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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 nice average
value problem for you

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using triple integration here.

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So what I'd like you to do is
to consider the tetrahedron that

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has vertices at the origin,
and at the points (1, 0, 0)

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and (0, 1, 0) and (0, 0, 1).

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So that's one point on
each of the positive axes,

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at distance 1 from the origin.

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So I've taken the liberty
of drawing it here for you.

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So consider that
solid tetrahedron.

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And what I'd like you to do
is find the average distance

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of the points in that
tetrahedron from the xy-plane.

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All right.

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So I'd like you to compute the
average value of the distance,

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as the point ranges over
the whole tetrahedron,

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of its distance
from the xy-plane.

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So why don't you
pause the video,

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spend some time working
that out, come back,

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and we can work it out together.

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Hopefully, you had some
luck with this problem.

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Let's get started.

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So the average value of a
function F over a region R

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is going to be 1 over
the volume of the region

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times the triple integral
over your whole region

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R of the function value f of
x, y, z with respect to volume,

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in any order you want.

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So I guess I'll write dV here,
and then to evaluate this,

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you set it up as an
iterated integral.

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So in our case,
the function we're

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trying to find the
average value of

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is the distance between
a point and the xy-plane.

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But that's an easy
function, right?

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That's just z.

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For any point in space, its
distance from the xy-plane

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is just its height.

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It's z-value.

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So the function
that we're seeking

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to find the average
value of is z,

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and so most of the
work of this problem

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then is going to be in figuring
out what the bounds are

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and then doing the actual
integral after that.

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So OK.

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So let's think about the bounds.

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This tetrahedron is a
nice, reasonably simple,

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geometric object.

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So in fact, it doesn't
matter too much which order

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you take your bounds.

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So I think I'm going to do
it in the order dz dy dx.

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You know, I'll do z first,
and then y, and then x.

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But it doesn't matter.

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If you do it a
different way, it'll

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probably work out very
similarly overall,

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and you'll still be able to
compare the overall process.

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So let's think about z.

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Yeah?

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So in this tetrahedron, we
want to integrate with respect

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to z first.

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So we look at this
tetrahedron and we say, OK,

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at a point x and
y-- you know, when

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we choose the x-
and y-values, what's

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the lowest surface-- what's
the smallest value z can take--

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and what's the upper
surface-- what's

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the largest value z can take?

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So the lowest surface
here is the xy-plane.

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That's the bottom face
of this tetrahedron.

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And for any choice of x
and y, the lowest value z

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can take is when
it's in the xy-plane.

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So when it's equal to 0.

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So this is going to be
equal, so in our case,

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so let's set this up.

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So it's going to be
an iterated integral.

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The function we're
integrating is z,

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and I said we'll do dz dy dx.

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And we just said that the lowest
value that z takes-- the dz--

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is 0.

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So the highest
value that z takes

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is when it hits
this top surface.

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This plane that passes through
the point (1, 0, 0), (0, 1, 0),

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and (0, 0, 1).

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All right.

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So we need to know the
equation of that plane.

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Luckily, that's a
pretty easy plane

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to write down the equation for.

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So this slanted plane passing
through the three vertices

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other than the origin is the
plane x plus y plus z equals 1.

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All right.

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So it's a nice, easy
plane to work with.

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And so what we want
to know is what's

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the value of z on that plane.

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So we isolate z on one side and
we bring everything else over.

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So we have that top value
of z, the largest value of z

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that z can take when x and y
are fixed, is 1 minus x minus 1.

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So that's what goes up here.

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So that's the biggest
value z can take.

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OK, good.

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So now we need to figure
out what the bounds on y

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are in terms of x.

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So what I like to
do in this case

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is I like to draw a
projection of your surface.

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So then you're in a
two-dimensional world,

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and then you can look at
that image more easily.

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So what we're going
to do is we're

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going to look at
this tetrahedron

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and we're going to
imagine projecting it down

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into the xy-plane.

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So for every point
in the tetrahedron,

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we're going to draw a dot
below it in the xy-plane.

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And then we're going to
look at that set of dots.

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So what that's going to
give us is this bottom face

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of the tetrahedron.

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Every point of the tetrahedron
is above its bottom face.

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That's not true for
every tetrahedron,

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but it's true for this one.

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So the region that
we're interested in

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is that bottom face.

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So I'm going to draw another
picture of it over on my left

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here.

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So that region is the region
that has vertices (0, 0),

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and (1, 0), and (0, 1).

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So it's this triangle.

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And this bottom
edge of the triangle

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is the line y equals 0.

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This left edge is
the line x equals 0.

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And this slanted line is
the line x plus y equals 1.

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OK, so this is that same
bottom face that we just drew.

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But now I've changed my axes
to our usual two-dimensional

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direction with x to
the right and y up.

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OK.

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So we're doing dy next.

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So we need to figure
out for a fixed value

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of x, what are the bounds on y?

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So we see from this picture
that for any fixed value of x, y

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goes from 0-- the
x-axis-- up to this line.

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OK?

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So on the x-axis, y takes the
value 0, and on this line,

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y takes the value 1 minus x.

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Finally, our outermost
variable of integration

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is x, and so we
need to know what

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are the absolute largest and
smallest values that x takes.

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So we can do that looking
either at this picture

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or looking at our
original picture.

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Either way, it's not
hard to see that x just

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goes between 0 and 1.

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The smallest value that x
takes in this tetrahedron is 0.

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The largest value it takes is 1.

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So that's our integral
that we have to compute.

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So that's not that bad at all.

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So now you have to go
through and you have

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to actually integrate this.

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Yeah?

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And so I'm going to look at
my notes just to make sure

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I don't make any big
arithmetic mistakes.

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So let's see.

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Now we do these one at a time.

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So this innermost integral
is an integral of z dz.

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OK, well that's easy.

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That's z squared over 2.

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And then we're taking z
squared over 2 between 0 and 1

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minus x minus y.

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So this innermost integral is z
squared over 2 between 0 and 1

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minus x minus y.

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So that's equal to-- so the
innermost integral gives us 1

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minus x minus y squared over 2.

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So that's what we get for
the innermost integral.

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So our integral that
we're looking at, then,

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is equal to the
integral, as x goes

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from 0 to 1, of the integral
as y goes from 0 to 1 minus x

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of this integrand.

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So this is the inner one.

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Let me write that, "inner."

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That's what I've got here.

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Just integrating z with respect
to z gives me z squared over 2.

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And then I evaluated at
the bounds of the integral.

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OK, so now I need to
do the middle one.

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So let's do that up here.

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So I need to compute
the integral.

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So now I take the bounds,
so the middle one is y,

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and the bounds are from 0 to 1
minus x, of the inner integral.

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This thing that I just computed.

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So that's of 1 minus x
minus y squared over 2, dy.

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OK.

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So All right.

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So this isn't that bad.

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This is a quadratic
polynomial in y.

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And so it's not
terribly hard to see.

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I'm running a little
bit out of board space.

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So I'm not going to give you
a full, detailed explanation.

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But it's not hard
to see, I think,

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that this integral of 1
minus x minus y squared

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over 2 with respect to y is 1
minus x minus y cubed over 3,

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but it's negative, because
the sign here is negative.

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And you could check by
differentiating this and seeing

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that you get that.

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And so we have to evaluate that
as y goes from 0 to 1 minus x.

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So what do we get?

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Well, when y is equal
to 1 minus x, this is 0.

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So we get 0 minus-- and when y
is equal to 0, this is minus 1

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minus x quantity cubed over
6-- so it's minus minus 1

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minus x cubed over 6, so that's
just 1 minus x cubed over 6.

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And so finally, the
outermost integral

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is we take the
inner two integrals

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and we integrate
them with respect

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to x as x goes from 0 to 1.

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So it's the integral from 0 to
1 of 1 minus x cubed over 6, dx.

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And that's going to equal-- I've
run out of space-- 1 over 24.

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All right.

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Except I've done something
wrong right at the beginning.

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I hope you all caught me.

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00:10:47,980 --> 00:10:48,540
Right?

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I had this 1 over V factor
here, and it disappeared.

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00:10:51,840 --> 00:10:52,340
Right?

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00:10:52,340 --> 00:10:55,370
I forgot about this 1
over V, so over here,

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I should have written
1 over V right

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in front of that integral.

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00:11:03,130 --> 00:11:05,570
I've correctly
computed 1 over 24

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00:11:05,570 --> 00:11:07,210
as the value of my
triple integral,

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but the average height
here isn't 1 over 24.

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It's 1 over 24V.

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All right.

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00:11:14,720 --> 00:11:18,007
So the average height is 1
over 24V, and so we need to go

223
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and we need to look
at our tetrahedron

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and figure out
what its volume is.

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So if we come over here
and see our tetrahedron.

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Now this is nice,
simple tetrahedron.

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The volume of a
tetrahedron is 1/3

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the area of the base
times the height, right?

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So this is a nice,
easy tetrahedron.

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Its height is 1.

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Its base is a right triangle
whose legs are both 1.

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So the base area is 1/2,
so the volume is 1/6.

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So if the volume
is 1/6, and we said

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the average value is 1 over 24V,
so that works out to 1 over 4.

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So let me write that
just in this space.

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So the average height
then is 1 over 4.

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So that's going to
be our final answer.

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00:12:05,860 --> 00:12:09,700
OK, so let's just recap
briefly what we did.

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00:12:09,700 --> 00:12:12,950
We had an average value
problem that we started with.

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00:12:12,950 --> 00:12:16,250
So we use this general formula
for average value problems.

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00:12:16,250 --> 00:12:18,500
When you have a
function f that you

242
00:12:18,500 --> 00:12:22,166
want to take its average
value of over a region R,

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00:12:22,166 --> 00:12:22,790
what do you do?

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Well, you take 1 over
the volume of the region

245
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times the triple integral of
your function f with respect

246
00:12:29,250 --> 00:12:31,610
to volume over that region.

247
00:12:31,610 --> 00:12:32,110
OK.

248
00:12:32,110 --> 00:12:34,850
So this is the average
value in general.

249
00:12:34,850 --> 00:12:37,845
In our particular case, the
function was the height.

250
00:12:37,845 --> 00:12:39,690
It was z.

251
00:12:39,690 --> 00:12:44,240
And then you have to set it
out choosing the proper bounds

252
00:12:44,240 --> 00:12:45,660
for your integrals.

253
00:12:45,660 --> 00:12:50,370
So in this case, you choose
some order of integration

254
00:12:50,370 --> 00:12:51,250
based on the region.

255
00:12:51,250 --> 00:12:53,270
In this particular case,
it's a nice, simple region.

256
00:12:53,270 --> 00:12:55,311
It doesn't matter too much
what order you choose.

257
00:12:57,760 --> 00:13:00,202
So I chose dz dy dx.

258
00:13:00,202 --> 00:13:01,410
And then what does that mean?

259
00:13:01,410 --> 00:13:04,580
So for the innermost one, you
look at your original solid.

260
00:13:04,580 --> 00:13:07,080
So I'm going to go back and
look at this picture again.

261
00:13:07,080 --> 00:13:10,340
So for your innermost variable
you say-- so if it's z,

262
00:13:10,340 --> 00:13:14,250
you say, so when I fix x and
y, what's the bottom surface

263
00:13:14,250 --> 00:13:18,090
and what's the top surface
when I solve that for z

264
00:13:18,090 --> 00:13:19,210
in terms of x and y?

265
00:13:19,210 --> 00:13:22,080
So here, that was
the plane z equals 0,

266
00:13:22,080 --> 00:13:25,250
and the plane z equals
1 minus x minus y.

267
00:13:25,250 --> 00:13:27,020
So that explains my
bounds over here.

268
00:13:27,020 --> 00:13:29,789
Why they were 0 and
1 minus x minus y.

269
00:13:29,789 --> 00:13:32,080
Then, when you go to your
next variable-- in this case,

270
00:13:32,080 --> 00:13:33,740
it was y-- what do you do?

271
00:13:33,740 --> 00:13:37,310
Well, first you project to
eliminate this first variable.

272
00:13:37,310 --> 00:13:40,190
So you project your region down.

273
00:13:40,190 --> 00:13:42,010
Down in this case,
because it's z.

274
00:13:42,010 --> 00:13:44,350
So you project in
the z-direction.

275
00:13:44,350 --> 00:13:47,600
And you draw this
shadow of your region.

276
00:13:47,600 --> 00:13:48,850
So this is what I drew here.

277
00:13:48,850 --> 00:13:51,220
This is the shadow of my
region in the xy-plane,

278
00:13:51,220 --> 00:13:52,339
after I projected it.

279
00:13:52,339 --> 00:13:53,630
And then you do the same thing.

280
00:13:53,630 --> 00:13:56,340
So now this just
like what you did

281
00:13:56,340 --> 00:13:59,650
when you had to find bounds
for double integrals, when

282
00:13:59,650 --> 00:14:04,550
you wrote them as iterated
integrals a few lectures ago.

283
00:14:06,964 --> 00:14:08,380
And so then you
do the same thing.

284
00:14:08,380 --> 00:14:10,040
So then, in this
case, I was next

285
00:14:10,040 --> 00:14:11,290
integrating with respect to y.

286
00:14:11,290 --> 00:14:14,180
So I needed to find the
bounds on y with respect to x.

287
00:14:14,180 --> 00:14:16,160
So I needed to look,
in this picture,

288
00:14:16,160 --> 00:14:20,050
at the bottom edge and the
top edge of this region.

289
00:14:20,050 --> 00:14:21,620
And then your
outermost variable,

290
00:14:21,620 --> 00:14:23,440
you look at its absolute bounds.

291
00:14:23,440 --> 00:14:26,460
So the largest and smallest
value it takes on the region.

292
00:14:26,460 --> 00:14:28,060
OK, then you have
an iterated integral

293
00:14:28,060 --> 00:14:32,330
and you evaluate it by
successive integrations.

294
00:14:32,330 --> 00:14:33,090
OK.

295
00:14:33,090 --> 00:14:34,110
So that was what we did.

296
00:14:34,110 --> 00:14:37,250
We just did the three integrals,
starting from the inside

297
00:14:37,250 --> 00:14:39,550
and working our way out.