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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 the portion of the

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solid sphere of radius 1 that's
centered at 0, 0, 1,

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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 in

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spherical coordinates.

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In other words, I want you to
determine the values for rho,

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theta, and phi that will give us
all of D. And then, I would

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like us to just set up the
integral for the average

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distance of a point in
D 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 is just set up

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the integral for the average
distance of a point in that

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region from the origin.

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So why don't you pause the
video, work on those, and then

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when you're ready to see my
solutions, you can bring the

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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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then part b.

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Once we know the bounds that
describe D, it's not too hard

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to set up this integral.

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So the hard part of this problem
is understanding how

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to write this solid region D
in spherical coordinates.

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But it's actually really not
that hard either, so I'm going

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to try and take us through
it in a reasonable way.

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So the first thing: I'm going
to draw a very rough picture

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of the region 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 recommend that

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next time you encounter such a
problem, you begin by drawing

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yourself a picture.

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Even if it's not a great
picture, it will give you some

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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--

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say here, is the point
(0, 0, 1)--

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I have a sphere that's looking
in the z-y 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

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z equals 1 plane.

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So actually it's going to be
all of the circular slices

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that are in the top
half in the upper

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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 rho

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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

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around 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, I want to

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only have a restricted
value of theta.

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But because I'm covering all
the way around and my

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restriction is only in the
bottom half, my theta values

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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 from the z-axis to any

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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 in this

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problem is determining
the rho value.

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Now, the rho value, to get the
outer boundary, we'll look at

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that part first.

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Well, the boundary of this
sphere here has a certain

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equation in x, y, and z
that we know, 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

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radius 1 centered
at (0, 0, 1).

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So I'm going to write that here,
and we're going to show

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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, you could make

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the substitution for x and y in
spherical coordinates, and

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it simplifies to this.

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So either way, if you didn't
know r squared was this, you

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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 is describing the

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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--

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I'm 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 is

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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 it is when it

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hits the plane z equals 1--

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and I'm going to stop
it when it hits the

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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 that if this is

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my plane z equals 1, and I look
at this triangle I make

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right here.

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This angle down here, the bottom
angle is phi, and this

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is a right angle, and the rho
value I'm interested in is

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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 that phi is

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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 I'm interested in--

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the blue part here-- and so rho
is equal to secant phi is

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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 is I'm not supposed

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to include--

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it won't matter for
the integral--

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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 and it's going

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to be less than or equal
to 2 cosine phi, right?

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Right?

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

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geometry mistake here,
just to be sure.

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This picture tells me
that cosine phi is

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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 and they go

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to the 2 cosine phi length.

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I know maybe I'm beating a dead
horse here, but I want to

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make sure we understand
where the rho

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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, and the

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rho bounds.

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Now you notice that theta and
phi don't depend on the other

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variables, but rho
depends on phi.

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So we're going to have to
integrate that first. So now

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we can deal with part b.

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Part b-- let's come back
over and remind

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ourselves what it said--

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said set up the integral for
the average distance of a

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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 is a great function to

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have in spherical coordinates,
because it's just rho.

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So the function I'm supposed
to average over is the

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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 is going to equal 1

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divided by the volume of D times
the triple integral over

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D of the function rho dV, OK?

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

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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. So

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

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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, after I have everything

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in order over here. dV is
rho squared sine phi d

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rho d theta d phi.

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So I'm going to end up with a
rho cubed sine phi d rho d

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theta d phi, right?

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The dV gave me an extra rho
squared and a sine phi.

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That whole part is dV. And then
I keep one rho from the

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fact that the distance
function is rho.

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And so I get a rho
cubed there.

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So hopefully that makes sense.

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Now for d rho, I know the bounds
are secant phi to 2

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cosine phi.

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For d theta, my bounds
are 0 to 2 pi.

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And for d phi, my bounds
were 0 to pi over 4.

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I didn't make you evaluate it,
I'm just making you set it up.

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That actually is the solution
we wanted for part b.

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I wanted to average the distance
from any point in D

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to the origin.

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So I just took the average value
of the function rho over

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that region D. And so that's
how you finish that up.

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And so in this problem,
basically we want you to get

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really familiar with how to do
some things in these spherical

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coordinates, which are sometimes
a little hard to do.

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But if you noticed, what we were
doing in trying to figure

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out the bounds-- in
particular, trying

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to figure out rho--

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we took what we knew in the x-,
y-, z-coordinates about

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certain relationships, and then
we replaced the x-, y-,

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z-values by the values in terms
of rho and theta and

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phi, and you can simplify to
figure out the relationships

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you have between rho and
theta and phi for

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the boundary value.

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So that was one of
the techniques

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we were using there.

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And hopefully, the geometric
understanding of why these

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00:12:28,430 --> 00:12:32,130
angles go from 0 to 2 pi and
0 to pi over 4 is clear.

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00:12:32,130 --> 00:12:35,580
And actually, the fact that this
is rho equals 2 cosine

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phi should remind you of the
two-dimensional case where you

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00:12:38,020 --> 00:12:41,510
had some problem like r
equals 2 cosine theta.

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And that drew a circle
off-center from the origin.

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00:12:46,140 --> 00:12:50,240
It's the analogous thing
happening here.

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Maybe I should stop there before
I say too many things.

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00:12:52,490 --> 00:12:54,600
But again, the object of this
was just to get really

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00:12:54,600 --> 00:12:56,640
comfortable with spherical
coordinates, and I hope it's

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00:12:56,640 --> 00:12:57,100
helped you.

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00:12:57,100 --> 00:12:58,700
I'll stop there.

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00:12:58,700 --> 00:12:59,379