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Welcome back to recitation.

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In this video I'd
like us to work

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on the following application
for spherical coordinates.

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So what I'd like
us to do is find

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the gravitational attraction
of an upper solid half

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sphere of radius a
and center O, that's

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the gravitational attraction
on a mass m naught at O.

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So the m naught is sitting at
O, we have a solid half sphere

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of radius a and
center O, and we want

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to find the gravitational
attraction of the half sphere

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on the mass.

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And we're going to
assume that the density

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delta is equal to
the square root of x

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squared plus y squared.

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So why don't you pause the
video, take a little while

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to work on it, and when you
feel good about your solution,

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bring the video back up,
I'll show you what I did.

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OK.

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Welcome back.

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Again what we're going
to do is we're going

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to do an application problem.

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And what we have is we
have the solid half sphere

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of radius a and
center O, and we want

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to know the gravitational
attraction of it

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on a mass m naught.

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So the first thing
I'm going to do

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is draw a picture, just so
I can get myself oriented.

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The easiest thing
to do-- obviously,

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when I'm going to try
and figure this out,

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I'm going to use
spherical coordinates.

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I have an upper
solid half sphere,

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so that should tell me spherical
coordinates will be good.

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And the easiest way to deal
with spherical coordinates

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is obviously to choose
O to be the origin.

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So I'm going to have m naught
sitting right at the origin.

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And then I'm going
to have-- oops,

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I should write that
somewhere else.

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Maybe, well, I'll
just leave it there--

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and then I'm going to
have the solid half

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sphere, which I'm going to
just draw sort of like this.

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Maybe that doesn't look
like a sphere to you.

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But we have the
solid half sphere

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and it's exerting a
force on m naught.

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And if I want to
know the force-- how

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we denote the force in
lecture was the components

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were the x-component
of the force,

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the y-component of the
force, and the z-component

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of the force.

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And so I want to point out these
are not partial derivatives.

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This is the standard notation
you use for force in this case,

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and so these are not
the partial derivatives.

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They're just the
x-component, the y-component,

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and the z-component.

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And if you notice,
this upper solid half

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sphere-- because of
where m naught is,

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the x-component and the
y-component of the force

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are 0 based on symmetry.

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So all I really need to worry
about is the z-component.

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I'm only really interested
in one component of the force

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and it's the z-component.

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Because the other two
are going to be 0, just

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by the symmetry of
this solid half sphere.

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So I really only
need the z-component.

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So what I have to do, in
order to calculate the force,

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is I have to figure out the
magnitude and the direction.

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So let's write down
the two components.

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These were given
to you in class.

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So the magnitude is going
to be equal to G m naught

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dm divided by rho squared.

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So this was given to
you in class already.

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So G is some constant; m
naught is the mass you have;

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dm, we have to determine
a little bit about it;

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and then we're dividing
by rho squared.

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And the direction--
because I'm only

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interested in the z-component--
the direction is just z,

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but then I want to make sure
that the direction component is

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z divided by rho.

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The full direction is
[x, y, z] divided by rho,

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but I'm only interested
in that last component.

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So the z-component
is just z over rho.

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If that confused
you, I just want

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to remind you that
you had [x,  y, z]

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over rho is the full
direction of the force,

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we're just pulling
off the z-component.

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So what we're going to
be trying to figure out--

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we need to integrate the
product of these two things,

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but I need to figure
out what dm actually is.

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So let me remind you also
what dm is and then we'll

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put it all together.

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dm, to remind you, is supposed
to be the density times dV.

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And the density in this
case was square root

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of x squared plus y squared,
that's just equal to r.

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So it's just r*dV.

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So there's a bunch
of little pieces

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and now I have to
put them together.

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So if I want to determine
the full attraction

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in the z-direction, because
the other two are 0,

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I just need to do
this triple integral.

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I'll figure out my
bounds momentarily.

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I'm going to leave a lot of room
to write my bounds, because I

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always run out of room.

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And then I just
need to integrate

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G m naught divided by rho
squared times z divided

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by rho times r dV.

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Actually I won't even
do the bounds yet

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because I have to
change everything

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into polar coordinates.

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So let's figure
out how to do that.

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What is z in polar
coordinates? z

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in polar coordinates we
know is rho cosine phi.

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So I'm going to
have a G m naught--

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I'm going to keep this
rho cubed right here--

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and then I'm going to
have a rho cosine phi.

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r is rho sine phi, so I'll
just put another rho, and then

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a sine phi.

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And then dV is rho squared
sine phi d rho d phi d theta.

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So what I'm going
to do here is--

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I have to add in a rho squared,
so I'm just going to cube this.

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And another sine phi, so
I'm going to square that.

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And then I'm going to have
a d rho d phi d theta.

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Hopefully that wasn't too scary.

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But the z is my rho cosine phi.

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The r is a single rho sine phi.

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And then dV includes a
rho squared sine phi,

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so I put a cubed here
and a squared here,

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and then d rho d phi d theta.

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So now I just have
to integrate, so I

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have to figure out my bounds.

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So we're integrating
first in rho.

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And rho, it's a sphere- a
half sphere of radius a.

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It's a solid half sphere.

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So I'm just integrating
from 0 up to a.

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And then phi-- because
it's a solid half sphere,

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phi is starting at 0
and it's going down

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until it hits the xy-plane.

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And so that's pi over 2.

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We know we start at 0, we go to
pi over 2 when we go a quarter

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turn, basically.

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When we go 90 degrees, right?

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So we go 0 to pi over
2, and then theta--

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once I've taken my rho and
I've gone up as far as I can,

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and I've gone down this way, I
have to rotate it all the way

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around-- and so theta is
actually going from 0 to 2*pi.

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All right.

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So this is my full
integral I have here.

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And now let's do
some simplifying

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before we integrate.

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Notice I have a rho cubed
in the denominator here

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and a rho cubed in
the numerator there.

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So when I integrate in rho,
I only have this single rho.

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When I integrate that rho,
everything else is fixed.

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So I integrate that rho, I
get a rho squared over 2,

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and I evaluate it
at 0 and a and so I

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get an a squared over 2,
as integrating this part.

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I'm going to pull
all that to the front

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and then write what I get next.

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Let me write what I get next.

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I'm going to have a G
m naught as a constant.

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I'm going to be multiplying
by a squared over 2

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from the rho integral.

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And then I have
2 more integrals.

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Integral from 0 to 2*pi.

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And the integral from 0 to
pi over 2 of cosine phi sine

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squared phi d phi d theta.

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So how do I integrate
this in phi?

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Looks like it's going to
be sine cubed phi over 3.

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Let me just double check.

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The derivative of that is 3
sine squared phi cosine phi.

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Yes.

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So, this is going to be
sine cubed phi over 3

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evaluated at 0 and pi over 2.

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This inner part is.

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And so let's see
what we get there.

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Sine of 0 is 0.

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Sine of pi over 2 is 1.

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So I just pick up a
1/3 from this integral.

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So now I'm going to
pull that out in front.

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So again, the thing
I've written down here

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is just the integral of this.

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So I just get a 1/3 out of that.

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Let me double check.

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Yep, sine of pi
over 2 is still 1.

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Good.

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So I get a 1/3 out of that.

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That's a constant.

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So I'll pull that to the front.

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And then notice
what I'm left with.

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I'm just integrating
0 to 2*pi of d theta.

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Well that's just 2*pi.

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So I've got a 1/3
from the inside one,

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times a 2*pi from
the outside one.

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So I just end up getting
2*pi times-- or 2*pi over 3,

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because there's my 1/3 also--
times G m naught a squared over

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2.

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And if I simplify
that, I just get

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G m naught a squared pi over 3.

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And so if I want to know
the force of this solid half

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sphere acting on the mass
m naught, where the mass is

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positioned at the center of the
base of the solid half sphere,

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it's going to be exactly
the following thing.

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By symmetry, the first
two components are 0

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and the last component is this
G m naught a squared pi over 3.

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And so that's my final solution.

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Let me just take us back and
remind us where we came from.

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So we wanted to find the force,
the gravitational attraction

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of this solid half sphere that
had a radius a, was centered

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at O, and had mass m
naught sitting at O.

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And so we were interested
in drawing the easiest

213
00:09:31,070 --> 00:09:31,800
picture possible.

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We're going to put
O at the origin,

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so m naught's at the origin.

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Then we have a solid
half sphere of radius a.

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So this is-- you know, if
I come all the way out,

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my radius length is a.

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And I gave you the
density was equal to r.

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So we know-- the force
we're interested in,

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F sub x, F sub y, F sub
z, from the picture alone,

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we see F sub x
and F sub y are 0.

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So we really only
need the z-component.

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So you know how to
find the magnitude

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and you know how to find the
direction of the vector field.

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The reason that you've
divided by rho here, just

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to remind you, is
that the direction,

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00:10:02,166 --> 00:10:04,790
you want it to be a unit vector.

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00:10:04,790 --> 00:10:06,680
And so this is what
makes it a unit vector.

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And so the direction part
that I'm interested in

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is the z-part.

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So it's z over rho.

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And so all I need to do
is take the z over rho,

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multiply it by the magnitude,
and replace the dm by what it

235
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is in terms of the volume form.

236
00:10:21,910 --> 00:10:24,910
And in terms of the volume
form, it's the density times dV

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and so in this case
it's just r*dV.

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So I made all those
substitutions.

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00:10:29,490 --> 00:10:31,701
And then all I had to do
is do a lot of replacement.

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And that was kind
of the messy part,

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and then the
integral's fairly easy.

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So the messy part is z I
replaced by rho cosine phi.

243
00:10:38,690 --> 00:10:40,810
r I replaced by rho sine phi.

244
00:10:40,810 --> 00:10:45,264
And dV I replaced by rho squared
sine phi d rho d phi d theta.

245
00:10:45,264 --> 00:10:46,680
And then I just
have to figure out

246
00:10:46,680 --> 00:10:49,265
what the bounds are for the
half sphere, solid half sphere.

247
00:10:49,265 --> 00:10:50,140
And then I integrate.

248
00:10:50,140 --> 00:10:51,690
And the integration
is all ultimately

249
00:10:51,690 --> 00:10:54,110
one step at a time,
single variable.

250
00:10:54,110 --> 00:10:56,870
And they were all
fairly simple integrals.

251
00:10:56,870 --> 00:10:59,790
So in the end, we see that
your mass m naught sitting

252
00:10:59,790 --> 00:11:04,490
at the origin, the force exerted
on it by this solid half sphere

253
00:11:04,490 --> 00:11:08,960
is 0, 0 comma G times m naught
times a squared pi divided

254
00:11:08,960 --> 00:11:09,870
by 3.

255
00:11:09,870 --> 00:11:12,150
All right, that's
where I'll stop.

256
00:11:12,150 --> 00:11:12,939