WEBVTT
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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
00:00:13.730 --> 00:00:15.610
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,
00:01:42.470 --> 00:01:43.540
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.
00:01:48.730 --> 00:01:50.160
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,
00:02:06.506 --> 00:02:08.130
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
00:02:10.421 --> 00:02:13.850
and then doing the actual
integral after that.
00:02:13.850 --> 00:02:14.830
So OK.
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So let's think about the bounds.
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This tetrahedron is a
nice, reasonably simple,
00:02:21.250 --> 00:02:24.810
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
00:02:38.860 --> 00:02:42.260
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
00:03:02.210 --> 00:03:04.010
we choose the x-
and y-values, what's
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the lowest surface-- what's
the smallest value z can take--
00:03:07.580 --> 00:03:09.460
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.
00:05:04.480 --> 00:05:06.210
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
00:06:40.160 --> 00:06:41.790
is x, and so we
need to know what
00:06:41.790 --> 00:06:48.251
are the absolute largest and
smallest values that x takes.
00:06:48.251 --> 00:06:50.250
So we can do that looking
either at this picture
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or looking at our
original picture.
00:06:51.708 --> 00:06:53.820
Either way, it's not
hard to see that x just
00:06:53.820 --> 00:06:55.250
goes between 0 and 1.
00:06:55.250 --> 00:06:57.680
The smallest value that x
takes in this tetrahedron is 0.
00:06:57.680 --> 00:07:01.320
The largest value it takes is 1.
00:07:01.320 --> 00:07:03.430
So that's our integral
that we have to compute.
00:07:03.430 --> 00:07:04.840
So that's not that bad at all.
00:07:04.840 --> 00:07:06.590
So now you have to go
through and you have
00:07:06.590 --> 00:07:08.300
to actually integrate this.
00:07:08.300 --> 00:07:08.800
Yeah?
00:07:08.800 --> 00:07:11.830
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.
00:07:14.620 --> 00:07:16.930
So let's see.
00:07:16.930 --> 00:07:18.920
Now we do these one at a time.
00:07:18.920 --> 00:07:23.435
So this innermost integral
is an integral of z dz.
00:07:23.435 --> 00:07:24.310
OK, well that's easy.
00:07:24.310 --> 00:07:26.020
That's z squared over 2.
00:07:26.020 --> 00:07:30.010
And then we're taking z
squared over 2 between 0 and 1
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minus x minus y.
00:07:32.250 --> 00:07:42.790
So this innermost integral is z
squared over 2 between 0 and 1
00:07:42.790 --> 00:07:45.270
minus x minus y.
00:07:45.270 --> 00:07:50.370
So that's equal to-- so the
innermost integral gives us 1
00:07:50.370 --> 00:07:53.999
minus x minus y squared over 2.
00:07:53.999 --> 00:07:56.040
So that's what we get for
the innermost integral.
00:07:56.040 --> 00:07:59.470
So our integral that
we're looking at, then,
00:07:59.470 --> 00:08:02.010
is equal to the
integral, as x goes
00:08:02.010 --> 00:08:06.240
from 0 to 1, of the integral
as y goes from 0 to 1 minus x
00:08:06.240 --> 00:08:09.190
of this integrand.
00:08:09.190 --> 00:08:10.250
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.
00:08:16.170 --> 00:08:19.420
Just integrating z with respect
to z gives me z squared over 2.
00:08:19.420 --> 00:08:23.010
And then I evaluated at
the bounds of the integral.
00:08:23.010 --> 00:08:25.080
OK, so now I need to
do the middle one.
00:08:25.080 --> 00:08:26.570
So let's do that up here.
00:08:31.980 --> 00:08:33.540
So I need to compute
the integral.
00:08:33.540 --> 00:08:36.310
So now I take the bounds,
so the middle one is y,
00:08:36.310 --> 00:08:44.937
and the bounds are from 0 to 1
minus x, of the inner integral.
00:08:44.937 --> 00:08:46.270
This thing that I just computed.
00:08:46.270 --> 00:08:56.820
So that's of 1 minus x
minus y squared over 2, dy.
00:08:56.820 --> 00:08:57.390
OK.
00:08:57.390 --> 00:08:58.982
So All right.
00:08:58.982 --> 00:08:59.940
So this isn't that bad.
00:08:59.940 --> 00:09:04.170
This is a quadratic
polynomial in y.
00:09:04.170 --> 00:09:06.440
And so it's not
terribly hard to see.
00:09:06.440 --> 00:09:08.990
I'm running a little
bit out of board space.
00:09:08.990 --> 00:09:14.090
So I'm not going to give you
a full, detailed explanation.
00:09:14.090 --> 00:09:16.820
But it's not hard
to see, I think,
00:09:16.820 --> 00:09:25.740
that this integral of 1
minus x minus y squared
00:09:25.740 --> 00:09:33.942
over 2 with respect to y is 1
minus x minus y cubed over 3,
00:09:33.942 --> 00:09:36.150
but it's negative, because
the sign here is negative.
00:09:36.150 --> 00:09:38.240
And you could check by
differentiating this and seeing
00:09:38.240 --> 00:09:39.030
that you get that.
00:09:39.030 --> 00:09:46.290
And so we have to evaluate that
as y goes from 0 to 1 minus x.
00:09:46.290 --> 00:09:47.040
So what do we get?
00:09:47.040 --> 00:09:52.050
Well, when y is equal
to 1 minus x, this is 0.
00:09:52.050 --> 00:09:58.070
So we get 0 minus-- and when y
is equal to 0, this is minus 1
00:09:58.070 --> 00:10:02.495
minus x quantity cubed over
6-- so it's minus minus 1
00:10:02.495 --> 00:10:09.760
minus x cubed over 6, so that's
just 1 minus x cubed over 6.
00:10:09.760 --> 00:10:16.210
And so finally, the
outermost integral
00:10:16.210 --> 00:10:20.060
is we take the
inner two integrals
00:10:20.060 --> 00:10:21.490
and we integrate
them with respect
00:10:21.490 --> 00:10:23.600
to x as x goes from 0 to 1.
00:10:23.600 --> 00:10:33.590
So it's the integral from 0 to
1 of 1 minus x cubed over 6, dx.
00:10:33.590 --> 00:10:44.261
And that's going to equal-- I've
run out of space-- 1 over 24.
00:10:44.261 --> 00:10:44.760
All right.
00:10:44.760 --> 00:10:47.080
Except I've done something
wrong right at the beginning.
00:10:47.080 --> 00:10:47.980
I hope you all caught me.
00:10:47.980 --> 00:10:48.540
Right?
00:10:48.540 --> 00:10:51.840
I had this 1 over V factor
here, and it disappeared.
00:10:51.840 --> 00:10:52.340
Right?
00:10:52.340 --> 00:10:55.370
I forgot about this 1
over V, so over here,
00:10:55.370 --> 00:10:57.087
I should have written
1 over V right
00:10:57.087 --> 00:10:58.170
in front of that integral.
00:11:03.130 --> 00:11:05.570
I've correctly
computed 1 over 24
00:11:05.570 --> 00:11:07.210
as the value of my
triple integral,
00:11:07.210 --> 00:11:10.750
but the average height
here isn't 1 over 24.
00:11:10.750 --> 00:11:13.252
It's 1 over 24V.
00:11:13.252 --> 00:11:14.720
All right.
00:11:14.720 --> 00:11:18.007
So the average height is 1
over 24V, and so we need to go
00:11:18.007 --> 00:11:19.590
and we need to look
at our tetrahedron
00:11:19.590 --> 00:11:21.430
and figure out
what its volume is.
00:11:21.430 --> 00:11:23.560
So if we come over here
and see our tetrahedron.
00:11:23.560 --> 00:11:25.960
Now this is nice,
simple tetrahedron.
00:11:25.960 --> 00:11:30.140
The volume of a
tetrahedron is 1/3
00:11:30.140 --> 00:11:34.170
the area of the base
times the height, right?
00:11:34.170 --> 00:11:36.540
So this is a nice,
easy tetrahedron.
00:11:36.540 --> 00:11:37.860
Its height is 1.
00:11:37.860 --> 00:11:40.610
Its base is a right triangle
whose legs are both 1.
00:11:40.610 --> 00:11:44.990
So the base area is 1/2,
so the volume is 1/6.
00:11:44.990 --> 00:11:47.250
So if the volume
is 1/6, and we said
00:11:47.250 --> 00:11:52.020
the average value is 1 over 24V,
so that works out to 1 over 4.
00:11:52.020 --> 00:11:55.170
So let me write that
just in this space.
00:11:55.170 --> 00:12:03.500
So the average height
then is 1 over 4.
00:12:03.500 --> 00:12:05.860
So that's going to
be our final answer.
00:12:05.860 --> 00:12:09.700
OK, so let's just recap
briefly what we did.
00:12:09.700 --> 00:12:12.950
We had an average value
problem that we started with.
00:12:12.950 --> 00:12:16.250
So we use this general formula
for average value problems.
00:12:16.250 --> 00:12:18.500
When you have a
function f that you
00:12:18.500 --> 00:12:22.166
want to take its average
value of over a region R,
00:12:22.166 --> 00:12:22.790
what do you do?
00:12:22.790 --> 00:12:25.630
Well, you take 1 over
the volume of the region
00:12:25.630 --> 00:12:29.250
times the triple integral of
your function f with respect
00:12:29.250 --> 00:12:31.610
to volume over that region.
00:12:31.610 --> 00:12:32.110
OK.
00:12:32.110 --> 00:12:34.850
So this is the average
value in general.
00:12:34.850 --> 00:12:37.845
In our particular case, the
function was the height.
00:12:37.845 --> 00:12:39.690
It was z.
00:12:39.690 --> 00:12:44.240
And then you have to set it
out choosing the proper bounds
00:12:44.240 --> 00:12:45.660
for your integrals.
00:12:45.660 --> 00:12:50.370
So in this case, you choose
some order of integration
00:12:50.370 --> 00:12:51.250
based on the region.
00:12:51.250 --> 00:12:53.270
In this particular case,
it's a nice, simple region.
00:12:53.270 --> 00:12:55.311
It doesn't matter too much
what order you choose.
00:12:57.760 --> 00:13:00.202
So I chose dz dy dx.
00:13:00.202 --> 00:13:01.410
And then what does that mean?
00:13:01.410 --> 00:13:04.580
So for the innermost one, you
look at your original solid.
00:13:04.580 --> 00:13:07.080
So I'm going to go back and
look at this picture again.
00:13:07.080 --> 00:13:10.340
So for your innermost variable
you say-- so if it's z,
00:13:10.340 --> 00:13:14.250
you say, so when I fix x and
y, what's the bottom surface
00:13:14.250 --> 00:13:18.090
and what's the top surface
when I solve that for z
00:13:18.090 --> 00:13:19.210
in terms of x and y?
00:13:19.210 --> 00:13:22.080
So here, that was
the plane z equals 0,
00:13:22.080 --> 00:13:25.250
and the plane z equals
1 minus x minus y.
00:13:25.250 --> 00:13:27.020
So that explains my
bounds over here.
00:13:27.020 --> 00:13:29.789
Why they were 0 and
1 minus x minus y.
00:13:29.789 --> 00:13:32.080
Then, when you go to your
next variable-- in this case,
00:13:32.080 --> 00:13:33.740
it was y-- what do you do?
00:13:33.740 --> 00:13:37.310
Well, first you project to
eliminate this first variable.
00:13:37.310 --> 00:13:40.190
So you project your region down.
00:13:40.190 --> 00:13:42.010
Down in this case,
because it's z.
00:13:42.010 --> 00:13:44.350
So you project in
the z-direction.
00:13:44.350 --> 00:13:47.600
And you draw this
shadow of your region.
00:13:47.600 --> 00:13:48.850
So this is what I drew here.
00:13:48.850 --> 00:13:51.220
This is the shadow of my
region in the xy-plane,
00:13:51.220 --> 00:13:52.339
after I projected it.
00:13:52.339 --> 00:13:53.630
And then you do the same thing.
00:13:53.630 --> 00:13:56.340
So now this just
like what you did
00:13:56.340 --> 00:13:59.650
when you had to find bounds
for double integrals, when
00:13:59.650 --> 00:14:04.550
you wrote them as iterated
integrals a few lectures ago.
00:14:06.964 --> 00:14:08.380
And so then you
do the same thing.
00:14:08.380 --> 00:14:10.040
So then, in this
case, I was next
00:14:10.040 --> 00:14:11.290
integrating with respect to y.
00:14:11.290 --> 00:14:14.180
So I needed to find the
bounds on y with respect to x.
00:14:14.180 --> 00:14:16.160
So I needed to look,
in this picture,
00:14:16.160 --> 00:14:20.050
at the bottom edge and the
top edge of this region.
00:14:20.050 --> 00:14:21.620
And then your
outermost variable,
00:14:21.620 --> 00:14:23.440
you look at its absolute bounds.
00:14:23.440 --> 00:14:26.460
So the largest and smallest
value it takes on the region.
00:14:26.460 --> 00:14:28.060
OK, then you have
an iterated integral
00:14:28.060 --> 00:14:32.330
and you evaluate it by
successive integrations.
00:14:32.330 --> 00:14:33.090
OK.
00:14:33.090 --> 00:14:34.110
So that was what we did.
00:14:34.110 --> 00:14:37.250
We just did the three integrals,
starting from the inside
00:14:37.250 --> 00:14:39.550
and working our way out.