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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 integration.

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And I have a
problem here for you

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on computing a volume of a
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;

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it's the volume inside
the paraboloid z

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equals x squared plus y squared
and bounded by the plane z

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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

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some chunk of that paraboloid,
and what I want to know

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is, what's the volume
of that piece that gets

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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,

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work out this
problem, come back,

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and we 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,

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so let's start to work
through it together.

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So sometimes you have a
problem 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 clear

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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--

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what 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,

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this paraboloid just
becomes a parabola.

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So that becomes the
parabola z equals

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y squared, which is a plane
section of the paraboloid z

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equals x squared 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

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of 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

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is what the 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

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is we need to figure out-- when
you project this region down,

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when you flatten it along z,
so you're disregarding z now,

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and then you're just looking
at its shadow, its footprint

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in the xy-plane-- you
want to figure out,

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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 R

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down here.

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So I'll call this region
R. And that region

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will be the projection of
this solid region down.

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And it has some
boundary curve-- C,

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say-- the boundary
curve of the region

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R. Just in case we need
to refer to them later,

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it's good to give them letters
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

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or 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

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of the curve of
intersection of this plane

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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

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and 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 equation

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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

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2y, that's a point (x, y) such
that directly above that point

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is a place 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,

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you can see that
we can rewrite this

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as x squared plus y
minus 1 squared equals 1.

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

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and 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, 1)

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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

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like in 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

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of radius 1 with center (0, 1).

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OK, great.

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So I'm just going to
shade that in again

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because I like 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 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

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going to be going from the
paraboloid up to the plane.

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That's going to be the
innermost integral,

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but then the middle
integral is going

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to be y in terms of
x or x in 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,

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what that region looks like
over which we'll be integrating

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for the outer two integrals.

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

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or we 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

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is tangent to one of
the axes at the origin.

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So this is a reasonably
nice situation

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to do polar coordinates in,
or cylindrical coordinates.

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You have to remember
from when you learned

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cylindrical and polar
coordinates what

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the equation of
such a circle is.

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

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and I'm going to 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
equation in polar--

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these are the x- and y-axes
here-- so this curve has,

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in polar coordinates, the
equation r 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,

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and I want to
integrate over it once.

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Remember, polar coordinates
are a little tricky

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because 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,

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and then as theta 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 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

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or equal to theta less
than or equal to pi here.

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So at pi over 2 at
the top, and at pi

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it comes back for
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 out

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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 this outer boundary, 2

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sine theta.

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So this describes
this region big R

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that we're trying
to 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

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in polar coordinates.

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You could also describe it
in rectangular coordinates,

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and 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 shot

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

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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,

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so let me write our
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, it's
the paraboloid z

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equals x squared plus
y squared-- but we're

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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,

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and it's going up to the
plane z equals 2y-- now y

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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-- just ignore
that-- from r squared less than

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or equal to z, and
it's going all the way

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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-plane,

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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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00:10:37,130 --> 00:10:40,840
So now, how do we get its volume
after we 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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00:10:44,580 --> 00:10:49,330
So V, the volume
of a region D, is

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equal to the triple integral
over that solid of dV.

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00:10:55,800 --> 00:10:56,400
OK?

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00:10:56,400 --> 00:10:59,370
And in our case, in
cylindrical coordinates,

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

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00:11:08,540 --> 00:11:09,040
OK?

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00:11:09,040 --> 00:11:21,030
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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So this triple integral
gives us precisely the volume

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of our region.

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And in order to figure
out what that volume is,

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we just have to
evaluate this integral.

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So let's start doing that.

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00:12:01,960 --> 00:12:05,160
I don't think I'm going
to go quite all the way,

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but I'll do most of the work.

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So OK.

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00:12:08,260 --> 00:12:10,110
So let's do the
innermost integral first.

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Whenever you have
a triple integral

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like this-- a 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

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integrating with
respect to z-- and r

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doesn't have any z's in it--
so this inner integral is

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00:12:22,830 --> 00:12:24,690
going to be easy.

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So I'm going to rewrite
this as equal to-- we keep

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our outer two bounds, so
it's still from 0 to pi,

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and it's still from 0 to 2
sine theta-- of 2r squared sine

245
00:12:42,980 --> 00:12:48,460
theta minus r cubed dr d theta.

246
00:12:48,460 --> 00:12:51,760
So what I've done here
is I've just integrated.

247
00:12:51,760 --> 00:12:56,516
I get the anti-derivative
of r dz is r*z.

248
00:12:56,516 --> 00:12:57,890
And so then I take
the difference

249
00:12:57,890 --> 00:12:58,931
between those two bounds.

250
00:12:58,931 --> 00:13:03,130
So I get r times 2r sine
theta minus r times r squared.

251
00:13:03,130 --> 00:13:07,190
So r times 2r sine theta
is 2r squared sine theta.

252
00:13:07,190 --> 00:13:09,052
Minus r times r squared
is minus r cubed.

253
00:13:09,052 --> 00:13:10,760
OK, so I've just done
the first integral.

254
00:13:10,760 --> 00:13:13,770
So now integrating
with respect to r.

255
00:13:13,770 --> 00:13:16,490
OK, this second one
isn't so bad either.

256
00:13:16,490 --> 00:13:19,240
As far as r is concerned,
this is just a polynomial.

257
00:13:19,240 --> 00:13:22,100
Theta is constant with
respect to r when we're

258
00:13:22,100 --> 00:13:23,320
doing an integral like this.

259
00:13:23,320 --> 00:13:23,980
So OK.

260
00:13:23,980 --> 00:13:26,140
So the second integral
is not too bad either.

261
00:13:26,140 --> 00:13:30,450
So this is the integral-- so
our outer integral from 0 to pi

262
00:13:30,450 --> 00:13:33,540
sticks around-- let's
not do this one in one

263
00:13:33,540 --> 00:13:35,715
fell swoop I think--
so it's going

264
00:13:35,715 --> 00:13:42,900
to become 2 r cubed
over 3, sine theta,

265
00:13:42,900 --> 00:13:46,360
minus r to the fourth over 4.

266
00:13:46,360 --> 00:13:49,290
And we're taking
that between r equals

267
00:13:49,290 --> 00:13:55,420
0 and r equals 2 sine theta.

268
00:13:55,420 --> 00:13:59,300
And then that whole thing is
going to be integrated d theta.

269
00:13:59,300 --> 00:14:01,160
So what do we get
when we plug this in?

270
00:14:01,160 --> 00:14:04,770
Well, at r equals zero, this
is just 0, so that's easy.

271
00:14:04,770 --> 00:14:08,090
And so we need the top
one, r equals 2 sine theta.

272
00:14:08,090 --> 00:14:11,850
So this is going to give
me something like 16/3 sine

273
00:14:11,850 --> 00:14:15,930
to the fourth theta minus
4 sine to the fourth theta,

274
00:14:15,930 --> 00:14:27,830
so I think that works out to
be 4/3 sine to the fourth theta

275
00:14:27,830 --> 00:14:30,250
d theta, between 0 and pi.

276
00:14:30,250 --> 00:14:33,950
So now you have to remember
how to do integrals like this.

277
00:14:33,950 --> 00:14:36,990
So this is something you
probably learned back

278
00:14:36,990 --> 00:14:44,170
in the trig integral section of
your Calculus I or 18.01 class.

279
00:14:44,170 --> 00:14:45,950
So when it's an
even power here, I

280
00:14:45,950 --> 00:14:50,110
think the thing that we do is
we use our half-angle formulas.

281
00:14:50,110 --> 00:14:55,040
So now I'm going to tell you
what your final steps are.

282
00:14:55,040 --> 00:14:58,310
So first, you're going to
use your half-angle formula.

283
00:14:58,310 --> 00:15:00,060
So what is that
half-angle formula?

284
00:15:00,060 --> 00:15:07,850
So it's sine squared theta
is equal to 1 minus cosine

285
00:15:07,850 --> 00:15:11,410
2 theta over 2.

286
00:15:11,410 --> 00:15:13,600
So you're going to have to
plug this in here, right?

287
00:15:13,600 --> 00:15:16,900
Sine to the fourth is sine
squared quantity squared.

288
00:15:16,900 --> 00:15:20,030
And then you're going to have
a cosine squared 2 theta,

289
00:15:20,030 --> 00:15:22,587
so you're going to have to
use the double-angle formula.

290
00:15:22,587 --> 00:15:25,170
This time you're going to have
to use the double-angle formula

291
00:15:25,170 --> 00:15:27,640
for cosine, which is very
similar, although not

292
00:15:27,640 --> 00:15:28,440
exactly the same.

293
00:15:28,440 --> 00:15:30,570
So you're going to
have to use those two

294
00:15:30,570 --> 00:15:31,646
double-angle formulas.

295
00:15:31,646 --> 00:15:33,020
After that, you'll
have something

296
00:15:33,020 --> 00:15:35,549
that is straightforward
to integrate.

297
00:15:35,549 --> 00:15:38,090
So you'll have something that's
straightforward to integrate.

298
00:15:38,090 --> 00:15:40,210
You'll integrate
it, and if I'm not

299
00:15:40,210 --> 00:15:45,360
mistaken, what
you get at the end

300
00:15:45,360 --> 00:15:49,960
is that you just get a fairly
nice and simple pi over 2

301
00:15:49,960 --> 00:15:51,040
as your answer.

302
00:15:51,040 --> 00:15:54,150
So you can check your
work there, and make sure

303
00:15:54,150 --> 00:15:56,210
that you've got out
pi over 2 at the end.

304
00:15:56,210 --> 00:15:58,400
And hopefully, if
you tried to do

305
00:15:58,400 --> 00:16:00,590
this using rectangular
coordinates,

306
00:16:00,590 --> 00:16:03,040
you also came out with
something like this as well.

307
00:16:03,040 --> 00:16:05,450
In that case, you would have
to do a trig substitution

308
00:16:05,450 --> 00:16:07,790
at some point to
evaluate your intervals,

309
00:16:07,790 --> 00:16:10,330
or you might have
an arcsine involved.

310
00:16:10,330 --> 00:16:13,230
Something like that will happen.

311
00:16:13,230 --> 00:16:15,380
But it should also give
you pi over 2, of course.

312
00:16:15,380 --> 00:16:17,380
Because it's the
same region, just

313
00:16:17,380 --> 00:16:18,970
described in a different way.

314
00:16:18,970 --> 00:16:21,930
So let me quickly
recap what we did.

315
00:16:21,930 --> 00:16:27,200
Way back over here, we had this
description of this region.

316
00:16:27,200 --> 00:16:32,310
So it was the region above our
paraboloid and below a plane.

317
00:16:32,310 --> 00:16:35,069
And so when we're
setting this up,

318
00:16:35,069 --> 00:16:37,360
we have to figure out, in
order to do a triple integral

319
00:16:37,360 --> 00:16:41,774
over this region, in
order to find its volume,

320
00:16:41,774 --> 00:16:43,440
we have to pick an
order of integration,

321
00:16:43,440 --> 00:16:46,510
and then we have to
know what the bounds are

322
00:16:46,510 --> 00:16:49,610
for the inside in terms of
the outer two variables,

323
00:16:49,610 --> 00:16:52,370
for the middle one in terms of
the outermost one, and so on.

324
00:16:52,370 --> 00:16:56,650
So in this case,
that means-- First,

325
00:16:56,650 --> 00:17:01,120
it was a natural choice to
make z the first variable--

326
00:17:01,120 --> 00:17:02,605
the innermost variable.

327
00:17:02,605 --> 00:17:03,980
And so then after
that, we needed

328
00:17:03,980 --> 00:17:08,220
to project to find the
relationship in the xy-plane

329
00:17:08,220 --> 00:17:09,580
between the other variables.

330
00:17:09,580 --> 00:17:17,230
Now in this case, we did that
by solving this little algebra

331
00:17:17,230 --> 00:17:17,780
problem here.

332
00:17:17,780 --> 00:17:23,151
We solved for z in the two
surfaces that we were given,

333
00:17:23,151 --> 00:17:24,650
and we set them
equal to each other.

334
00:17:24,650 --> 00:17:27,080
And so this gives us a
description for the boundary

335
00:17:27,080 --> 00:17:28,310
curve for our region.

336
00:17:28,310 --> 00:17:31,550
And because it's a
nice circle, this

337
00:17:31,550 --> 00:17:34,630
suggested that one possibility
was cylindrical coordinates.

338
00:17:34,630 --> 00:17:38,090
So we went ahead, and we found
in cylindrical coordinates

339
00:17:38,090 --> 00:17:39,840
the description of this shadow.

340
00:17:39,840 --> 00:17:42,940
And then we used the
knowledge we previously

341
00:17:42,940 --> 00:17:46,140
had to describe the whole region
in cylindrical coordinates.

342
00:17:46,140 --> 00:17:48,550
So we had this description
of our entire region.

343
00:17:48,550 --> 00:17:50,320
And then to compute
its volume, we just

344
00:17:50,320 --> 00:17:52,200
set up the triple
integral volume

345
00:17:52,200 --> 00:17:55,330
is equal to a
triple integral dV.

346
00:17:55,330 --> 00:17:57,170
In our case, dV-- since
we're in cylindrical

347
00:17:57,170 --> 00:17:59,890
coordinates-- that's
r dz dr d theta.

348
00:17:59,890 --> 00:18:04,240
We put in our bounds, and then
we evaluated the integral.

349
00:18:04,240 --> 00:18:05,854
I'll stop there.