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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've been
learning about triple

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integration.

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And I have a problem here for
you on computing a volume of a

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region using a triple
integral.

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So let's look at this.

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So I have a volume and I'm
describing it to you; it's the

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volume inside the paraboloid
z equals x squared plus y

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squared and bounded by the
plane z equals 2y.

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So I've drawn a little
picture here for you.

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So this is the paraboloid
here.

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And we're just taking
a plane cut of it.

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And so this is going to slice
off some chunk of that

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paraboloid, and what I want to
know is what the volume is of

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that piece that gets cut off
by that plane there.

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So below the plane and
above the paraboloid.

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So, why don't you pause the
video, take some time, work

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out this problem, come
back, and we

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can work on it together.

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I hope you had some luck
with this problem.

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I think it's a bit of a tricky
one, so let's start to work

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through it together.

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So sometimes you have a problem

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with a triple integral.

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And you need to set up your
bounds of integration.

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And sometimes you can look
at it and it's just

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clear what they are.

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If you're integrating over a
cube, life is really easy.

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But in this case, this region
that we want to integrate over

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is kind of more complicated
to understand.

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Right?

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So it's easy to see-- well,
relatively easy to see-- what

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the bounds on z are.

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So let me draw a couple of
two-dimensional pictures.

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So I'm going to draw the
yz-plane cross section here.

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So in the yz-plane cross
section, this paraboloid just

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becomes a parabola.

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So that becomes the parabola z
equals y squared, which is a

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plane section of the paraboloid
z equals x squared

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plus y squared.

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And this plane z equals 2y
becomes the line z equals 2y.

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And this little sliver
is a plane section of

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the region in question.

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So we see that z is going from
the paraboloid to the plane.

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And over here, we see that z is
going from the paraboloid

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to the plane.

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But what we really need to
understand then is what the

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relationship between
x and y is.

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So what is the shadow
of this region?

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How are x and y related
to each other?

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How can we bound x in terms
of y or y in terms of x?

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Or should we use cylindrical
coordinates or what?

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And so in order to that what
we need to do is we need to

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figure out--

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when you project this region
down, when you flatten it

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along z, so you're disregarding
z now, and then

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you're just looking
at its shadow, its

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footprint, in the xy plane--

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you want to figure out,
what is that region?

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What does it look like?

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So somehow we'll project down
and there will be some region

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R down here.

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So I'll call this region R. And
that region will be the

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projection of this solid
region down.

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And it has some boundary
curve--

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C, say-- the boundary curve of
the region R. Just in case we

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need to refer to them later,
it's good to give them letters

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so that they have names.

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So what we need to figure out
now is what is this region R?

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Now this is tough to do by just
intuitive reasoning or

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just by looking at this
picture I've drawn.

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So in this case, we're kind of
forced to use some algebra.

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All right.

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So what do we know about this
region R and this curve C?

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Well, C is the projection
downwards of the curve of

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intersection of this plane with
this paraboloid, right?

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So it's the projection down at
this curve intersection.

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So what does that mean
about its equation?

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Well, it means it's what we get
if we solve for z in one

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of the two equations
of the surfaces and

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plug it into the other.

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And that will give us an
equation with just x and y,

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and that will be the equation
of this curve C. OK.

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So in our case, that means
that C is given by this

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equation x squared plus
y squared equals 2y.

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All right.

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So whenever x squared plus y
squared equals 2y, that's a

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point x, y such that directly
above that point is a place

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where the plane intersects
the paraboloid.

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Well, OK.

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So what is this curve?

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Well, a little bit of algebra
can help us sort that out.

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If you bring the 2y over here
and complete the square, you

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can see that we can rewrite
this as x squared plus (y

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minus 1) squared equals 1.

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I brought the 2y over, I've
added 1 to both sides, and

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I've factored the y part.

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And so this is an easy equation
to recognize.

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This is the equation of a
circle with center (0,

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1) and radius 1.

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So let's draw that.

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And so here is a picture of what
the shadow looks like in

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the xy plane.

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So the center is at height 1,
and then it's this circle.

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That's almost a circle.

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It looks enough like a circle
for my purposes.

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So this is the region R. It's a
circular region of radius 1

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with center (0, 1).

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OK, great.

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So I'm just going to shade that
in again because I like

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doing that.

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OK.

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So that's the region R.

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So what is this region R?

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Let's look back over here.

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It's the shadow of our solid
region in the xy plane.

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So when you project
down, that's the

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region that you get.

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So why do we need that?

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So we know when we set up this
triple integral, z is going to

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be going from the paraboloid
up to the plane.

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That's going to be the innermost
integral, but then

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the middle integral is going to
be y in terms of x or x in

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terms of y.

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Or if we do polar coordinates
or cylindrical coordinates,

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it's going to be R in
terms of theta.

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So we need to figure out what
the boundary is, what that

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region looks like over which
we'll be integrating for the

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outer two integrals.

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OK, so now I've been saying we
could use cylindrical or we

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could use rectangular.

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What do we want to use?

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Well, so this is a circle.

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It's not centered at the origin,
but it is tangent to

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one of the axes at the origin.

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So this is a reasonably nice
situation to do polar

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coordinates in, or cylindrical
coordinates.

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You have to remember from when
you learned cylindrical and

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polar coordinates what the
equation of such a circle is.

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And so I'm going to write it
down here, and I'm going to

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invite you to go look
up why this is true

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if you don't remember.

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This curve has an equation
in polar--

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these are the x- and
y-axes here--

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so this curve has in polar
coordinates the equation r

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equals 2 sine theta.

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All right.

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So that gives me this
curve here.

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The outer boundary.

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And now what I want is, I don't
just want the curve.

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I want to integrate over the
whole region, and I want to

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integrate over it once.

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Remember, polar coordinates are
a little tricky because

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you have to worry about are
you overlapping and so on.

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So how does this work?

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At theta equals 0, or at the
origin, and then as theta

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grows, we get further
and further away.

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So this is our radius
growing out.

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And then at pi over 2,
we're at the top

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point of the circle.

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And then as it comes back into
pi, it comes back in.

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So we want theta going from 0
less than or equal to theta

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less than or equal to pi here.

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So at pi over 2 at the top, and
at pi it comes back for

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the first time.

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And what about r?

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Well, it looks like r has
to go all the way

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out to 2 sine theta.

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And in fact, we always want
it to start at the origin.

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So we always want r
to go from 0 to--

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this outer boundary--

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2 sine theta.

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So this describes this region
big R that we're trying to

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integrate over.

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This circular region in
polar coordinates.

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So OK.

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So it's a fairly easy
description of polar

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coordinates.

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You could also describe it in
rectangular coordinates, and

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you could try to solve
the problem that way.

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I'm not going to do it for you,
but you could give it a

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shot and see if you can come
out with the same answer in

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the end that we do.

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So OK.

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So now, what have we done?

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Well, I haven't written our
bounds, so let me write our

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bounds on z right here.

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So we know that z is going
from the paraboloid.

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If we look at the paraboloid
z equals x

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squared plus y squared--

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but we're working in cylindrical
coordinates now,

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so we need to write this in
terms of r and theta--

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so that's z is going from r
squared, and it's going up to

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the plane z equals 2y--

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now y in cylindrical coordinates
is r sine theta.

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So z is going from r squared
to 2r sine theta.

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So let's go write that
down over here.

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So z is going from--

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just ignore that--

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from r squared less than or
equal to z, and it's going all

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the way up to 2r sine theta.

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So these three equations
describe our region.

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Yeah?

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0 less than theta less than
pi: that just says theta.

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OK?

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Then when theta is going
from 0 to pi--

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r going from 0 to 2 sine theta--
that says in the xy

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plane we're tracing out
this circular shadow.

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And then as z goes from r
squared to 2r sine theta, that

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says above this shadow we're
above the paraboloid

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and below the plane.

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So that's exactly the
region that we want.

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So OK.

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So now how do we get it
to volume after we

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figured this out?

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00:10:40,840 --> 00:10:44,580
Well, we write down the
triple integral.

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So V, the volume of a region
D, is equal to the triple

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integral over that
solid of dV. OK?

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And in our case, in cylindrical
coordinates, dV is

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going to be dz times r dr d
theta, or r dz dr d theta.

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OK?

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So this is equal to, if we're
integrating, r dz dr d theta.

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And now we need to put
in our bounds.

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If we look over on this side
of me, here they are.

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And these are our bounds that
we're going to be using.

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So theta is going
from 0 to pi.

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And r is going from
0 to 2 sine theta.

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And z is going from r squared
to 2r sine theta.

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00:11:48,200 --> 00:11:50,780

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So this triple integral
gives us precisely the

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volume of our region.

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And in order to figure out what
that volume is, we just

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have to evaluate
this integral.

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So let's start doing that.

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I don't think I'm going to go
quite all the way, but I'll do

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most of the work.

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00:12:07,680 --> 00:12:08,260
So OK.

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00:12:08,260 --> 00:12:10,480
So let's do the innermost
integral first. Whenever you

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have a triple integral
like this-- a

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00:12:12,120 --> 00:12:13,290
nice iterated integral--

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you always start at the inside
and work your way out.

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Well here, our integrand is r,
and we're integrating with

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respect to z-- and r doesn't
have any z's in it--

250
00:12:21,850 --> 00:12:24,690
so this inner integral
is going to be easy.

251
00:12:24,690 --> 00:12:28,340
So I'm going to rewrite
this as equal to--

252
00:12:28,340 --> 00:12:31,696
we keep our outer two bounds--
so it's still from 0 to pi,

253
00:12:31,696 --> 00:12:42,980
and it's still from 0 to 2 sine
theta, of 2r squared sine

254
00:12:42,980 --> 00:12:48,460
theta minus r cubed
dr d theta.

255
00:12:48,460 --> 00:12:51,760
So what I've done here is
I've just integrated.

256
00:12:51,760 --> 00:12:56,580
I get the anti-derivative
of r dz is rz.

257
00:12:56,580 --> 00:12:57,890
And so then I take
the difference

258
00:12:57,890 --> 00:12:58,840
between those two bounds.

259
00:12:58,840 --> 00:13:03,130
So I get r times 2r sine theta
minus r times r squared.

260
00:13:03,130 --> 00:13:07,190
So r times 2r sine theta is
2r squared sine theta.

261
00:13:07,190 --> 00:13:09,080
Minus r times r squared
is minus r cubed.

262
00:13:09,080 --> 00:13:10,760
OK, so I've just done
the first integral.

263
00:13:10,760 --> 00:13:13,770
So now integrating with
respect to r.

264
00:13:13,770 --> 00:13:16,490
OK, this second one isn't
so bad either.

265
00:13:16,490 --> 00:13:19,240
As far as r is concerned, this
is just a polynomial.

266
00:13:19,240 --> 00:13:22,440
Theta is constant with respect
to r when we're doing an

267
00:13:22,440 --> 00:13:23,320
integral like this.

268
00:13:23,320 --> 00:13:23,980
So OK.

269
00:13:23,980 --> 00:13:26,140
So the second integral is
not too bad either.

270
00:13:26,140 --> 00:13:28,620
So this is the integral
for our outer

271
00:13:28,620 --> 00:13:30,450
integral from 0 to pi.

272
00:13:30,450 --> 00:13:31,210
Sticks around--

273
00:13:31,210 --> 00:13:35,350
lets not do this one in one
fell swoop I think--

274
00:13:35,350 --> 00:13:42,900
so it's going to become 2r
cubed over 3, sine theta,

275
00:13:42,900 --> 00:13:46,360
minus r to the fourth over 4.

276
00:13:46,360 --> 00:13:51,450
And we're taking that between
r equals 0 and r

277
00:13:51,450 --> 00:13:55,420
equals 2 sine theta.

278
00:13:55,420 --> 00:13:56,730
And then that whole thing
is going to be

279
00:13:56,730 --> 00:13:59,300
integrated d theta.

280
00:13:59,300 --> 00:14:01,160
So what do we get when
we plug this in?

281
00:14:01,160 --> 00:14:04,770
Well, at r equals zero, this
is just 0, so that's easy.

282
00:14:04,770 --> 00:14:08,090
And so we need the top one,
r equals 2 sine theta.

283
00:14:08,090 --> 00:14:11,950
So this is going to give me
something like 16/3 sine to

284
00:14:11,950 --> 00:14:16,100
the fourth theta minus 4 sine
to the fourth theta, so I

285
00:14:16,100 --> 00:14:27,950
think that works out to be 4/3
sine to the fourth theta d

286
00:14:27,950 --> 00:14:30,250
theta, between 0 and pi.

287
00:14:30,250 --> 00:14:33,950
So now you have to remember how
to do integrals like this.

288
00:14:33,950 --> 00:14:38,190
So this is something you
probably learned back in the

289
00:14:38,190 --> 00:14:44,170
trig integral section of your
Calculus I or 18.01 class.

290
00:14:44,170 --> 00:14:46,540
So when it's an even power here,
I think the thing that

291
00:14:46,540 --> 00:14:50,110
we do is we use our half-angle
formulas.

292
00:14:50,110 --> 00:14:55,040
So now I'm going to tell you
what your final steps are.

293
00:14:55,040 --> 00:14:58,310
So first, you're going to use
your half-angle formula.

294
00:14:58,310 --> 00:15:00,060
So what is that half-angle
formula?

295
00:15:00,060 --> 00:15:08,120
So it's sine squared theta is
equal to 1 minus cosine 2

296
00:15:08,120 --> 00:15:11,410
theta over 2.

297
00:15:11,410 --> 00:15:13,600
So you're going to have to
plug this in here, right?

298
00:15:13,600 --> 00:15:16,900
Sine to the fourth is sine
squared quantity squared.

299
00:15:16,900 --> 00:15:20,190
And then you're going to have
a cosine squared 2 theta, so

300
00:15:20,190 --> 00:15:22,860
you're going to have to use
the double-angle formula.

301
00:15:22,860 --> 00:15:24,820
This time you're going to have
to use the double-angle

302
00:15:24,820 --> 00:15:27,640
formula for cosine, which is
very similar, although not

303
00:15:27,640 --> 00:15:28,440
exactly the same.

304
00:15:28,440 --> 00:15:30,570
So you're going to have
to use those two

305
00:15:30,570 --> 00:15:31,920
double-angle formulas.

306
00:15:31,920 --> 00:15:33,210
After that, you'll have
something that is

307
00:15:33,210 --> 00:15:35,980
straightforward to integrate.

308
00:15:35,980 --> 00:15:37,290
So you'll have something that's

309
00:15:37,290 --> 00:15:38,090
straightforward to integrate.

310
00:15:38,090 --> 00:15:43,090
You'll integrate it, and if I'm
not mistaken, what you get

311
00:15:43,090 --> 00:15:48,400
at the end is that you just get
a fairly nice and simple

312
00:15:48,400 --> 00:15:51,040
pi over 2 as your answer.

313
00:15:51,040 --> 00:15:54,300
So you can check your work
there, and make sure that

314
00:15:54,300 --> 00:15:56,210
you've got out pi over
2 at the end.

315
00:15:56,210 --> 00:15:59,970
And hopefully, if you tried to
do this using rectangular

316
00:15:59,970 --> 00:16:02,270
coordinates, you also came
out with something

317
00:16:02,270 --> 00:16:03,040
like this as well.

318
00:16:03,040 --> 00:16:05,530
In that case, you would have to
do a trig substitution at

319
00:16:05,530 --> 00:16:08,630
some point to evaluate your
intervals, or you might have

320
00:16:08,630 --> 00:16:10,330
an arcsine involved.

321
00:16:10,330 --> 00:16:13,230
Something like that
will happen.

322
00:16:13,230 --> 00:16:15,380
But it should also give you
pi over 2, of course.

323
00:16:15,380 --> 00:16:17,900
Because it's the same region,
just described

324
00:16:17,900 --> 00:16:18,970
in a different way.

325
00:16:18,970 --> 00:16:21,930
So let me quickly recap
what we did.

326
00:16:21,930 --> 00:16:24,610
Way back over here, we had this

327
00:16:24,610 --> 00:16:27,200
description of this region.

328
00:16:27,200 --> 00:16:32,310
So it was the region above our
paraboloid and below a plane.

329
00:16:32,310 --> 00:16:35,910
And so when we're setting this
up, we have to figure out in

330
00:16:35,910 --> 00:16:38,520
order to do a triple integral
over this region in order to

331
00:16:38,520 --> 00:16:43,440
find its volume, we have to pick
an order of integration,

332
00:16:43,440 --> 00:16:47,900
and then we have to know what
the bounds are for the inside

333
00:16:47,900 --> 00:16:50,170
in terms of the outer two
variables, for the middle one

334
00:16:50,170 --> 00:16:52,370
in terms of the outermost
one, and so on.

335
00:16:52,370 --> 00:17:00,310
So in this case, that means it
was a natural choice to make z

336
00:17:00,310 --> 00:17:02,800
the first variable-- the
innermost variable.

337
00:17:02,800 --> 00:17:05,310
And so then after that, we
needed to project to find the

338
00:17:05,310 --> 00:17:09,580
relationship in the xy plane
between the other variables.

339
00:17:09,580 --> 00:17:17,230
Now in this case, we did that by
solving this little algebra

340
00:17:17,230 --> 00:17:17,780
problem here.

341
00:17:17,780 --> 00:17:23,280
We solved for z in the two
surfaces that we were given,

342
00:17:23,280 --> 00:17:24,650
and we set them equal
to each other.

343
00:17:24,650 --> 00:17:27,080
And so this gives us a
description for the boundary

344
00:17:27,080 --> 00:17:28,310
curve for our region.

345
00:17:28,310 --> 00:17:32,690
And because it's a nice circle,
this suggested that

346
00:17:32,690 --> 00:17:34,630
one possibility was cylindrical
coordinates.

347
00:17:34,630 --> 00:17:38,210
So we went ahead, and we found
in cylindrical coordinates the

348
00:17:38,210 --> 00:17:39,840
description of this shadow.

349
00:17:39,840 --> 00:17:44,090
And then we used the knowledge
we previously had to describe

350
00:17:44,090 --> 00:17:46,140
the whole region in cylindrical
coordinates.

351
00:17:46,140 --> 00:17:48,550
So we had this description
of our entire region.

352
00:17:48,550 --> 00:17:51,030
And then to compute its volume,
we just set up the

353
00:17:51,030 --> 00:17:55,330
triple integral volume is equal
to a triple integral dV.

354
00:17:55,330 --> 00:17:56,310
In our case, dV--

355
00:17:56,310 --> 00:17:57,720
since we're in cylindrical
coordinates--

356
00:17:57,720 --> 00:17:59,890
that's r dz dr d theta.

357
00:17:59,890 --> 00:18:04,240
We put in our bounds, and then
we evaluated the integral.

358
00:18:04,240 --> 00:18:05,960
I'll stop there.

359
00:18:05,960 --> 00:18:06,353