WEBVTT

00:00:06.916 --> 00:00:07.540
JOEL LEWIS: Hi.

00:00:07.540 --> 00:00:08.894
Welcome back to recitation.

00:00:08.894 --> 00:00:11.310
In lecture, you've been learning
about triple integration,

00:00:11.310 --> 00:00:13.730
and I have a nice average
value problem for you

00:00:13.730 --> 00:00:15.610
using triple integration here.

00:00:15.610 --> 00:00:19.810
So what I'd like you to do is
to consider the tetrahedron that

00:00:19.810 --> 00:00:23.335
has vertices at the origin,
and at the points (1, 0, 0)

00:00:23.335 --> 00:00:25.750
and (0, 1, 0) and (0, 0, 1).

00:00:25.750 --> 00:00:28.880
So that's one point on
each of the positive axes,

00:00:28.880 --> 00:00:30.360
at distance 1 from the origin.

00:00:30.360 --> 00:00:34.920
So I've taken the liberty
of drawing it here for you.

00:00:34.920 --> 00:00:38.620
So consider that
solid tetrahedron.

00:00:38.620 --> 00:00:42.400
And what I'd like you to do
is find the average distance

00:00:42.400 --> 00:00:45.900
of the points in that
tetrahedron from the xy-plane.

00:00:45.900 --> 00:00:46.640
All right.

00:00:46.640 --> 00:00:49.820
So I'd like you to compute the
average value of the distance,

00:00:49.820 --> 00:00:53.830
as the point ranges over
the whole tetrahedron,

00:00:53.830 --> 00:00:56.580
of its distance
from the xy-plane.

00:00:56.580 --> 00:00:59.577
So why don't you
pause the video,

00:00:59.577 --> 00:01:01.410
spend some time working
that out, come back,

00:01:01.410 --> 00:01:02.743
and we can work it out together.

00:01:11.292 --> 00:01:13.250
Hopefully, you had some
luck with this problem.

00:01:13.250 --> 00:01:14.140
Let's get started.

00:01:19.660 --> 00:01:25.510
So the average value of a
function F over a region R

00:01:25.510 --> 00:01:29.040
is going to be 1 over
the volume of the region

00:01:29.040 --> 00:01:32.730
times the triple integral
over your whole region

00:01:32.730 --> 00:01:42.470
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.

00:01:43.540 --> 00:01:46.900
So I guess I'll write dV here,
and then to evaluate this,

00:01:46.900 --> 00:01:48.730
you set it up as an
iterated integral.

00:01:48.730 --> 00:01:50.160
So in our case,
the function we're

00:01:50.160 --> 00:01:51.730
trying to find the
average value of

00:01:51.730 --> 00:01:55.070
is the distance between
a point and the xy-plane.

00:01:55.070 --> 00:01:56.770
But that's an easy
function, right?

00:01:56.770 --> 00:01:58.310
That's just z.

00:01:58.310 --> 00:02:01.820
For any point in space, its
distance from the xy-plane

00:02:01.820 --> 00:02:02.770
is just its height.

00:02:02.770 --> 00:02:03.534
It's z-value.

00:02:03.534 --> 00:02:04.950
So the function
that we're seeking

00:02:04.950 --> 00:02:06.506
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

00:02:08.130 --> 00:02:10.421
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.

00:02:14.830 --> 00:02:17.890
So let's think about the bounds.

00:02:17.890 --> 00:02:21.250
This tetrahedron is a
nice, reasonably simple,

00:02:21.250 --> 00:02:24.810
geometric object.

00:02:24.810 --> 00:02:28.160
So in fact, it doesn't
matter too much which order

00:02:28.160 --> 00:02:29.320
you take your bounds.

00:02:29.320 --> 00:02:33.460
So I think I'm going to do
it in the order dz dy dx.

00:02:33.460 --> 00:02:36.320
You know, I'll do z first,
and then y, and then x.

00:02:36.320 --> 00:02:37.350
But it doesn't matter.

00:02:37.350 --> 00:02:38.860
If you do it a
different way, it'll

00:02:38.860 --> 00:02:42.260
probably work out very
similarly overall,

00:02:42.260 --> 00:02:46.330
and you'll still be able to
compare the overall process.

00:02:46.330 --> 00:02:48.501
So let's think about z.

00:02:48.501 --> 00:02:49.000
Yeah?

00:02:49.000 --> 00:02:55.750
So in this tetrahedron, we
want to integrate with respect

00:02:55.750 --> 00:02:57.290
to z first.

00:02:57.290 --> 00:03:00.620
So we look at this
tetrahedron and we say, OK,

00:03:00.620 --> 00:03:02.210
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

00:03:04.010 --> 00:03:07.580
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

00:03:09.460 --> 00:03:11.060
the largest value z can take?

00:03:11.060 --> 00:03:13.500
So the lowest surface
here is the xy-plane.

00:03:13.500 --> 00:03:15.490
That's the bottom face
of this tetrahedron.

00:03:15.490 --> 00:03:19.580
And for any choice of x
and y, the lowest value z

00:03:19.580 --> 00:03:21.570
can take is when
it's in the xy-plane.

00:03:21.570 --> 00:03:22.570
So when it's equal to 0.

00:03:25.386 --> 00:03:27.260
So this is going to be
equal, so in our case,

00:03:27.260 --> 00:03:28.230
so let's set this up.

00:03:28.230 --> 00:03:31.241
So it's going to be
an iterated integral.

00:03:31.241 --> 00:03:33.320
The function we're
integrating is z,

00:03:33.320 --> 00:03:38.500
and I said we'll do dz dy dx.

00:03:38.500 --> 00:03:42.970
And we just said that the lowest
value that z takes-- the dz--

00:03:42.970 --> 00:03:43.660
is 0.

00:03:43.660 --> 00:03:45.870
So the highest
value that z takes

00:03:45.870 --> 00:03:48.250
is when it hits
this top surface.

00:03:48.250 --> 00:03:53.210
This plane that passes through
the point (1, 0, 0), (0, 1, 0),

00:03:53.210 --> 00:03:54.531
and (0, 0, 1).

00:03:54.531 --> 00:03:55.030
All right.

00:03:55.030 --> 00:03:56.992
So we need to know the
equation of that plane.

00:03:56.992 --> 00:03:58.450
Luckily, that's a
pretty easy plane

00:03:58.450 --> 00:04:00.470
to write down the equation for.

00:04:00.470 --> 00:04:09.050
So this slanted plane passing
through the three vertices

00:04:09.050 --> 00:04:16.465
other than the origin is the
plane x plus y plus z equals 1.

00:04:16.465 --> 00:04:17.160
All right.

00:04:17.160 --> 00:04:19.239
So it's a nice, easy
plane to work with.

00:04:19.239 --> 00:04:20.780
And so what we want
to know is what's

00:04:20.780 --> 00:04:22.450
the value of z on that plane.

00:04:22.450 --> 00:04:25.200
So we isolate z on one side and
we bring everything else over.

00:04:25.200 --> 00:04:29.590
So we have that top value
of z, the largest value of z

00:04:29.590 --> 00:04:33.534
that z can take when x and y
are fixed, is 1 minus x minus 1.

00:04:33.534 --> 00:04:34.700
So that's what goes up here.

00:04:37.830 --> 00:04:40.360
So that's the biggest
value z can take.

00:04:40.360 --> 00:04:41.460
OK, good.

00:04:41.460 --> 00:04:44.690
So now we need to figure
out what the bounds on y

00:04:44.690 --> 00:04:46.880
are in terms of x.

00:04:46.880 --> 00:04:49.260
So what I like to
do in this case

00:04:49.260 --> 00:04:53.110
is I like to draw a
projection of your surface.

00:04:53.110 --> 00:04:55.980
So then you're in a
two-dimensional world,

00:04:55.980 --> 00:04:58.370
and then you can look at
that image more easily.

00:04:58.370 --> 00:04:59.150
So what we're going
to do is we're

00:04:59.150 --> 00:05:00.730
going to look at
this tetrahedron

00:05:00.730 --> 00:05:03.330
and we're going to
imagine projecting it down

00:05:03.330 --> 00:05:04.480
into the xy-plane.

00:05:04.480 --> 00:05:06.210
So for every point
in the tetrahedron,

00:05:06.210 --> 00:05:08.800
we're going to draw a dot
below it in the xy-plane.

00:05:08.800 --> 00:05:11.070
And then we're going to
look at that set of dots.

00:05:11.070 --> 00:05:16.000
So what that's going to
give us is this bottom face

00:05:16.000 --> 00:05:17.060
of the tetrahedron.

00:05:17.060 --> 00:05:19.477
Every point of the tetrahedron
is above its bottom face.

00:05:19.477 --> 00:05:21.060
That's not true for
every tetrahedron,

00:05:21.060 --> 00:05:23.160
but it's true for this one.

00:05:23.160 --> 00:05:24.890
So the region that
we're interested in

00:05:24.890 --> 00:05:25.930
is that bottom face.

00:05:25.930 --> 00:05:29.070
So I'm going to draw another
picture of it over on my left

00:05:29.070 --> 00:05:29.780
here.

00:05:29.780 --> 00:05:39.065
So that region is the region
that has vertices (0, 0),

00:05:39.065 --> 00:05:41.370
and (1, 0), and (0, 1).

00:05:41.370 --> 00:05:44.050
So it's this triangle.

00:05:44.050 --> 00:05:48.280
And this bottom
edge of the triangle

00:05:48.280 --> 00:05:49.860
is the line y equals 0.

00:05:49.860 --> 00:05:52.770
This left edge is
the line x equals 0.

00:05:52.770 --> 00:05:58.490
And this slanted line is
the line x plus y equals 1.

00:05:58.490 --> 00:06:03.310
OK, so this is that same
bottom face that we just drew.

00:06:03.310 --> 00:06:06.970
But now I've changed my axes
to our usual two-dimensional

00:06:06.970 --> 00:06:11.550
direction with x to
the right and y up.

00:06:11.550 --> 00:06:13.560
OK.

00:06:13.560 --> 00:06:14.890
So we're doing dy next.

00:06:14.890 --> 00:06:16.640
So we need to figure
out for a fixed value

00:06:16.640 --> 00:06:19.270
of x, what are the bounds on y?

00:06:19.270 --> 00:06:22.860
So we see from this picture
that for any fixed value of x, y

00:06:22.860 --> 00:06:27.760
goes from 0-- the
x-axis-- up to this line.

00:06:27.760 --> 00:06:28.330
OK?

00:06:28.330 --> 00:06:32.590
So on the x-axis, y takes the
value 0, and on this line,

00:06:32.590 --> 00:06:36.640
y takes the value 1 minus x.

00:06:36.640 --> 00:06:40.160
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

00:06:50.250 --> 00:06:51.708
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

00:07:11.830 --> 00:07:14.620
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

00:07:30.010 --> 00:07:32.250
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.

00:08:10.250 --> 00:08:15.070
Let me write that, "inner."

00:08:15.070 --> 00:08:16.170
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.