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

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(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

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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

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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,

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right?

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Right?

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So let me double-check and make
sure I didn't make a geometry

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mistake here, just to be sure.

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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,

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but I want to make sure
we understand where

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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

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what it said-- said
set up the integral

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for the average distance of
a point in D from the origin.

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So I'm taking the average
value of a function.

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What is that function
I'm averaging?

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How do I find the
distance from the origin?

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Well, the distance
from the origin

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is a great function to have
in spherical coordinates,

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because it's just rho.

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So the function I'm
supposed to average over

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is the function rho.

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In spherical coordinates,
that's the function.

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So let me write down what
we're going to have here.

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So in part b, the
average distance

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is going to equal 1
divided by the volume of D

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times the triple integral over
D of the function rho dV, OK?

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So now I have to write dV in
the spherical coordinates,

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and I have to write D in the
spherical coordinates bounds.

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And then I know I have to
figure out the volume of D.

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So we're going to figure
out each of these things,

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and then we'll be done.

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All right.

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So first, what is
the volume of D?

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Well, the volume of D, let's
think about what it is.

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It's a sphere of radius 1.

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And so the volume of a sphere
of radius 1 is 4/3 pi r cubed.

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And I want half of that.

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So I want 2/3.

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Since my radius is 1, I
just have to do 2/3 pi.

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So the first part is
1 divided by 2/3 pi.

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That's the volume of a
half-sphere of radius 1.

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And now let's integrate.

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I'll leave a little
space to write my bounds.

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I'm going to write
the bounds last,

00:10:37.960 --> 00:10:40.080
after I have everything
in order over here.

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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.