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All right, so the past few
weeks,
we've been looking at double
integrals and the plane,
line integrals in the plane,
and will we are going to do now
from now on basically until the
end of the term,
will be very similar stuff,
but in space.
So, we are going to learn how
to do triple integrals in space,
flux in space,
work in space,
divergence, curl,
all that.
So,
that means, basically,
if you were really on top of
what we've been doing these past
few weeks,
then it will be just the same
with one more coordinate.
And, you will see there are
some differences.
But, conceptually,
it's pretty similar.
There are a few tricky things,
though.
Now, that also means that if
there is stuff that you are not
sure about in the plane,
then I encourage you to review
the material that we've done
over the past few weeks to make
sure that everything in the
plane is completely clear to you
because it will be much harder
to understand stuff in space if
things are still shaky in the
plane.
OK, so the plan is we're going
to basically go through the same
stuff, but in space.
So, it shouldn't be surprising
that we will start today with
triple integrals.
OK, so the way triple integrals
work is if I give you a function
of three variables,
x, y, z,
and I give you some region in
space,
so, some solid,
then I can take the integral
over this region over function f
dV where dV stands for the
volume element.
OK, so what it means is we will
just take every single little
piece of our solid,
take the value of f there,
multiply by the small volume of
each little piece,
and sum all these things
together.
And,
so this volume element here,
well, for example,
if you are doing the integral
in rectangular coordinates,
that will become dx dy dz or
any permutation of that because,
of course, we have lots of
possible orders of integration
to choose from.
So, rather than bore you with
theory and all sorts of
complicated things,
let's just do examples.
And, you will see, basically,
if you understand how to set up
iterated integrals into
variables,
that you basically understand
how to do them in three
variables.
You just have to be a bit more
careful.
And, there's one more step.
OK, so let's take our first
triple integral to be on the
region.
So, of course,
there's two different things as
always.
There is the region of
integration and there's the
function we are integrating.
Now, the function we are
integrating, well,
it will come in handy when you
actually try to evaluate the
integral.
But, as you can see,
probably, the new part is
really hard to set it up.
So, the function won't really
matter that much for me.
So, in the examples I'll do
today, functions will be kind of
silly.
So, for example,
let's say that we want to look
at the region between two
paraboloids, one given by z = x
^2 y ^2.
The other is z = 4 - x ^2 - y
^2.
And, so, I haven't given you,
yet, the function to integrate.
OK, this is not the function to
integrate.
This is what describes the
region where I will integrate my
function.
And, let's say that I just want
to find the volume of this
region, which is the triple
integral of just one dV.
OK, similarly,
remember, when we try to find
the area of the region in the
plane, we are just integrating
one dA.
Here we integrate one dV.
that will give us the volume.
Now, I know that you can
imagine how to actually do this
one as a double integral.
But, the goal of the game is to
set up the triple integral.
It's not actually to find the
volume.
So, what does that look like?
Well, z = x ^2 y ^2,
that's one of our favorite
paraboloids.
That's something that looks
like a parabola with its bottom
at the origin that you spin
about the z axis.
And, z equals four minus x
squared minus y squared,
well, that's also a paraboloid.
But, this one is pointing down,
and when you take x equals y
equals zero, you get z equals
four.
So, it starts at four,
and it goes down like that.
OK, so the solid that we'd like
to consider is what's in between
in here.
So, it has a curvy top which is
this downward paraboloid,
a curvy bottom which is the
other paraboloid.
And, what about the sides?
Well, do you have any idea what
we get here?
Yeah, it's going to be a circle
because entire picture is
invariant by rotation about the
z axis.
So, if you look at the picture
just, say, in the yz plane,
you get this point and that
point.
And, when you rotate everything
around the z axis,
you will just get a circle
here.
OK, so our goal is to find the
volume of this thing,
and there's lots of things I
could do to simplify the
calculation,
or even not do it as a triple
integral at all.
But, I want to actually set it
up as a triple integral just to
show how we do that.
OK, so the first thing we need
to do is choose an order of
integration.
And, here, well,
I don't know if you can see it
yet, but hopefully soon that
will be intuitive to you.
I claim that I would like to
start by integrating first over
z.
What's the reason for that?
Well, the reason is if I give
you x and y, then you can find
quickly, what's the bottom and
top values of z for that choice
of x and y?
OK, so if I have x and y given,
then I can find above that:
what is the bottom z and the
top z corresponding to the
vertical line above that point?
The portion of it that's inside
our solid, so somehow,
there's a bottom z and a top z.
And, so the top z is actually
on the downward paraboloid.
So, it's four minus x squared
minus y squared.
The bottom value of z is x
squared plus y squared.
OK, so if I want to start to
set this up, I will write the
triple integral.
And then, so let's say I'm
going to do it dz first,
and then, say,
dy dx.
It doesn't really matter.
So then, for a given value of x
and y, I claim z goes from the
bottom surface.
The bottom face is z equals x
squared plus y squared.
The top face is four minus x
squared minus y squared.
OK, is that OK with everyone?
Yeah?
Any questions so far?
Yes?
Why did I start with z?
That's a very good question.
So, I can choose whatever order
I want, but let's say I did x
first .
Then, to find the inner
integral bounds,
I would need to say, OK,
I've chosen values of,
see, in the inner integral,
you've fixed the two other
variables,
and you're just going to vary
that one.
And, you need to find bounds
for it.
So, if I integrate over x
first, I have to solve,
answer the following question.
Say I'm given values of y and z.
What are the bounds for x?
So, that would mean I'm slicing
my solid by lines that are
parallel to the x axis.
And, see, it's kind of hard to
find, what are the values of x
at the front and at the back?
I mean, it's possible,
but it's easier to actually
first look for z at the top and
bottom.
Yes?
dy dx, or dx dy?
No, it's completely at random.
I mean, you can see x and y
play symmetric roles.
So, if you look at it,
it's reasonably clear that z
should be the easiest one to set
up first for what comes next.
xy or yx, it's the same.
Yes?
Yes, it will be easier to use
cylindrical coordinates.
I'll get to that just as soon
as I'm done with this one.
OK, so let's continue a bit
with that.
And, as you mentioned,
actually we don't actually want
to do it with xy in the end.
In a few minutes,
we will actually switch to
cylindrical coordinates.
But, for now,
we don't even know what they
are.
OK, so I've done the inner
integral by looking at,
you know, if I slice by
vertical lines,
what is the top?
What is the bottom for a given
value of x and y?
So, the bounds in the inner
integral depend on both the
middle and outer variables.
Next, I need to figure out what
values of x and y I will be
interested in.
And, the answer for that is,
well, the values of x and y
that I want to look at are all
those that are in the shade of
my region.
So, in fact,
to set up the middle and outer
bounds, what I want to do is
project my solid.
So, my solid looks like this
kind of thing.
And, I don't really know how to
call it.
But, what's interesting now is
I want to look at the shadow
that it casts in the xy plane.
OK, and, of course,
that shadow will just be the
disk that's directly below this
disk here that's separating the
two halves of the solid.
And so, now I will want to
integrate over,
I want to look at all the xy's,
x and y, in the shadow.
So, now I'm left with,
actually, something we've
already done,
namely setting up a double
integral over x and y.
So, if it helps,
here, we don't strictly need
it, but if it helps,
it could be useful to actually
draw a picture of this shadow in
the xy plane.
So, here it would just look,
again, like a disk,
and set it up.
Now, the question is,
how do we find the size of this
disk, the size of the shadow?
Well, basically we have to
figure out where our two
paraboloids intersect.
There's nothing else.
OK, so, one way how to find the
shadow in the xy plane -- --
well,
here we actually know the
answer a priori,
but even if we didn't,
we could just say,
well, our region lives wherever
the bottom surface is below the
top surface,
OK, so we want to look at
things wherever bottom value of
z is less than the top value of
z,
I mean, less or less than or
equal, that's the same thing.
So, if the bottom value of z is
x squared plus y squared should
be less than four minus x
squared minus y squared,
and if you solve for that,
then you will get,
well, so let's move these guys
over here.
You'll get two x squared plus
two y squared less than four.
That becomes x squared plus y
squared less than two.
So, that means that's a disk of
radius square root of two,
OK?
So, we kind of knew in advance
it was going to be a disk,
but what we've learned now is
that this radius is square root
of two.
So, if we want to set up,
if we really want to set it up
using dy dx like they started,
then we can do it because we
know,
so, for the middle integral,
now,
we want to fix a value of x.
And, for that fixed value of x,
we want to figure out the
bounds for y.
Well, the answer is y goes from
here to here.
What's here?
Well, here, y is square root of
two minus x squared.
And, here it's negative square
root of two minus x squared.
So, y will go from negative
square root of two minus x
squared to positive square root.
And then, x will go from
negative root two to root two.
OK, if that's not completely
clear to you,
then I encourage you to go over
how we set up double integrals
again.
OK, does that make sense,
kind of?
Yeah?
Well, so, when we set up,
remember, we are setting up a
double integral,
dy dx here.
So, when we do it dy dx,
it means we slice this region
of a plane by vertical line
segments.
So, this middle guy would be
what used to be the inner
integral.
So, in the inner,
remember, you fix the value of
x, and you ask yourself,
what is the range of values of
y in my region?
So, y goes from here to here,
and what here and here are
depends on the value of x.
How?
Well, we have to find the
relation between x and y at
these points.
These points are on the circle
of radius root two.
So, if you want this circle
maybe I should have written,
is x squared plus y squared
equals two.
And, if you solve for y,
given x, you get plus minus
root of two minus x squared,
OK?
Yes?
Is there a way to compute this
with symmetry?
Well, certainly,
yeah, this solid looks
sufficiently symmetric,
but actually you could
certainly,
if you don't want to do the
whole disk,
you could just do quarter
disks,
and multiply by four.
You could even just look at the
lower half of the solid,
and multiply them by two,
so, total by eight.
So, yeah, certainly there's
lots of ways to make it slightly
easier by using symmetry.
Now, the most spectacular way
to use symmetry here,
of course, is to use that we
have this rotation symmetry and
switch,
actually, not do this guy in xy
coordinates but instead in polar
coordinates.
So -- So, the smarter thing to
do would be to use polar
coordinates instead of x and y.
Of course, we want to keep z.
I mean, we are very happy with
z the way it is.
But, we'll just change x and y
to R cos theta,
R sine theta,
OK, because,
well, let's see actually how we
would evaluate this guy.
So, well actually, let's not.
It's kind of boring.
So, let me just point out one
small thing here,
sorry, before I do that.
So, if you start computing the
inner integral,
OK, so let me not do that yet,
sorry,
so if you try to compute the
inner integral,
you'll be integrating from x
squared plus y squared to four
minus x squared minus y squared
dz.
Well, that will integrate to z
between these two bounds.
So, you will get four minus two
x squared minus two y squared.
Now, when you put that into the
remaining ones,
you'll get something that's
probably not very pleasant of
four minus two x squared minus
two y squared dy dx.
And here, you see that to
evaluate this,
you would switch to polar
coordinates.
Oh, by the way,
so if your initial instincts
had been to,
given that you just want the
volume,
you could also have found the
volume just by doing a double
integral of the height between
the top and bottom.
Well, you would just have
gotten this, right,
because this is the height
between top and bottom.
So, it's all the same.
It doesn't really matter.
But with this,
of course, we will be able to
integrate all sorts of
functions, not just one over the
solid.
So, we will be able to do much
more than just volumes.
OK, so let's see,
how do we do it with polar
coordinates instead?
Well, so -- Well,
that would become,
so let's see.
So, I want to keep dz.
But then, dx dy or dy dx would
become r dr d theta.
And, if I try to set up the
bounds, well,
I probably shouldn't keep this
x squared plus y squared around.
But, x squared plus y squared
is easy in terms of r and theta.
That's just r squared.
OK, I mean, in general I could
have something that depends also
on theta.
That's perfectly legitimate.
But here, it simplifies,
and this guy up here,
four minus x squared minus y
squared becomes four minus r
squared.
And now, the integral that we
have to do over r and theta,
well, we look again at the
shadow.
The shadow is still a disk of
radius root two.
That hasn't changed.
And now, we know how to set up
this integral in polar
coordinates.
r goes from zero to root two,
and theta goes from zero to two
pi.
OK, and now it becomes actually
easier to evaluate.
OK, so now we have actually a
name for this because we're
doing it in space.
So, these are called,
actually, cylindrical
coordinates.
So, in fact,
you already knew about
cylindrical coordinates even if
you did not know the name.
OK, so the idea of cylindrical
coordinates is that instead of
x, y, and z, to locate a point
in space, you will use three
coordinates.
One of them is basically how
high it is above the xy plane.
So, that will be z.
And then, you will use polar
coordinates for the projection
of your point on the xy plane.
So, r will be the distance from
the z axis.
And theta will be the angle
from the x axis
counterclockwise.
So, the one thing to be careful
about is because of the usual
convention, that we make the x
axis point toward us.
Theta equals zero is no longer
to the right.
Now, theta equals zero is to
the front, and the angel is
measured from the front
counterclockwise.
OK, so,
and of course,
if you want to know how to
convert between x,
y, z and r theta z,
well, the formulas are just the
same as in usual polar
coordinates.
R cos theta,
r sine theta,
and z remain z.
OK, so why are these called
cylindrical coordinates,
by the way?
Well, let's say that I gave you
the equation r equals a,
where a is some constant.
Say r equals one, for example.
So, r equals one in 2D,
that used to be just a circle
of radius one.
Now, in space,
a single equation actually
defines a surface,
not just a curve anymore.
And, the set of points where r
is a, well, that's all the
points that are distance a from
the z axis.
So, in fact,
what you get this way is a
cylinder of radius a centered on
the z axis.
OK, so that's why they are
called cylindrical coordinates.
By the way, so now,
similarly, if you look at the
equation theta equals some given
value, well, so that used to be
just a ray from the origin.
Now, that becomes a vertical
half plane.
For example,
if I set the value of theta and
let r and z vary,
well, r is always positive,
but basically that means I am
taking a vertical plane that
comes out in this direction.
OK, any questions about
cylindrical coordinates?
Yes?
Yeah, so I'm saying when you
fix theta, you get only a half
plane, not a full plane.
I mean, it goes all the way up
and down, but it doesn't go back
to the other side of the z axis.
Why?
That's because r is always
positive by convention.
So, for example,
here, we say theta is zero.
At the back, we say theta is pi.
We don't say theta is zero and
r is negative.
We say r is positive and theta
is pi.
It's a convention, largely.
But, sticking with this
convention really will help you
to set up the integrals
properly.
I mean, otherwise there is just
too much risk for mistakes.
Yes?
Well, so the question is if I
were to use symmetry to do this
one, would I multiply by four or
by two?
Well, it depends on how much
symmetry you are using.
So, I mean, it's your choice.
You can multiply by two,
by four, by eight depending on
how much you cut it.
So, it depends on what symmetry
you use, if you use symmetry
between top and bottom you'd
say, well, the volume is twice
the lower half.
If you use the left and right
half, you would say it's twice
each half.
If you cut it into four pieces,
and so on.
So, and again,
you don't have to use the
symmetry.
If you don't think of using
polar coordinates,
then it can save you from
doing,
you know, you can just start at
zero here and here,
and simplify things a tiny bit.
But, OK, yes?
So, to define a vertical full
plane, well, first of all it
depends on whether it passes
through the z axis or not.
If it doesn't,
then you'd have to remember how
you do in polar coordinates.
I mean, basically the answer
is, if you have a vertical
plane, so, it doesn't depend on
z.
The equation does not involve z.
It only involves r and theta.
And, how it involves r and
theta is exactly the same as
when you do a line in polar
coordinates in the plane.
So, if it's a line passing
through the origin,
you say, well,
theta is either some value or
the other one.
If it's a line that doesn't
passes to the origin,
but it's more tricky.
But hopefully you've seen how
to do that.
OK, let's move on a bit.
So, one thing to know,
I mean, basically,
the important thing to remember
is that the volume element in
cylindrical coordinates,
well, dx dy dz becomes r dr d
theta dz.
And, that shouldn't be
surprising because that's just
dx dy becomes r dr d theta.
And, dz remains dz.
I mean, so, the way to think
about it,
if you want,
is that if you take a little
piece of solid in space,
so it has some height, delta z,
and it has a base which has
some area delta A,
then the small volume, delta v,
is equal to the area of a base
times the height.
So, now, when you make the
things infinitely small,
you will get dV is dA times dz,
and you can use whichever
formula you want for area in the
xy plane.
OK, now in practice,
you choose which order you
integrate in.
As you have probably seen,
a favorite of mine is z first
because very often you'll know
what the top and bottom of your
solid look like,
and then you will reduce to
just something in the xy plane.
But, there might be situations
where it's actually easier to
start first with dx dy or r dr d
theta, and then save dz for
last.
I mean, if you seen how to,
in single variable calculus,
the disk and shell methods for
finding volumes,
that's exactly the dilemma of
shells versus disks.
One of them is you do z first.
The other is you do z last.
OK, so what are things we can
do now with triple integrals?
Well, we can find the volume of
solids by just integrating dV.
And, we've seen that.
We can find the mass of a solid.
OK, so if we have a density,
delta, which,
remember, delta is basically
the mass divided by the volume.
OK, so the small mass element,
maybe I should have written
that as dm, the mass element,
is density times dV.
So now, this is the real
physical density.
If you are given a material,
usually, the density will be in
grams per cubic meter or cubic
inch, or whatever.
I mean, there is tons of
different units.
But, so then,
the mass of your solid will be
just the triple integral of
density, dV because you just sum
the mass of each little piece.
And, of course,
if the density is one,
then it just becomes the
volume.
OK,
now, it shouldn't be surprising
to you that we can also do
classics that we had seen in the
plane such as the average value
of a function,
the center of mass,
and moment of inertia.
OK, so the average value of the
function f of x,
y, z in the region,
r,
that would be f bar,
would be one over the volume of
the region times the triple
integral of f dV.
Or, if we have a density,
and we want to take a weighted
average -- Then we take one over
the mass where the mass is the
triple integral of the density
times the triple integral of f
density dV.
So, as particular cases,
there is, again,
the notion of center of mass of
the solid.
So, that's the point that
somehow right in the middle of
the solid.
That's the point mass by which
there is a point at which you
should put point mass so that it
would be equivalent from the
point of view of dealing with
forces and translation effects,
of course, not for rotation.
But, so the center of mass of a
solid is just given by taking
the average values of x,
y, and z.
OK, so there is a special case
where, so, x bar is one over the
mass times triple integral of x
density dV.
And, same thing with y and z.
And, of course,
very often, you can use
symmetry to not have to compute
all three of them.
For example,
if you look at this solid that
we had, well,
I guess I've erased it now.
But, if you remember what it
looked, well,
it was pretty obvious that the
center of mass would be in the z
axis.
So, no need to waste time
considering x bar and y bar.
And, in fact,
you can also find z bar by
symmetry between the top and
bottom, and let you figure that
out.
Of course, symmetry only works,
I should say,
symmetry only works if the
density is also symmetric.
If I had taken my guy to be
heavier at the front than at the
back, then it would no longer be
true that x bar would be zero.
OK, next on the list is moment
of inertia.
Actually, in a way,
moment of inertia in 3D is
easier conceptually than in 2D.
So, why is that?
Well, because now the various
flavors that we had come
together in a nice way.
So, the moment of inertia of an
axis,
sorry, with respect to an axis
would be,
again, given by the triple
integral of the distance to the
axis squared times density,
times dV.
And, in particular,
we have our solid.
And, we might skewer it using
any of the coordinate axes and
then try to rotate it about one
of the axes.
So, we have three different
possibilities,
of course, the x,
y, or z axis.
And, so now,
rotating about the z axis
actually corresponds to when we
were just doing things for flat
objects in the xy plane.
That corresponded to rotating
about the origin.
So, secretly,
we were saying we were rotating
about the point.
But actually,
it was just rotating about the
z axis.
Just I didn't want to introduce
the z coordinate that we didn't
actually need at the time.
So -- [APPLAUSE]
OK, so moment of inertia about
the z axis, so,
what's the distance to the z
axis?
Well, we've said that's exactly
r.
That's the cylindrical
coordinate, r.
So, the square of a distance is
just r squared.
Now, if you didn't want to do
it in cylindrical coordinates
then, of course,
r squared is just x squared
plus y squared.
Square of distance from the z
axis is just x squared plus y
squared.
Similarly, now,
if you want the distance from
the x axis, well,
that will be y squared plus z
squared.
OK, try to convince yourselves
of the picture,
or else just argue by symmetry:
you know, if you change the
positions of the axis.
So, moment of inertia about the
x axis is the double integral of
y squared plus z squared delta
dV.
And moment of inertia about the
y axis is the same thing,
but now with x squared plus z
squared.
And so, now,
if you try to apply these
things for flat solids that are
in the xy plane,
so where there's no z to look
at,
well, you see these formulas
become the old formulas that we
had.
But now, they all fit together
in a more symmetric way.
OK, any questions about that?
No?
OK, so these are just formulas
to remember.
So, OK, let's do an example.
Was there a question that I
missed?
No?
OK, so let's find the moment of
inertia about the z axis of a
solid cone -- -- between z
equals a times r and z equals b.
So, just to convince you that
it's a cone, so,
z equals a times r means the
height is proportional to the
distance from the z axis.
So, let's look at what we get
if we just do it in the plane of
a blackboard.
So, if I go to the right here,
r is just the distance from the
x axis.
The height should be
proportional with
proportionality factor A.
So, that means I take a line
with slope A.
If I'm on the left,
well, it's the same story
except distance to the z axis is
still positive.
So, I get the symmetric thing.
And, in fact,
it doesn't matter which
vertical plane I do it in.
This is the same if I rotate
about.
See, there's no theta in here.
So, it's the same in all
directions.
So, I claim it's a cone where
the slope of the rays is A.
OK, and z equals b.
Well, that just means we stop
in our horizontal plane at
height b.
OK, so that's solid cone really
just looks like this.
That's our solid.
OK, so it has a flat top,
that circular top,
and then the point is at v.
The tip of it is at the origin.
So, let's try to compute its
moment of inertia about the z
axis.
So, that means maybe this is
like the top that you are going
to spin.
And, it tells you how hard it
is to actually spin that top.
Actually, that's also useful if
you're going to do mechanical
engineering because if you are
trying to design gears,
and things like that that will
rotate,
you might want to know exactly
how much effort you'll have to
put to actually get them to
spin,
and whether you're actually
going to have a strong enough
engine, or whatever,
to do it.
OK, so what's the moment of
inertia of this guy?
Well, that's the triple
integral of, well,
we have to choose x squared
plus y squared or r squared.
Let's see, I think I want to
use cylindrical coordinates to
do that, given the shape.
So, we use r squared.
I might have a density that
let's say the density is one.
So, I don't have density.
I still have dV.
Now, it will be my choice to
choose between doing the dz
first or doing r dr d theta
first.
Just to show you how it goes
the other way around,
let me do it r dr d theta dz
this time.
Then you can decide on a
case-by-case basis which one you
like best.
OK, so if we do it in this
direction, it means that in the
inner and middle integrals,
we've fixed a value of z.
And, for that particular value
of z, we'll be actually slicing
our solid by a horizontal plane,
and looking at what we get,
OK?
So, what does that look like?
Well, I fixed a value of z,
and I slice my solid by a
horizontal plane.
Well, I'm going to get a circle
certainly.
What's the radius,
well, a disk actually,
what's the radius of the disk?
Yeah, the radius of the disk
should be z over a because the
equation of that cone,
we said it's z equals ar.
So, if you flip it around,
so, maybe I should switch to
another blackboard.
So, the equation of a cone is z
equals ar, or equivalently r
equals z over a.
So, for a given value of z,
I will get, this guy will be a
disk of radius z over a.
OK, so, moment of inertia is
going to be, well,
we said r squared,
r dr d theta dz.
Now, so, to set up the inner
and middle integrals,
I just set up a double integral
over this disk of radius z over
a.
So, it's easy.
r goes from zero to z over a.
Theta goes from zero to 2pi.
OK, and then,
well, if I set up the bounds
for z, now it's my outer
variable.
So, the question I have to ask
is what is the first slice?
What is the last slice?
So, the bottommost value of z
would be zero,
and the topmost would be b.
And so, that's it I get.
So, exercise,
it's not very hard.
Try to set it up the other way
around with dz first and then r
dr d theta.
It's pretty much the same level
of difficulty.
I'm sure you can do both of
them.
So, and also,
if you want to practice
calculations,
you should end up getting pi b
to the five over 10a to the four
if I got it right.
OK, let me finish with one more
example.
I'm trying to give you plenty
of practice because in case you
haven't noticed,
Monday is a holiday.
So, you don't have recitation
on Monday, which is good.
But it means that there will be
lots of stuff to cover on
Wednesday.
So -- Thank you.
OK, so third example,
let's say that I want to just
set up a triple integral for the
region where z is bigger than
one minus y inside the unit ball
centered at the origin.
So, the unit ball is just,
you know, well,
stay inside of the unit sphere.
So, its equation,
if you want,
would be x squared plus y
squared plus z squared less than
one.
OK, so that's one thing you
should remember.
The equation of a sphere
centered at the origin is x
squared plus y squared plus z
squared equals radius squared.
And now, we are going to take
this plane, z equals one minus
y.
So, if you think about it,
it's parallel to the x axis
because there's no x in its
coordinate in its equation.
At the origin,
the height is one.
So, it starts right here at one.
And, it slopes down with y with
slope one.
OK, so it's a plane that comes
straight out here,
and it intersects the sphere,
so here and here,
but also at other points in
between.
Any idea what kind of shape
this is?
Well, it's an ellipse,
but it's even more than that.
It's also a circle.
If you slice a sphere by a
plane, you always get a circle.
But, of course,
it's a slanted circle.
So, if you look at it in the xy
plane, if you project it to the
xy plane, that you will get an
ellipse.
OK, so we want to look at this
guy in here.
So, how do we do that?
Well, so maybe I should
actually draw quickly a picture.
So, in the yz plane,
it looks just like this,
OK?
But, if I look at it from above
in the xy plane,
then its shadow,
well, see, it will sit entirely
where y is positive.
So, it sits entirely above
here, and it goes through here
and here.
And, in fact,
when you project that slanted
circle, now you will get an
ellipse.
And, well, I don't really know
how to draw it well,
but it should be something like
this.
OK, so now if you want to try
to set up that double integral,
sorry, the triple integral,
well, so let's say we do it in
rectangular coordinates because
we are really evil.
[LAUGHTER]
So then, the bottom surface,
OK, so we do it with z first.
So, the bottom surface is the
slanted plane.
So, the bottom value would be z
equals one minus y.
The top value is on the sphere.
So, the sphere corresponds to z
equals square root of one minus
x squared minus y squared.
So, you'd go from the plane to
the sphere.
And then, to find the bounds
for x and y, you have to figure
out what exactly,
what the heck is this region
here?
So, what is this region?
Well, we have to figure out,
for what values of x and y the
plane is below the ellipse.
So, the condition is that,
sorry, the plane is below the
sphere.
OK, so, that's when the plane
is below the sphere.
That means one minus y is less
than square root of one minus x
squared minus y squared.
So, you have to somehow
manipulate this to extract
something simpler.
Well, probably the only way to
do it is to square both sides,
one minus y squared should be
less than one minus x squared
minus y squared.
And, if you work hard enough,
you'll find quite an ugly
equation.
But, you can figure out what
are, then, the bounds for x
given y, and then set up the
integral?
So, just to give you a hint,
the bounds on y will be zero to
one.
The bounds on x,
well, I'm not sure you want to
see them,
but in case you do,
it will be from negative square
root of 2y minus 2y squared to
square root of 2y minus 2y
squared.
So, exercise,
figure out how I got these by
starting from that.
Now, of course,
if we just wanted the volume of
this guy, we wouldn't do it this
way.
We do symmetry,
and actually we'd rotate the
thing so that our spherical cap
was actually centered on the z
axis because that would be a
much easier way to set it up.
But, depending on what function
we are integrating,
we can't always do that.
