JOEL LEWIS: Hi.
Welcome back to recitation.
In lecture, you've been learning
about triple integration.
And I have a
problem here for you
on computing a volume of a
region using a triple integral.
So let's look at this.
So I have a volume and
I'm describing it to you;
it's the volume inside
the paraboloid z
equals x squared plus y squared
and bounded by the plane z
equals 2y.
So I've drawn a little
picture here for you.
So this is the paraboloid here.
And we're just taking
a plane cut of it.
And so this is
going to slice off
some chunk of that paraboloid,
and what I want to know
is, what's the volume
of that piece that gets
cut off by that plane there?
So below the plane and
above the paraboloid.
So, why don't you pause
the video, take some time,
work out this
problem, come back,
and we can work on it together.
I hope you had some
luck with this problem.
I think it's a bit
of a tricky one,
so let's start to work
through it together.
So sometimes you have a
problem with a triple integral.
And you need to set up
your bounds of integration.
And sometimes you can look
at it and it's just clear
what they are.
If you're integrating over
a cube, life is really easy.
But in this case, this region
that we want to integrate over
is kind of more
complicated to understand.
Right?
So it's easy to see-- well,
relatively easy to see--
what the bounds on z are.
So let me draw a couple of
two-dimensional pictures.
So I'm going to draw the
yz-plane cross section here.
So in the yz-plane
cross section,
this paraboloid just
becomes a parabola.
So that becomes the
parabola z equals
y squared, which is a plane
section of the paraboloid z
equals x squared plus y squared.
And this plane z equals 2y
becomes the line z equals 2y.
And this little sliver
is a plane section
of the region in question.
So we see that z is going from
the paraboloid to the plane.
And over here, we see that z
is going from the paraboloid
to the plane.
But what we really
need to understand then
is what the relationship
between x and y is.
So what is the shadow
of this region?
How are x and y
related to each other?
How can we bound x in terms
of y or y in terms of x?
Or should we use cylindrical
coordinates or what?
And so in order to
that what we need to do
is we need to figure out-- when
you project this region down,
when you flatten it along z,
so you're disregarding z now,
and then you're just looking
at its shadow, its footprint
in the xy-plane-- you
want to figure out,
what is that region?
What does it look like?
So somehow we'll project down
and there will be some region R
down here.
So I'll call this region
R. And that region
will be the projection of
this solid region down.
And it has some
boundary curve-- C,
say-- the boundary
curve of the region
R. Just in case we need
to refer to them later,
it's good to give them letters
so that they have names.
So what we need to figure out
now is what is this region R?
Now this is tough to do by
just intuitive reasoning
or just by looking at
this picture I've drawn.
So in this case, we're kind
of forced to use some algebra.
All right.
So what do we know about this
region R and this curve C?
Well, C is the
projection downwards
of the curve of
intersection of this plane
with this paraboloid, right?
So it's the projection down
at this curve intersection.
So what does that mean
about its equation?
Well, it means it's what we
get if we solve for z in one
of the two equations
of the surfaces
and plug it into the other.
And that will give us an
equation with just x and y,
and that will be the
equation of this curve C. OK.
So in our case, that means that
C is given by this equation
x squared plus y
squared equals 2y.
All right.
So whenever x squared
plus y squared equals
2y, that's a point (x, y) such
that directly above that point
is a place where the plane
intersects the paraboloid.
Well, OK.
So what is this curve?
Well, a little bit of algebra
can help us sort that out.
If you bring the 2y over
here and complete the square,
you can see that
we can rewrite this
as x squared plus y
minus 1 squared equals 1.
I brought the 2y over,
I've added 1 to both sides,
and I've factored the y part.
And so this is an easy
equation to recognize.
This is the equation of a
circle with center (0, 1)
and radius 1.
So let's draw that.
And so here is a picture
of what the shadow looks
like in the xy-plane.
So the center is at height
1, and then it's this circle.
That's almost a circle.
It looks enough like a
circle for my purposes.
So this is the region R.
It's a circular region
of radius 1 with center (0, 1).
OK, great.
So I'm just going to
shade that in again
because I like doing that.
OK.
So that's the region R.
So what is this region R?
Let's look back over here.
It's the shadow of our solid
region in the xy-plane.
So when you project down,
that's the region that you get.
So why do we need that?
So we know when we set up
this triple integral, z is
going to be going from the
paraboloid up to the plane.
That's going to be the
innermost integral,
but then the middle
integral is going
to be y in terms of
x or x in terms of y.
Or if we do polar coordinates
or cylindrical coordinates,
it's going to be R
in terms of theta.
So we need to figure out
what the boundary is,
what that region looks like
over which we'll be integrating
for the outer two integrals.
OK, so now I've been saying
we could use cylindrical
or we could use rectangular.
What do we want to use?
Well, so this is a circle.
It's not centered at
the origin, but it
is tangent to one of
the axes at the origin.
So this is a reasonably
nice situation
to do polar coordinates in,
or cylindrical coordinates.
You have to remember
from when you learned
cylindrical and polar
coordinates what
the equation of
such a circle is.
And so I'm going to
write it down here,
and I'm going to invite you
to go look up why this is true
if you don't remember.
This curve has
equation in polar--
these are the x- and y-axes
here-- so this curve has,
in polar coordinates, the
equation r equals 2 sine theta.
All right.
So that gives me
this curve here.
The outer boundary.
And now what I want is, I
don't just want the curve.
I want to integrate
over the whole region,
and I want to
integrate over it once.
Remember, polar coordinates
are a little tricky
because you have to worry about
are you overlapping and so on.
So how does this work?
At theta equals 0,
or at the origin,
and then as theta grows, we
get further and further away.
So this is our
radius growing out.
And then at pi over 2, we're
at the top point of the circle.
And then as it comes back
into pi, it comes back in.
So we want theta
going from 0 less than
or equal to theta less
than or equal to pi here.
So at pi over 2 at
the top, and at pi
it comes back for
the first time.
And what about r?
Well, it looks like r
has to go all the way out
to 2 sine theta.
And in fact, we always want
it to start at the origin.
So we always want r to go from
0 to this outer boundary, 2
sine theta.
So this describes
this region big R
that we're trying
to integrate over.
This circular region
in polar coordinates.
So OK.
So it's a fairly
easy description
in polar coordinates.
You could also describe it
in rectangular coordinates,
and you could try to solve
the problem that way.
I'm not going to do it for you,
but you could give it a shot
and see if you can come out
with the same answer in the end
that we do.
So OK.
So now, what have we done?
Well, I haven't
written our bounds,
so let me write our
bounds on z right here.
So we know that z is
going from the paraboloid.
If we look, it's
the paraboloid z
equals x squared plus
y squared-- but we're
working in cylindrical
coordinates now,
so we need to write this
in terms of r and theta--
so that's z is going
from r squared,
and it's going up to the
plane z equals 2y-- now y
in cylindrical coordinates
is r sine theta.
So z is going from r
squared to 2r sine theta.
So let's go write
that down over here.
So z is going from-- just ignore
that-- from r squared less than
or equal to z, and
it's going all the way
up to 2r sine theta.
So these three equations
describe our region.
Yeah?
0 less than theta less than
pi: that just says theta.
OK?
Then when theta is
going from 0 to pi--
r going from 0 to 2 sine theta--
that says in the xy-plane,
we're tracing out
this circular shadow.
And then as z goes from r
squared to 2r sine theta, that
says above this shadow
we're above the paraboloid
and below the plane.
So that's exactly the
region that we want.
So OK.
So now, how do we get its volume
after we figured this out?
Well, we write down
the triple integral.
So V, the volume
of a region D, is
equal to the triple integral
over that solid of dV.
OK?
And in our case, in
cylindrical coordinates,
dV is going to be dz times r
dr d theta, or r dz dr d theta.
OK?
So this is equal to, if we're
integrating, r dz dr d theta.
And now we need to
put in our bounds.
If we look over on this
side of me, here they are.
And these are our bounds
that we're going to be using.
So theta is going from 0 to pi.
And r is going from
0 to 2 sine theta.
And z is going from r
squared to 2r sine theta.
So this triple integral
gives us precisely the volume
of our region.
And in order to figure
out what that volume is,
we just have to
evaluate this integral.
So let's start doing that.
I don't think I'm going
to go quite all the way,
but I'll do most of the work.
So OK.
So let's do the
innermost integral first.
Whenever you have
a triple integral
like this-- a nice
iterated integral--
you always start at the
inside and work your way out.
Well here, our integrand
is r, and we're
integrating with
respect to z-- and r
doesn't have any z's in it--
so this inner integral is
going to be easy.
So I'm going to rewrite
this as equal to-- we keep
our outer two bounds, so
it's still from 0 to pi,
and it's still from 0 to 2
sine theta-- of 2r squared sine
theta minus r cubed dr d theta.
So what I've done here
is I've just integrated.
I get the anti-derivative
of r dz is r*z.
And so then I take
the difference
between those two bounds.
So I get r times 2r sine
theta minus r times r squared.
So r times 2r sine theta
is 2r squared sine theta.
Minus r times r squared
is minus r cubed.
OK, so I've just done
the first integral.
So now integrating
with respect to r.
OK, this second one
isn't so bad either.
As far as r is concerned,
this is just a polynomial.
Theta is constant with
respect to r when we're
doing an integral like this.
So OK.
So the second integral
is not too bad either.
So this is the integral-- so
our outer integral from 0 to pi
sticks around-- let's
not do this one in one
fell swoop I think--
so it's going
to become 2 r cubed
over 3, sine theta,
minus r to the fourth over 4.
And we're taking
that between r equals
0 and r equals 2 sine theta.
And then that whole thing is
going to be integrated d theta.
So what do we get
when we plug this in?
Well, at r equals zero, this
is just 0, so that's easy.
And so we need the top
one, r equals 2 sine theta.
So this is going to give
me something like 16/3 sine
to the fourth theta minus
4 sine to the fourth theta,
so I think that works out to
be 4/3 sine to the fourth theta
d theta, between 0 and pi.
So now you have to remember
how to do integrals like this.
So this is something you
probably learned back
in the trig integral section of
your Calculus I or 18.01 class.
So when it's an
even power here, I
think the thing that we do is
we use our half-angle formulas.
So now I'm going to tell you
what your final steps are.
So first, you're going to
use your half-angle formula.
So what is that
half-angle formula?
So it's sine squared theta
is equal to 1 minus cosine
2 theta over 2.
So you're going to have to
plug this in here, right?
Sine to the fourth is sine
squared quantity squared.
And then you're going to have
a cosine squared 2 theta,
so you're going to have to
use the double-angle formula.
This time you're going to have
to use the double-angle formula
for cosine, which is very
similar, although not
exactly the same.
So you're going to
have to use those two
double-angle formulas.
After that, you'll
have something
that is straightforward
to integrate.
So you'll have something that's
straightforward to integrate.
You'll integrate
it, and if I'm not
mistaken, what
you get at the end
is that you just get a fairly
nice and simple pi over 2
as your answer.
So you can check your
work there, and make sure
that you've got out
pi over 2 at the end.
And hopefully, if
you tried to do
this using rectangular
coordinates,
you also came out with
something like this as well.
In that case, you would have
to do a trig substitution
at some point to
evaluate your intervals,
or you might have
an arcsine involved.
Something like that will happen.
But it should also give
you pi over 2, of course.
Because it's the
same region, just
described in a different way.
So let me quickly
recap what we did.
Way back over here, we had this
description of this region.
So it was the region above our
paraboloid and below a plane.
And so when we're
setting this up,
we have to figure out, in
order to do a triple integral
over this region, in
order to find its volume,
we have to pick an
order of integration,
and then we have to
know what the bounds are
for the inside in terms of
the outer two variables,
for the middle one in terms of
the outermost one, and so on.
So in this case,
that means-- First,
it was a natural choice to
make z the first variable--
the innermost variable.
And so then after
that, we needed
to project to find the
relationship in the xy-plane
between the other variables.
Now in this case, we did that
by solving this little algebra
problem here.
We solved for z in the two
surfaces that we were given,
and we set them
equal to each other.
And so this gives us a
description for the boundary
curve for our region.
And because it's a
nice circle, this
suggested that one possibility
was cylindrical coordinates.
So we went ahead, and we found
in cylindrical coordinates
the description of this shadow.
And then we used the
knowledge we previously
had to describe the whole region
in cylindrical coordinates.
So we had this description
of our entire region.
And then to compute
its volume, we just
set up the triple
integral volume
is equal to a
triple integral dV.
In our case, dV-- since
we're in cylindrical
coordinates-- that's
r dz dr d theta.
We put in our bounds, and then
we evaluated the integral.
I'll stop there.
