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So, last week we learned how to
do triple integrals in
rectangular and cylindrical
coordinates.
And, now we have to learn about
spherical coordinates,
which you will see are a lot of
fun.
So, what's the idea of
spherical coordinates?
Well, you're going to represent
a point in space using the
distance to the origin and two
angles.
So, in a way, 
you can think of these as a
space analog of polar
coordinates because you just use
distance to the origin,
and then you have to use angles
to determine in which direction
you're going.
So, somehow they are more polar
than cylindrical coordinates.
So, how do we do that?
So, let's say that you have a
point in space at coordinates x,
y, z.
Then, instead of using x,
y, z, you will use,
well, one thing you'll use is
the distance from the origin.
OK, and that is denoted by the
Greek letter which looks like a
curly p, but actually it's the
Greek R.
So -- That's the distance from
the origin.
And so, that can take values
anywhere between zero and
infinity.
Then, we have to use two other
angles.
And, so for that,
let me actually draw the
vertical half plane that
contains our point starting from
the z axis.
OK, so then we have two new
angles.
Well, one of them is not really
new.
One is new.
That's phi is the angle
downwards from the z axis.
And the other one,
theta, is the angle
counterclockwise from the x
axis.
OK, so phi, let me do it better.
So, there's two ways to draw
the letter phi,
by the way.
And, I recommend this one
because it doesn't look like a
rho.
So, that's easier.
That's the angle that you have
to go down from the positive z
axis.
And, 
so that angle varies from zero
when you're on the z axis,
increase to pi over two when
you are on the xy plane all the
way to pi or 180� when you are
on the negative z axis.
It doesn't go beyond that.
OK, so -- Phi is always between
zero and pi.
And, finally,
the last one,
theta, is just going to be the
same as before.
So, it's the angle after you
project to the xy plane.
That's the angle
counterclockwise from the x
axis.
OK, so that's a little bit
overwhelming not just because of
the new letters,
but also because there is a lot
of angles in there.
So, let me just try to,
you know, suggest two things
that might help you a little
bit.
So, one is, these are called
spherical coordinates because if
you fix the value of rho,
then you are moving on a sphere
centered at the origin.
OK, so let's look at what
happens on a sphere centered at
the origin, so,
with equation rho equals a.
Well, then phi measures how far
south you are going,
measures the distance from the
North Pole.
So, 
if you've learned about
latitude and longitude in
geography,
well, phi and theta you can
think of as latitude and
longitude except with slightly
different conventions.
OK, so, phi is more or less the
same thing as latitude in the
sense that it measures how far
north or south you are.
The only difference is in
geography,
latitude is zero on the equator
and becomes something north,
something south, 
depending on how far you go
from the equator.
Here, you measure a latitude
starting from the North Pole
which is zero,
increasing all the way to the
South Pole, which is at pi.
And, theta or you can think of
as longitude,
which measures how far you are
east or west.
So, the Greenwich Meridian
would be here,
now, the one on the x axis.
That's the one you use as the
origin for longitude,
OK?
Now, if you don't like
geography, here's another way to
think about it.
So -- Let's start again from
cylindrical coordinates,
which hopefully you're kind of
comfortable with now.
OK, so you know about
cylindrical coordinates where we
have the z coordinates stay z,
and the xy plane we do R and
theta polar coordinates.
And now, let's think about what
happens when you look at just
one of these vertical planes
containing the z axis.
So, you have the z axis,
and then you have the direction
away from the z axis,
which I will call r,
just because that's what r
measures.
Of course, r goes all around
the z axis, but I'm just doing a
slice through one of these
vertical half planes,
fixing the value of theta.
Then, r of course is a polar
coordinate seen from the point
of view of the xy plane.
But here, it looks more like
you have rectangular coordinates
again.
So the idea of spherical
coordinate is you're going to
polar coordinates again in the
rz plane.
OK, so if I have a point here,
then rho will be the distance
from the origin.
And phi will be the angle,
except it's measured from the
positive z axis,
not from the horizontal axis.
But, the idea in here, 
see, 
let me put that between quotes
because I'm not sure how correct
that is,
but in a way, 
you can think of this as polar
coordinates in the rz plane.
So, in particular, 
that's the key to understanding
how to switch between spherical
coordinates and cylindrical
coordinates,
and then all the way to x,
y, z if you want,
right, 
because this picture here tells
us how to express z and r in
terms of rho and phi.
So, let's see how that works.
If I project here or here,
so, this line is z.
But, it's also rho times cosine
phi.
So, I get z equals rho cos phi.
And, if I look at r,
it's the same thing,
but on the other side.
So, r will be rho sine phi.
OK, so you can use this to
switch back and forth between
spherical and cylindrical.
And of course,
if you remember what x and y
were in terms of r and theta,
you can also keep doing this to
figure out, oops.
So, x is r cos theta.
That becomes rho sine phi cos
theta.
Y is r sine theta.
So, that becomes rho sine phi
sine theta.
And z is rho cos phi.
But, basically you don't really
need to remember these formulas
as long as you remember how to
express r in terms of rho sine
phi,
and x equals r cos theta.
So, now, of course, 
we're going to use spherical
coordinates in situations where
we have a lot of symmetry,
and in particular, 
where the z axis plays a
special role.
Actually, that's the same with
cylindrical coordinates.
Cylindrical and secure
coordinates are set up so that
the z axis plays a special role.
So, that means whenever you
have a geometric problem,
and you are not told how to
choose your coordinates,
it's probably wiser to try to
center things on the z axis.
That's where these coordinates
are the best adapted.
And, 
in case you ever need to switch
backwards,
I just want to point out, 
so, rho is the square root of r
squared plus z squared,
which means it's the square
root of x squared plus y squared
plus z squared.
OK, so that's basically all the
formulas about spherical
coordinates.
OK, any questions about that?
OK, let's see,
who had seen spherical
coordinates before just to see?
OK, that's not very many.
So, I'm sure for,
one of you saw it twice.
That's great.
Sorry, oops,
OK, so let's just look quickly
at equations of some of the
things.
So, as I've said,
if I set rho equals a,
that will be just a sphere of
radius a centered at the origin.
More interesting things:
let's say I give you phi equals
pi over four.
What do you think that looks
like?
Actually, let's take a quick
poll on things.
OK, yeah, everyone seems to be
saying it's a cone,
and that's indeed the correct
answer.
So, how do we see that?
Well, remember,
phi is the angle downward from
the z axis.
So, let's say that I'm going to
look first at what happens if
I'm in the right half of a plane
of a blackboard,
so, in the yz plane.
Then, phi is the angle downward
from here.
So, if I want to get pi over
four, that's 45�.
That means I'm going to go
diagonally like this.
Of course, if I'm in the left
half of a plane of a blackboard,
it's going to be the same.
I also take pi over four.
And, I get the other half.
And, because the equation does
not involve theta,
it's all the same if I rotate
my vertical plane around the z
axis.
So, I get the same picture in
any of these vertical half
planes, actually.
OK, now, so this is phi equals
pi over four.
And, just in case,
to point out to you what's
going on, when phi equals pi
over four, cosine and sine are
equal to each other.
They are both one over root two.
So, you can find,
again, the equation of this
thing in cylindrical
coordinates, which I'll remind
you was z equals r.
OK, in general,
phi equals some given number,
or z equals some number times
r.
That will be a cone centered on
the z axis.
OK, a special case:
what if I say phi equals pi
over two?
Yeah, it's just going to be the
xy plane.
OK, that's the flattest of all
cones.
OK, so phi equals pi over two
is going to be just the xy
plane.
And, in general,
if phi is less than pi over
two, then you are in the upper
half space.
If phi is more than pi over
two, you'll be in the lower half
space.
OK, so that's pretty much all
we need to know at this point.
So, what's next?
Well, remember we were trying
to do triple integrals.
So now we're going to triple
integrals in spherical
coordinates.
And, for that,
we first need to understand
what the volume element is.
What will be dV?
OK, so dV will be something,
d rho, d phi,
d theta, or in any order that
you want.
But, this one is usually the
most convenient.
So, to find out what it is,
well, we should look at how we
are going to be slicing things
now.
OK, so if you integrate d rho,
d phi, d theta,
it means that you are actually
slicing your solid into little
pieces that live,
somehow, 
if you set an interval of rows, 
OK, 
sorry, maybe I should, 
so, if you first integrate over
rho,
it means that you will actually
choose first the direction from
the origin even by phi and
theta.
And, in that direction,
you will try to figure out,
how far does your region
extend?
And, of course,
how far that goes might depend
on phi and theta.
Then, you will vary phi.
So, you have to know,
for a given value of theta,
how far down does your solid
extend?
And, finally,
the value of theta will
correspond to,
in which directions around the
z axis do we go?
So, we're going to see that in
examples.
But before we can do that,
we need to get the volume
element.
So, what I would like to
suggest is that we need to
figure out,
what is the volume of a small
piece of solid which corresponds
to a certain change,
delta rho, 
delta phi, 
and delta theta?
So, delta rho means that you
have two concentric spheres,
and you are looking at a very
thin shell in between them.
And then, you would be looking
at a piece of that spherical
shell corresponding to small
values of phi and theta.
So, because I am stretching the
limits of my ability to draw on
the board, here's a picture.
I'm going to try to reproduce
on the board,
but so let's start by looking
just at what happens on the
sphere of radius a,
and let's try to figure out the
surface area elements on the
sphere in terms of phi and
theta.
And then, we'll add the rho
direction.
OK, so -- So,
let me say, let's start by
understanding surface area on a
sphere of radius a.
So, that means we'll be looking
at a little piece of the sphere
corresponding to angles delta
phi and in that direction here
delta theta.
OK, so when you draw a map of
the world on a globe,
that's exactly what the grid
lines form for you.
So, what's the area of this guy?
Well, of course,
all the sides are curvy.
They are all on the sphere.
None of them are straight.
But still, if it's small enough
and it looks like a rectangle,
so let's just try to figure
out, what are the sides of your
rectangle?
OK, so, let's see,
well, I think I need to draw a
bigger picture of this guy.
OK, so this guy,
so that's a piece of what's
called a parallel in geography.
That's a circle that goes
east-west.
So now, 
this parallel as a circle of
radius,
well, the radius is less than a
because if your vertical is to
the North Pole,
it will be actually much
smaller.
So, that's why when you say
you're going around the world it
depends on whether you do it at
the equator or the North Pole.
It's much easier at the North
Pole.
So, anyway, this is a piece of
a circle of radius,
well, the radius is what I
would call r because that's the
distance from the z axis.
OK, that's actually pretty hard
to see now.
So if you can see it better on
this one, then so this guy here,
this length is r.
And, r is just rho,
well, what was a times sine
phi.
Remember, we have this angle
phi in here.
I should use some color.
It's getting very cluttered.
So, we have this phi,
and so r is going to be rho
sine phi.
That rho is a.
So, let me just put a sine phi.
OK, and the corresponding angle
is going to be measured by
theta.
So, the length of this is going
to be a sine phi delta theta.
That's for this side.
Now, what about that side,
the north-south side?
Well, if you're moving
north-south, it's not like
east-west.
You always have to go all the
way from the North Pole to the
South Pole.
So, that's actually a great
circle meridian of length,
well, I mean,
well, the radius is the radius
of the sphere.
Total length is 2pi a.
So, this is a piece of a circle
of radius a.
And so, now,
the length of this one is going
to be a delta phi.
OK, so, just to recap,
this is a sine phi delta theta.
And, this guy here is a delta
phi.
So, you can't read it because
it's -- And so,
that tells us if I take a small
piece of the sphere,
then its surface area, 
delta s, 
is going to be approximately a
sine phi delta theta times a
delta phi,
which I'm going to rewrite as a
squared sine phi delta phi delta
theta.
So, what that means is,
say that I want to integrate
something just on the surface of
a sphere.
Well, I would use phi and theta
as my coordinates.
And then, to know how big a
piece of a sphere is,
I would just take a squared
sine phi d phi d theta.
OK, so that's the surface
element in a sphere.
And now, what about going back
into the third dimension,
so, adding some depth to these
things?
Well, I'm not going to try to
draw a picture because you've
seen that's slightly tricky.
Well, let me try anyway just
you can have fun with my
completely unreadable diagrams.
So anyway, if you look at,
now, something that's a bit
like that piece of sphere,
but with some thickness to it.
The thickness will be delta
rho, and so the volume will be
roughly the area of the thing on
the sphere times the thickness.
So, I claim that we will get
basically the volume element
just by multiplying things by d
rho.
So, let's see that.
So now, if I have a sphere of
radius rho, and another one
that's slightly bigger of radius
rho plus delta rho,
and then I have a little box in
here.
Then, 
I know that the volume of this
thing will be essentially,
well, its thickness, 
the thickness is going to be
delta rho times the area of its
base,
although it doesn't really
matter,
which is what we've called
delta s.
OK, so we will get,
sorry, a becomes rho now.
Square sine phi, 
delta rho, 
delta phi, 
delta theta, 
and so out of that we get the
volume element and spherical
coordinates,
which is rho squared sine phi d
rho,
d phi, 
d theta.
And, that's a formula that you
should remember.
OK, so whenever we integrate a
function,
and we decide to switch to
spherical coordinates,
then dx dy dz or r dr d theta
dz will become rho squared sine
phi d rho d phi d theta.
OK, any questions on that?
No?
OK, so let's -- Let's see how
that works.
So, as an example,
remember at the end of the last
lecture, I tried to set up an
example where we were looking at
a sphere sliced by a slanted
plane.
And now, we're going to try to
find the volume of that
spherical cap again,
but using spherical coordinates
instead.
So, I'm going to just be
smarter than last time.
So, last time,
we had set up these things with
a slanted plane that was cutting
things diagonally.
And, 
if I just want to find the
volume of this cap,
then maybe it makes more sense
to rotate things so that my
plane is actually horizontal,
and things are going to be
centered on the z axis.
So, in case you see that it's
the same, then that's great.
If not, then it doesn't really
matter.
You can just think of this as a
new example.
So, I'm going to try to find
the volume of a portion of the
unit sphere -- -- that lies
above the horizontal plane,
z equals one over root two.
OK, one over root two was the
distance from the origin to our
slanted plane.
So, after you rotate,
that say you get this value.
Anyway, it's not very important.
You can just treat that as a
good example if you want.
OK, so we can compute this in
actually pretty much any
coordinate system.
And also, of course,
we can set up not only the
volume, but we can try to find
the moment of inertia about the
central axis,
or all sorts of things.
But, we are just doing the
volume for simplicity.
So, actually,
this would go pretty well in
cylindrical coordinates.
But let's do it in spherical
coordinates because that's the
topic of today.
A good exercise:
do it in cylindrical and see if
you get the same thing.
So, how do we do that?
Well, we have to figure out how
to set up our triple integral in
spherical coordinates.
So, remember we'll be
integrating one dV.
So, dV will become rho squared
sign phi d rho d phi d theta.
And, now as we start,
we're already facing some
serious problem.
We want to set up the bounds
for rho for a given,
phi and theta.
So, that means we choose
latitude/longitude.
We choose which direction we
want to aim for,
you know, which point of the
sphere we want to aim at.
And, we are going to shoot a
ray from the origin towards this
point, and we want to know what
portion of the ray is in our
solid.
So -- We are going to choose a
value of phi and theta.
And, we are going to try to
figure out what part of our ray
is inside this side.
So, what should be clear is at
which point we leave the solid,
right?
What's the value of rho here?
It's just one.
The sphere is rho equals one.
That's pretty good.
The question is,
where do we enter the region?
So, we enter the region when we
go through this plane.
And, the plane is z equals one
over root two.
So, what does that tell us
about rho?
Well, it tells us,
so remember,
z is rho cosine phi.
So, the plane is z equals one
over root two.
That means rho cosine phi is
one over root two.
That means rho equals one over
root two cosine phi or,
as some of you know it,
one over root two times second
phi.
OK, so if we want to set up the
bounds, then we'll start with
one over root two second phi all
the way to one.
Now, what's next?
Well, so we've done,
I think that's basically the
hardest part of the job.
Next, we have to figure out,
what's the range for phi?
So, the range for phi,
well, we have to figure out how
far to the north and to the
south our region goes.
Well, the lower bound for phi
is pretty easy,
right, because we go all the
way to the North Pole direction.
So, phi starts at zero.
The question is,
where does it stop?
To find out where it stops,
we have to figure out,
what is the value of phi when
we hit the edge of the region?
OK, so maybe you see it.
Maybe you don't.
One way to do it geometrically
is to just, it's always great to
draw a slice of your region.
So, if you slice these things
by a vertical plane,
or actually even better,
a vertical half plane,
something to delete one half of
the picture.
So, I'm going to draw these r
and z directions as before.
So, my sphere is here.
My plane is here at one over
root two.
And, my solid is here.
So now, the question is what is
the value of phi when I'm going
to stop hitting the region?
And, if you try to figure out
first what is this direction
here, that's also one over root
two.
And so, this is actually 45�,
also known as pi over four.
The other way to think about it
is at this point,
well, rho is equal to one
because you are on the sphere.
But, you are also on the plane.
So, rho cos phi is one over
root two.
So, if you plug rho equals one
into here, you get cos phi
equals one over root two which
gives you phi equals pi over
four.
That's the other way to do it.
You can do it either by
calculation or by looking at the
picture.
OK, so either way,
we've decided that phi goes
from zero to pi over four.
So, this is pi over four.
Finally, what about theta?
Well, because we go all around
the z axis we are going to go
just zero to 2pi.
OK, any questions about these
bounds?
OK, so note how the equation of
this horizontal plane in
spherical coordinates has become
a little bit weird.
But, 
if you remember how we do
things,
say that you have a line in
polar coordinates,
and that line does not pass
through the origin,
then you also end up with
something like that.
You get something like r equals
a second theta or a cos second
theta for horizontal or vertical
lines.
And so, it's not surprising you
should get this.
That's in line with the idea
that we are just doing again,
polar coordinates in the rz
directions.
So of course,
in general, things can be very
messy.
But, generally speaking, 
the kinds of regions that we
will be setting up things for
are no more complicated or no
less complicated than what we
would do in the plane in polar
coordinates.
OK, so there's,
you know, a small list of
things that you should know how
to set up.
But, you won't have some
really, really strange thing.
Yes?
D rho?
Oh, you mean the bounds for rho?
Yes.
So, in the inner integral,
we are going to fix values of
phi and theta.
So, that means we fix in
advance the direction in which
we are going to shoot a ray from
the origin.
So now, as we shoot this ray,
we are going to hit our region
somewhere.
And, we are going to exit,
again, somewhere else.
OK, so first of all we have to
figure out where we enter,
where we leave.
Well, we enter when the ray
hits the flat face,
when we hit the plane.
And, we would leave when we hit
the sphere.
So, the lower bound will be
given by the plane.
The upper bound will be given
by the sphere.
So now, you have to get
spherical coordinate equations
for both the plane and the
sphere.
For the sphere, that's easy.
That's rho equals one.
For the plane,
you start with z equals one
over root two.
And, you switch it into
spherical coordinates.
And then, you solve for rho.
And, that's how you get these
bounds.
Is that OK?
All right, so that's the setup
part.
And, of course,
the evaluation part goes as
usual.
 
And, since I'm running short of
time, I'm not going to actually
do the evaluation.
I'm going to let you figure out
how it goes.
Let me just say in case you
want to check your answers,
so, at the end you get 2pi over
three minus 5pi over six root
two.
Yes, it looks quite complicated.
That's basically because you
get one over,
well, you get a second square
when you integrate C.
When you integrate rho squared,
you will get rho cubed over
three.
But that rho cubed will give
you a second cube for the lower
bound.
And, when you integrate sine
phi second cubed phi,
you do a substitution.
You see that integrates to one
over second squared with a
factor in front.
So, in the second square,
when you plug in,
no, that's not quite all of it.
Yeah, well, the second square
is one thing,
and also the other bound you
get sine phi which integrates to
cosine phi.
So, anyways,
you get lots of things.
OK, enough about it.
So, next, I have to tell you
about applications.
And, of course, 
well, there's the same
applications that we've seen
that last time,
finding volumes, 
finding masses, 
finding average values of
functions.
In particular,
now, we could say to find the
average distance of a point in
this solid to the origin.
Well, 
spherical coordinates become
appealing because the function
you are averaging is just rho
while in other coordinate
systems it's a more complicated
function.
So, if you are asked to find
the average distance from the
origin, spherical coordinates
can be interesting.
Also, 
well, there's moments of
inertia,
preferably the one about the z
axis because if you have to
integrate something that
involves x or y,
then your integrand will
contain that awful rho sine phi
sine theta or rho sine phi
cosine theta,
and then it won't be much fun
to evaluate.
So, that anyway,
there's the usual ones.
And then there's a new one.
So, in physics,
you've probably seen things
about gravitational attraction.
If not, well,
it's what causes apples to fall
and other things like that as
well.
So, anyway, physics tells you
that if you have two masses,
then they attract each other
with a force that's directed
towards each other.
And in intensity,
it's proportional to the two
masses, and inversely
proportional to the square of
the distance between them.
So, 
if you have a given solid with
a certain mass distribution,
and you want to know how it
attracts something else that you
will put nearby,
then you actually have to, 
the first approximation will be
to say,
well, let's just put a point
mass at its center of mass.
But, if you're solid is
actually not homogenous,
or has a weird shape,
then that's not actually the
exact answer.
So, in general,
you would have to just take
every single piece of your
object and figure out how it
attracts you,
and then compute the sum of
these.
So, for example, 
if you want to understand why
anything that you drop in this
room will fall down,
you have to understand that
Boston is actually attracting it
towards Boston.
And, Somerville's attracting it
towards Somerville,
and lots of things like that.
And, China, which is much
further on the other side is
going to attract towards China.
But, there's a lot of stuff on
the other side of the Earth.
And so, overall,
it's supposed to end up just
going down.
OK, so now, how to find this
out, well, you have to just
integrate over the entire Earth.
OK, so let's try to see how
that goes.
So, the setup that's going to
be easiest for us to do
computations is going to be that
we are going to be the test mass
that's going to be falling.
And, we are going to put
ourselves at the origin.
And, the solid that's going to
attract us is going to be
wherever we want in space.
You'll see, putting yourself at
the origin is going to be
better.
Well, you have to put something
at the origin.
And, the one that will stay a
point mass, I mean,
in my case not really a point,
but anyway, let's say that I'm
a point.
And then, I have a solid
attracting me.
Well, 
so then if I take a small piece
of it with the mass delta M,
then that portion of the solid
exerts a force on me,
which is going to be directed
towards it,
and we'll have intensity.
So, the gravitational force --
-- exerted by the mass delta M
at the point of x,
y, z in space on a mass at the
origin.
Well, we know how to express
that.
Physics tells us that the
magnitude of this force is going
to be, well, G is just a
constant.
It's the gravitational
constant, and its value depends
on which unit system you use.
Usually it's pretty small,
times the mass delta M,
times the test mass little m,
divided by the square of the
distance.
And, the distance from U to
that thing is conveniently
called rho since we've been
introducing spherical
coordinates.
So, that's the size,
that's the magnitude of the
force.
We also need to know the
direction of the force.
And, the direction is going to
be towards that point.
So, the direction of the force
is going to be that of x,
y, z.
But if I want a unit vector,
then I should scale this down
to length one.
So, let me divide this by rho
to get a unit vector.
So, that means that the force
I'm getting from this guy is
actually going to be G delta M m
over rho cubed times x,
y, z.
I'm just multiplying the
magnitude by the unit vector in
the correct direction.
OK, so now if I have not just
that little p is delta M,
but an entire solid,
then I have to sum all these
guys together.
And, I will get the vector that
gives me the total force
exerted, OK?
So, of course,
there's actually three
different calculations in one
because you have to sum the x
components to get the x
components of a total force.
Same with the y,
and same with the z.
So, let me first write down the
actual formula.
So, if you integrate over the
entire solid,
oh, and I have to remind you,
well, what's the mass,
delta M of a small piece of
volume delta V?
Well, it's the density times
the volume.
So, the mass is going to be,
sorry, density is delta.
There is a lot of Greek letters
there, times the volume element.
So, you will get that the force
is the triple integral over your
solid of G m x,
y, z over rho cubed,
delta dV.
Now, two observations about
that.
So, the first one,
well, of course,
these are just constants.
So, they can go out.
The second observation,
so here, we are integrating a
vector quantity.
So, what does that mean?
I just mean the x component of
a force is given by integrating
G m x over rho cubed delta dV.
The y components,
same thing with y.
The z components,
same thing with z.
OK, there's no,
like, you know,
just integrate component by
component to get each component
of the force.
So, now we could very well to
this in rectangular coordinates
if we want.
But the annoying thing is this
rho cubed.
Rho cubed is going to be x
squared plus y squared plus z
squared to the three halves.
That's not going to be a very
pleasant thing to integrate.
So, it's much better to set up
these integrals in spherical
coordinates.
And, if we're going to do it in
spherical coordinates,
then probably we don't want to
bother too much with x and y
components because those would
be unpleasant.
It would give us rho sine phi
cos theta or sine theta.
So, the actual way we will set
up things, set things up,
is to place the solid so that
the z axis is an axis of
symmetry.
And, of course,
that only works if the solid
has some axis of symmetry.
Like, if you're trying to find
the gravitational attraction of
the Pyramid of Giza,
then you won't be able to set
up so that it has rotation
symmetry.
Well, that's a tough fact of
life, and you have to actually
do it in x, y,
z coordinates.
But, if at all possible,
then you're going to place
things.
Well, I guess even then,
you could center it on the z
axis.
But anyway, so you're going to
mostly place things so that your
solid is actually centered on
the z-axis.
And, what you gain by that is
that by symmetry,
the gravitational force will be
directed along the z axis.
So, you will just have to
figure out the z component.
So, then the force will be
actually, you know in advance
that it will be given by zero,
zero, and some z component.
And then, you just need to
compute that component.
And, that component will be
just G times m times triple
integral of z over rho cubed
delta dV.
OK, so that's the first
simplification we can try to do.
The second thing is,
well, we have to choose our
favorite coordinate system to do
this.
But, I claim that actually
spherical coordinates are the
best -- -- because let's see
what happens.
So, G times mass times triple
integral, well,
a z in spherical coordinates
becomes rho cosine phi over rho
cubed.
Density, well,
we can't do anything about
density.
And then, dV becomes rho
squared sine phi d rho d phi d
theta.
Well, so, what happens with
that?
Well, you see that you have a
rho, a rho squared,
and a rho cubed that cancel
each other.
So, in fact,
it simplifies quite a bit if
you do it in spherical
coordinates.
 
 
OK, so the z component of the
force, sorry,
I'm putting a z here to remind
you it's the z component.
That is not a partial
derivative, OK?
Don't get things mixed up,
just the z component of the
force becomes Gm triple integral
of delta cos phi sine phi d rho
d phi d theta.
And, so this thing is not dV,
of course.
dV is much bigger,
but we've somehow canceled out
most of dV with stuff that was
in the integrand.
And see, that's actually
suddenly much less scary.
OK, so just to give you an
example of what you can prove it
this way, you can prove Newton's
theorem, which says the
following thing.
It says the gravitational
attraction -- -- of a spherical
planet,
I should say with uniform
density,
or actually it's enough for the
density to depend just on
distance to the center.
But we just simplify the
statement is equal to that of a
point mass -- -- with the same
total mass at its center.
OK, so what that means is that,
so the way we would set it up
is u would be sitting here and
your planet would be over here.
Or, if you're at the surface of
it, then of course you just put
it tangent to the xy plane here.
And, you would compute that
quantity.
Computation is a little bit
annoying if a sphere is sitting
up there because,
of course, you have to find
bounds,
and that's not going to be very
pleasant.
The case that we actually know
how to do fairly well is if you
are just at the surface of the
planet.
But then, 
what the theorem says is that
the force that you're going to
feel is exactly the same as if
you removed all of the planet
and you just put an equivalent
point mass here.
So, if the earth collapsed to a
black hole at the center of the
earth with the same mass,
well, you wouldn't notice the
difference immediately,
or, rather, you would, 
but at least not in terms of
your weight.
OK, that's the end for today.
 
 
