.
So this is going to be 1 minus, what's
the whole thingy computed at 0?
STUDENT: [INAUDIBLE].
MAGDALENA TODA: It's going to be minus 1, but minus minus 1
is plus 1.
So I have 2.
So only this chunk of the integral would be easy.
Minus 2/3.
OK?
So it can be done by hand, but why waste the time when
you can do it with Maple?
Yes, sir?
STUDENT: Where did you get rid of 8?
On the second, after the 8--
MAGDALENA TODA: No, I didn't.
That's exactly what we were talking.
Alex says, but you just talked about integral of 8r,
but you didn't want to do it.
I said, I didn't want to do it.
This is just the second chunk of this integral.
So I know that I can do integral of integral of 8r in no time.
Then I would need to take this and add that,
and get the number.
I don't care about the number.
I just care about the method.
Yes, sir?
STUDENT: Why are the limits from 0 to 1 instead of like 0
to r squared?
Because didn't we say earlier the domain
is x squared plus y squared?
Wouldn't that be r squared?
MAGDALENA TODA: No.
No, wait.
This is r squared.
STUDENT: Right.
Why didn't we plug r squared into the 1 again.
MAGDALENA TODA: And that means r is between 0 and 1, right?
STUDENT: Oh, OK.
MAGDALENA TODA: r squared being less than 1.
That means r is between 0 and 1.
OK?
And one last problem-- no.
No last problem.
We have barely 10 minutes.
So you read from the book some.
I will come back to this section, and I'll do review.
Have a wonderful spring break, and I'm
going to see you after spring break on Tuesday.
[INTERPOSING VOICES]