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 the volume is 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 an 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 of 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 at 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 it to 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 rz.
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 for our outer
integral from 0 to pi.
Sticks around--
lets not do this one in one fell swoop I think--
so it's going to become 2r 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 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.