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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've been learning
about surface integrals, flux,

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and the divergence theorem.

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And I have a nice
problem here that'll

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put your knowledge of
those things to the test.

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So in this situation,
I have a hemisphere.

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So this is a hemisphere.

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It's the right half of
a sphere, so the part

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where y is positive, or
non-negative, I guess.

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So just the surface of that
hemisphere, and it has radius R

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and it's centered at the origin.

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So I have this hemisphere
and I have the field F, which

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is given by y times j hat.

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So it's just always
in the y-direction.

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So I have this field and
I have this hemisphere,

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and what I'd like
to do is I'd like

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you to compute the flux of the
field through this hemisphere.

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But as a hint,
rather than doing it

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by computing the surface
integral by parameterizing,

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what I'd like you to do is
to use the divergence theorem

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to make your life a
little bit easier.

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So why don't you pause
the video, take some time

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to work that out, come back
and we can work on it together.

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Hopefully, you had some luck
working on this problem.

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Let's see how we
can go about it.

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So the thing that we're
asked to compute--

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let's give this a name,
just so we can refer to it.

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So let's call this surface
of the sphere S. All right.

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So we want the flux through
S. And so the flux through S

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is the double integral
over S of our field F

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dotted with the normal, with
respect to surface area,

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so with respect to surface area.

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So I guess this was a kind
of bad choice, this S. We've

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got S that means-- this
S means the sphere,

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this S just means surface area.

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OK.

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Hopefully, you can
keep those two sorted.

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This is the thing that we want.

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And we'd like to compute it--
rather than by parameterizing

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this hemisphere, we'd like
to compute it by using

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the divergence theorem.

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So the divergence
theorem relates

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the integral over a surface to
a triple integral over a region.

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So what we'd like to find is
a solid region for which this

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is all or part of the boundary.

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Now this half
hemisphere-- rather,

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half-sphere, this
hemisphere-- doesn't actually

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enclose a region, but
there's an obvious region

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that's closely
related to it, which

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would be the solid sphere.

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If you just filled in
that half-sphere there.

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So then, for that solid
sphere-- so the right half

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of the solid sphere, with radius
R, centered at the origin,

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that hemisphere, that
solid hemisphere,

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has a surface that
consists of S,

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and it also consists of a
disc centered at the origin.

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So it consists of--
I'm going to draw it in

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right here-- it's this disc.

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This one.

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That disc right there,
together with our hemisphere

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that interests us,
make up the boundary

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of this solid hemisphere.

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So I'm going to
call that disc-- I'm

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going to give it a name--
I'm going to call it S_2.

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So we want this
integral and we're going

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to use the divergence theorem.

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So the way we're going to
use the divergence theorem

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is we're going to use the
divergence theorem to relate

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this double integral
over a surface

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to that triple integral.

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But I'm going to have to get
S_2 involved because I need--

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for the divergence
theorem to apply,

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I need surfaces which
completely bound a solid region.

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So what we know by
the divergence theorem

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is that the double integral
over both S and S_2 of F

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dot n hat with respect
to surface area

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is equal to the
triple integral--

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OK, so I guess I need
another letter here.

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So I'm going to call this D
for that whole half-hemisphere,

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the solid hemisphere.

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So D, this is the
solid hemisphere.

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So this D is the
solid hemisphere

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that they bound--
of-- OK, so it's

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the triple integral over
D of the divergence of F

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with respect to volume.

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So this is what the
divergence theorem tells us

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and now we-- so OK.

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So good.

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So now what's going to be
nice is that in this case

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this triple integral is going
to be very easy to compute.

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And the surface
integral over S_2

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is going to be easy to compute.

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And so what we're going to be
left with-- after we compute

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those two things, we'll be
able to subtract and get

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just the surface
integral over S.

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So lets think about
doing those things.

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So first let's do the
surface integral over S_2.

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The surface integral
over S_2 of F

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dot n with respect
to surface area.

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Well, what is F
on that disc S_2?

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So let's go look at
our picture again.

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So this disc is the disc
of radius R centered

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at the origin in the xz-plane.

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So it's in the xz-plane.

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So that means it's in
the plane y equals 0.

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So F on that disc is just 0.

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It's the zero field.

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It's 0 everywhere on that disc.

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And if you take a surface
integral of a zero field

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dot the normal vector, well,
0 dot the normal is just 0.

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So if we come back over
here, this integrand

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is 0 dot the normal.

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So it's always 0.

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So we're integrating 0.

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So we just get 0.

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Definite integral of 0 is 0.

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Double integral of 0 is 0.

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Great.

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So this S_2, the
surface integral of F

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is really easy to compute.

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That's the first one we needed.

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Now we need the triple integral.

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So let's compute
the triple integral

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over the solid
hemisphere of div F dV.

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Well, what is div F?

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Well, you know, so
div F is just the sum

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of the partial derivatives
of the three components.

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So in our case if
we go back and look,

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well, F is 0 i hat plus
y j hat plus 0 k hat.

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So when you take the
partial derivatives

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you get 0 plus-- the
partial derivative of y

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with respect to y is just 1,
plus-- from the third component

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we also get 0.

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So the divergence--
let's go back over here--

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the divergence is just 1.

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It's just a constant 1.

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That's great.

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So this triple integral
is equal to-- I'm

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going to bring it
down here-- so it's

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equal to the triple
integral over D of 1 dV.

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But, of course,
when you integrate 1

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over a solid region
what you get is just

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the volume of that region.

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So whatever this number is, it's
the volume of that hemisphere.

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Well, OK, so what's
the volume of a sphere?

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It's 4/3 pi R cubed.

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So the volume of a
hemisphere is half of that.

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So that's 2 pi R
cubed divided by 3.

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So that's the triple
integral of the divergence

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over the solid region.

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So what does that mean?

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So our integral, the integral
over S of F dot n with respect

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to surface area is equal
to this triple integral

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over D of div F dV minus
the double integral over S2

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of F dot n d surface area.

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And so we just saw that
this is equal to 2 pi

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R cubed over 3 minus 0.

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But 0 is 0.

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So it's just 2 pi
R cubed over 3.

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So let's quickly recap.

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To start off with, we
just had this hemisphere.

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And we're asked to
compute a surface integral

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over this hemisphere of
a flux of this field F.

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So rather than going ahead
and calculating it directly,

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what we realized is that
by the divergence theorem,

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we could consider this to be a
difference of a triple integral

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minus another
surface integral such

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that those two surfaces together
bounded the region that you

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were triple integrating over.

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So in principle, there
were many possible regions

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that we could have chosen.

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Many possible solid
regions that we could

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have chosen to do that
triple integral over

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and to use the other
half of its boundary.

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But there's one
particularly nice one,

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which is just the
solid hemisphere

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of which this hemisphere
S was the boundary,

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or part of the boundary.

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And so then that gave us this
other part of its boundary

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was this disc.

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So we introduced this
new solid region, D,

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and the rest of
its boundary, which

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was this other
surface, this disc S_2.

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And so then, by the
divergence theorem,

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rather than computing
the integral

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over our original
region S we could just

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compute the triple integral
over the solid region

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and the surface integral over
the other half of the boundary.

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So sometimes that
won't be helpful.

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Sometimes you'll
have an ugly field,

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you'll have an ugly region,
things won't work out.

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But in this case, it
worked out really nicely.

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The triple integral
was easy to compute,

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as we saw over here, because
the divergence was constant.

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Because the divergence
was just equal to 1,

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the triple integral
just gave the volume.

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And the other surface
integral was also

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easy to compute because F dot n
on that surface was equal to 0.

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So integrating it was very easy.

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We just were integrating
0 and we got 0.

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So in this case, these choices
worked out very, very nicely.

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They made our life
nice and simple.

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And so in the end,
all we had to do

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were these two almost
trivial integrals.

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They didn't really require any
computation at all on our part.

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And then a little
bit of subtraction,

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except we were subtracting 0 so
even the subtraction was easy.

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So that was what we did,
worked out very nicely,

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2 pi R cubed over
3 was our answer.

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And I'll end there.