WEBVTT
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CHRISTINE BREINER: Welcome
back to recitation.
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In this video I'd like us to
work on the following problem.
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We're going to let
capital D denote
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the portion of the solid sphere
of radius 1 that's centered at
00:00:18.470 --> 00:00:23.110
(0, 0, 1), which also lies
about the plane z equal 1.
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And then I'd like us to
first supply the limits for D
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in spherical coordinates.
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In other words, I want you
to determine the values
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for rho, theta, and phi
that will give us all of D.
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And then, I would
like us to just set up
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the integral for the average
distance of a point in D
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from the origin.
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So there are two
parts to this problem.
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The first is to determine what
values of rho, theta, and phi
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describe this solid
region D. And then second
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is just set up the integral for
the average distance of a point
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in that region from the origin.
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So why don't you pause
the video, work on those,
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and then when you're
ready to see my solutions,
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you can bring the video back up.
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OK, welcome back.
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Well, I would say from looking
at this problem, actually part
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a is potentially a little bit
more hazardous for some of us
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than part b.
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Once we know the
bounds that describe D,
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it's not too hard to
set up this integral.
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So the hard part
of this problem is
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understanding how to write this
solid region D in spherical
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coordinates.
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But it's actually really
not that hard either,
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so I'm going to try and take us
through it in a reasonable way.
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So the first thing:
I'm going to draw
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a very rough picture
of the region
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so we understand
what it looks like.
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So in order to do part a.
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And if you did not do
this, I would highly
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recommend that next time you
encounter such a problem,
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you begin by drawing
yourself a picture.
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Even if it's not
a great picture,
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it will give you some intuition
about what's happening.
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So let me first draw the axes.
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And ultimately
what I have-- say,
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here is the point (0, 0, 1).
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I have a sphere that's
looking-- in the zy-plane,
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it looks something like this.
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And so it has this depth here.
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And that's the solid sphere
I want to be considering.
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And then I'm going to be
removing the bottom half.
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I'm only going to be looking
at the part that is above the z
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equals 1 plane.
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So actually it's going to be
all of the circular slices that
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are in the top half, the upper
hemisphere of this sphere.
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And so it's this
solid region there.
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That's D.
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And what I want to
do is to determine
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rho and theta and phi.
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Actually, the theta is
the easy one, right?
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Because theta, at each value
here, I go all the way around
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in the theta direction.
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At any given height or radius,
I want to go all the way around
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in the theta direction
from 0 to 2*pi.
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So my theta bounds
are the easy ones.
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I'm covering 0 to 2*pi in theta.
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Because if I cut off the
back half of the sphere,
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I want to only have a
restricted value of theta.
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But because I'm covering
all the way around
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and my restriction is
only in the bottom half,
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my theta values haven't changed.
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So theta is easy: 0 and 2*pi.
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Now the harder ones are
going to be rho and phi.
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But in fact, actually, phi
is not that hard either.
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I notice phi is the
angle that I make
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from the z-axis to
any given point.
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So I notice that I certainly
am including the point where
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phi is 0, and then I'm going
all the way down to a point
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out here, which is a 45
degree angle with the z-axis.
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So phi is also easy.
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It's actually just
between 0 and pi over 4.
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And the rho will be the
slightly harder part.
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So that's really the
only really tricky part
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in this problem is
determining the rho value.
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Now, the rho value, to
get the outer boundary,
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we'll look at that part first.
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Well, the boundary
of this sphere
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here has a certain equation
in x, y, and z that we know,
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right?
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It's x squared plus y squared
plus the quantity z minus 1
00:04:38.710 --> 00:04:41.360
squared equals 1.
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I mean, that's just the
equation for a sphere of radius
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1 centered at (0, 0, 1).
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So I'm going to write
that here, and we're
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going to show how we
can manipulate that.
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Right?
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x squared plus y
squared is r squared.
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r squared is rho squared
sine squared phi.
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So I can replace this by rho
squared sine squared phi.
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If you didn't know
that immediately,
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you could make the
substitution for x and y
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in spherical coordinates,
and it simplifies to this.
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So either way, if you didn't
know r squared was this,
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you can get it from just
doing the substitution.
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And then z is going
to be rho cosine phi.
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So here, I'm going to have
a rho cosine phi minus 1
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quantity squared equals 1.
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This, in the
spherical coordinates,
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is describing the boundary
of this entire sphere, right?
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And so I can actually
simplify this.
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It's not too hard.
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If I square this, I get
a little cancellation.
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And then because I
want my rho to be-- I'm
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assuming in this region,
rho is greater than 0--
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I can do a little
simplification.
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I come up with the fact that
rho is equal to 2 cosine phi.
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And let's describe
exactly where that is.
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That's the entire boundary
of this entire sphere
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is described by rho is
equal to 2 cosine phi.
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And so I want to think about
what my bounds are for rho.
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Actually, I'm going to grab
a piece of colored chalk.
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If I start at the origin,
I think about what is rho?
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So say this is a point on
the boundary of the sphere.
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I am going to start my
rho value-- whatever
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it is when it hits
the plane z equals 1--
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and I'm going to
stop it when it hits
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the boundary of this sphere.
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So my outer boundary for rho
is going to be this value.
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It's going to be
determined by phi, right?
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And now I have to determine
my inner boundary, right?
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And my inner boundary is
actually quite simple.
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It's a very simple
geometric thing.
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And so my inner boundary
deals with the fact
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that if this is my
plane z equals 1,
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and I look at this
triangle I make right here.
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This angle down here,
the bottom angle is phi,
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and this is a right angle, and
the rho value I'm interested in
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is this hypotenuse, right?
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I need to figure out what the
length of this is right here.
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And you can see it right
away from just the fact
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that phi is this angle here,
you get rho is secant phi.
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So the bottom boundary comes
from just simple geometry.
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You get this length is 1 here.
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So you get rho is equal
to secant phi, right?
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This length here is
1, this is the rho
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I'm interested in--
the blue part here--
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and so rho is equal to secant
phi is the lower bound.
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And it's equal to 2 cosine
phi at the upper bound, OK?
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And the thing I want
to be careful of
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is I'm not supposed to
include-- it won't matter
00:07:43.820 --> 00:07:47.680
for the integral-- but I'm not
supposed to include the plane.
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Let me write this,
and make sure.
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Rho is going to be
greater than secant phi
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and it's going to be less
than or equal to 2 cosine phi,
00:08:00.540 --> 00:08:02.990
right?
00:08:02.990 --> 00:08:03.490
Right?
00:08:03.490 --> 00:08:06.031
So let me double-check and make
sure I didn't make a geometry
00:08:06.031 --> 00:08:07.650
mistake here, just to be sure.
00:08:07.650 --> 00:08:10.710
This picture tells
me that cosine phi
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is equal to 1 over rho.
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That's good.
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So rho is equal to
1 over cosine phi.
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So I get secant phi there.
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So my rho values start
at the secant phi length
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and they go to the
2 cosine phi length.
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I know maybe I'm beating
a dead horse here,
00:08:25.980 --> 00:08:27.771
but I want to make sure
we understand where
00:08:27.771 --> 00:08:29.070
the rho values are coming from.
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So actually, I have all
the bounds I need now.
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I have the theta bounds,
and I have the phi bounds,
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and the rho bounds.
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Now you notice
that theta and phi
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don't depend on the
other variables,
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but rho depends on phi.
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So we're going to have
to integrate that first.
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So now we can deal with part b.
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Part b-- let's come back
over and remind ourselves
00:08:50.950 --> 00:08:53.350
what it said-- said
set up the integral
00:08:53.350 --> 00:08:57.600
for the average distance of
a point in D from the origin.
00:08:57.600 --> 00:09:00.080
So I'm taking the average
value of a function.
00:09:00.080 --> 00:09:02.190
What is that function
I'm averaging?
00:09:02.190 --> 00:09:04.474
How do I find the
distance from the origin?
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Well, the distance
from the origin
00:09:05.890 --> 00:09:08.098
is a great function to have
in spherical coordinates,
00:09:08.098 --> 00:09:09.540
because it's just rho.
00:09:09.540 --> 00:09:12.180
So the function I'm
supposed to average over
00:09:12.180 --> 00:09:13.880
is the function rho.
00:09:13.880 --> 00:09:15.880
In spherical coordinates,
that's the function.
00:09:15.880 --> 00:09:18.010
So let me write down what
we're going to have here.
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So in part b, the
average distance
00:09:27.050 --> 00:09:34.330
is going to equal 1
divided by the volume of D
00:09:34.330 --> 00:09:44.000
times the triple integral over
D of the function rho dV, OK?
00:09:44.000 --> 00:09:47.520
So now I have to write dV in
the spherical coordinates,
00:09:47.520 --> 00:09:50.802
and I have to write D in the
spherical coordinates bounds.
00:09:50.802 --> 00:09:53.010
And then I know I have to
figure out the volume of D.
00:09:53.010 --> 00:09:54.640
So we're going to figure
out each of these things,
00:09:54.640 --> 00:09:56.012
and then we'll be done.
00:09:56.012 --> 00:09:56.600
All right.
00:09:56.600 --> 00:09:58.650
So first, what is
the volume of D?
00:09:58.650 --> 00:10:01.080
Well, the volume of D, let's
think about what it is.
00:10:01.080 --> 00:10:06.180
It's a sphere of radius 1.
00:10:06.180 --> 00:10:13.400
And so the volume of a sphere
of radius 1 is 4/3 pi r cubed.
00:10:13.400 --> 00:10:14.740
And I want half of that.
00:10:14.740 --> 00:10:16.260
So I want 2/3.
00:10:16.260 --> 00:10:19.700
Since my radius is 1, I
just have to do 2/3 pi.
00:10:19.700 --> 00:10:24.980
So the first part is
1 divided by 2/3 pi.
00:10:24.980 --> 00:10:28.790
That's the volume of a
half-sphere of radius 1.
00:10:28.790 --> 00:10:29.855
And now let's integrate.
00:10:33.220 --> 00:10:36.502
I'll leave a little
space to write my bounds.
00:10:36.502 --> 00:10:37.960
I'm going to write
the bounds last,
00:10:37.960 --> 00:10:40.080
after I have everything
in order over here.
00:10:40.080 --> 00:10:45.490
dV is rho squared sine
phi d rho d theta d phi.
00:10:45.490 --> 00:10:51.680
So I'm going to end up with
a rho cubed sine phi d rho d
00:10:51.680 --> 00:10:55.750
theta d phi, right?
00:10:55.750 --> 00:11:01.030
The dV gave me an extra
rho squared and a sine phi.
00:11:01.030 --> 00:11:02.830
That whole part is dV.
00:11:02.830 --> 00:11:05.660
And then I keep one
rho from the fact
00:11:05.660 --> 00:11:08.100
that the distance
function is rho.
00:11:08.100 --> 00:11:10.181
And so I get a rho cubed there.
00:11:10.181 --> 00:11:11.430
So hopefully that makes sense.
00:11:11.430 --> 00:11:19.025
Now for d rho, I know the bounds
are secant phi to 2 cosine phi.
00:11:19.025 --> 00:11:22.820
For d theta, my
bounds are 0 to 2 pi.
00:11:22.820 --> 00:11:25.930
And for d phi, my bounds
were 0 to pi over 4.
00:11:28.660 --> 00:11:32.180
I didn't make you evaluate it,
I'm just making you set it up.
00:11:32.180 --> 00:11:35.510
That actually is the solution
we wanted for part b.
00:11:35.510 --> 00:11:38.340
I wanted to average the
distance from any point in D
00:11:38.340 --> 00:11:39.520
to the origin.
00:11:39.520 --> 00:11:42.600
So I just took the average
value of the function rho
00:11:42.600 --> 00:11:46.460
over that region D. And so
that's how you finish that up.
00:11:46.460 --> 00:11:49.190
And so in this
problem, basically we
00:11:49.190 --> 00:11:51.030
want you to get really
familiar with how
00:11:51.030 --> 00:11:54.800
to do some things in these
spherical coordinates, which
00:11:54.800 --> 00:11:56.860
are sometimes a
little hard to do.
00:11:56.860 --> 00:11:59.450
But if you noticed, what we were
doing in trying to figure out
00:11:59.450 --> 00:12:02.150
the bounds-- in particular,
trying to figure out rho--
00:12:02.150 --> 00:12:04.210
we took what we
knew in the x-, y-,
00:12:04.210 --> 00:12:06.940
z-coordinates about
certain relationships,
00:12:06.940 --> 00:12:11.640
and then we replaced the x-, y-,
z-values by the values in terms
00:12:11.640 --> 00:12:16.730
of rho and theta and phi, and
you can simplify to figure out
00:12:16.730 --> 00:12:19.880
the relationships you have
between rho and theta and phi
00:12:19.880 --> 00:12:21.410
for the boundary value.
00:12:21.410 --> 00:12:24.390
So that was one of the
techniques we were using there.
00:12:24.390 --> 00:12:28.430
And hopefully, the geometric
understanding of why these
00:12:28.430 --> 00:12:32.130
angles go from 0 to 2*pi
and 0 to pi over 4 is clear.
00:12:32.130 --> 00:12:35.080
And actually, the
fact that this is rho
00:12:35.080 --> 00:12:36.650
equals 2 cosine phi
should remind you
00:12:36.650 --> 00:12:39.230
of the two-dimensional case
where you had some problem
00:12:39.230 --> 00:12:41.510
like r equals 2 cosine theta.
00:12:41.510 --> 00:12:46.140
And that drew a circle
off-center from the origin.
00:12:46.140 --> 00:12:50.199
It's the analogous
thing happening here.
00:12:50.199 --> 00:12:52.490
Maybe I should stop there
before I say too many things.
00:12:52.490 --> 00:12:54.070
But again, the object
of this was just
00:12:54.070 --> 00:12:55.880
to get really comfortable
with spherical coordinates,
00:12:55.880 --> 00:12:57.100
and I hope it's helped you.
00:12:57.100 --> 00:12:58.700
I'll stop there.