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Welcome back to recitation.

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In this video I would like us
to do the following problem.

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We're going to let z equal x
squared plus y and we want S

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to be the graph of z above the
unit square in the x-y plane.

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And what we'd like to do is
for F equal to zi plus xk,

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find the upward flux of F
through S. So S is our surface

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that's a graph over the x-y
plane and the unit square of z

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equal to x squared plus y.

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And we want to compute
the upward flux of F

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through that surface.

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So why don't you work on this
problem, pause the video, and

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then when you're ready to
see my solution, bring

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the video back up.

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OK, welcome back.

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So again what we want to do is
we want to find the upward

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flux of F through this surface
S. And let's think about

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first, how do we describe
the surface S?

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S is the graph of z equal
x squared plus y.

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So we can think of it as, z is
really a function of x and y

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over the unit square.

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So we can say F of x, y is equal
to z which is equal to x

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squared plus y.

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And then we know how to
compute the normal--

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well, we know how to compute n
dS, which also in class is

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sometimes written notationally
as dS with the vector dS.

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So we know how to compute
this form right here.

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And it is-- you were shown
in class that if F is--

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or sorry.

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If you have a graph, if your
surface is a graph, then this

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is exactly equal to the vector
minus f sub x comma minus f

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sub y comma 1 dx dy.

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So that's exactly
what this n dS--

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so n is the vector, and dS is
the surface form we have here.

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So n dS is exactly equal to
the vector minus f sub x,

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minus f sub y, 1, dx dy.

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So what do we have here with
f sub x? f sub x--

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because f is equal to z-- f sub
x is 2x and f sub y is 1.

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So in our case we get exactly
minus 2x, comma minus

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1, comma 1, dx dy.

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And now to compute the surface
integral what we do--

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or sorry, to compute the flux
along the surface--

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what we do is we integrate
over the surface--

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which I guess we should remember
that's a double

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integral, because it's
over a surface--

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of F dotted with dS.

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But that's the same as
integrating over the region.

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So we have this surface, we
know the region below that

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defines the surface in
the x-y coordinates.

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So it's integrating over the
region of F dotted with this

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vector here.

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Because n dS, dS is n dS and
in the x-y components it's

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exactly equal to this.

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Minus 2x, minus 1, 1, dx dy.

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So now we're integrating.

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We've gone from looking
at a surface integral.

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Now we're integrating--

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we were integrating F dot dS on
the surface, to now taking

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F dotted with this vector on
the region in the x-y plane

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over which we can define S.

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So the region we're
interested in,

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remember, is the unit square.

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So we have the unit square which
is x goes from 0 to 1

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and y goes from 0 to 1.

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And then what we're doing is
we're looking at F as a

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function of x, y, and z.

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And we want to dot that
with this vector.

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And it's all being done in
the variables x and y.

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So we should be able to change
everything to x and y

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ultimately.

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So let's look at what we
get when we do that.

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So F--

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I'm going to remind myself--

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F was equal to zi plus xk.

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Which, if I write that in the
component form, it's z comma,

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0 comma, x.

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So F dotted with our minus
f sub x minus f sub y 1--

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which was minus 2x,
minus 1, 1--

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we see we get--

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minus 2x dotted with z-- we get
minus 2xz and then we get

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0 and then we get x.

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So we get minus 2xz plus x.

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That's exactly what f dotted
with the vector we have is.

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So now also we know that z was
equal to x squared plus y.

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So we actually get negative
2x times x squared

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plus y plus x again.

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So I'm going to just
expand that so it's

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easier to deal with.

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So we get negative 2x cubed
minus 2xy plus x.

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And now we have exactly what--

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if we look over here-- we have
exactly this entire part here

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written as a function
of x and y.

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Which is good.

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Why is that good?

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Because everything we're
integrating is in x and y.

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We're doing dx and dy so we just
need to figure out the

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bounds and compute
the integral.

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So let's come over here.

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So the flux then is going to
be equal to-- well, we know

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the region.

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We know the region is y and x
are both going from 0 to 1.

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So the order doesn't matter
because nothing depends on

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another function.

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And then we're integrating
exactly this function.

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Negative 2x to the third
minus 2xy plus x dy dx.

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So when we integrate
in y, we should be

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careful what we get here.

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We're going to have the integral
from 0 to 1, and then

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we're going to have--

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this we get a negative 2x
cubed times y, and then

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evaluate it at 0 and 1.

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So we just get a negative
2x cubed again.

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We integrate this we have a
negative 2xy squared over 2.

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So at 0 we get nothing
and at 1 we get 1/2.

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And so we get minus 2x.

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And then here when we integrate
in y, we get x times

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y and we evaluate that at 1 and
0, and we got just plus x.

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So let me just make
sure I didn't make

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any mistakes there.

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So this one, I'm integrating
it in y and so I get a

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negative 2x cubed y, evaluated
at 0 and 1.

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So at 1 I just get a negative
2x cubed, at 0 I get 0.

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In this one, I have
a negative 2xy.

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When I integrate that I get
a y squared over 2.

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The 2s kill off.

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So I'm left with a negative
of xy squared.

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Evaluating that at 0 and 1, at
0 I get 0 and at 1 I get

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negative x.

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Oh, there.

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So there shouldn't
be a 2 there.

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And then here when I integrate
that I get xy evaluated at y

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equals 0 and y equals 1, and
take that difference.

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And at 1 I get just x and
at 0 I get nothing.

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Hopefully that one
is correct now.

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Because I forgot to kill off
the 2 there first. So those

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subtract off and I'm left with
minus the integral from 0 to 1

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of 2x cubed dx.

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Well, that's going to be minus
of x cubed, it's going to be x

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to the fourth over 4.

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And then I have the
2 still here.

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So that will divide out.

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Evaluate at 0 and 1.

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At 0 I obviously get nothing.

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At 1 I get negative 1/2.

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And so the flux of F across
the surface is equal to

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negative 1/2.

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And that's the upward flux.

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So obviously if I wanted to know
the downward flux, that

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would be positive 1/2.

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It doesn't have anything
to do with what F is.

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It has to do with the direction
of the normal that

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I'm dotting F with.

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So since I was dotting F with
the upward normal--

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which is the ndS that I showed
you was the upward normal--

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then I know that this
is the upward flux.

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So let me just remind you
what we did here.

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Let's come back to the
very beginning.

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So the object was that we had
z as a function of x and y.

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So we knew we had a surface
sitting over some region in

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the x-y plane.

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And we wanted to compute
the flux of a

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certain vector field--

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the upward flux of a certain
vector field--

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across that surface.

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And so all we had to do to
solve this problem was

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ultimately understand what n
dS was-- which you actually

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did in class.

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You saw what n dS is,
this is the upward

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normal through the surface.

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And then recognize
that the flux--

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again, we saw this from class--
that the flux is equal

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to the double integral over the
surface of F dot dS, which

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is the same as the double
integral over the region of F

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dotted with n dS.

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Where ndS now I'm referring to
as n is the vector and dS is--

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this whole component is ndS--

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that's what we found.

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And so then we know F. It's
in terms of z, x, and y.

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But then we can find it
in terms of x and y.

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When we take that dot product we
end up with exactly just a

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function of x and y, when we
replace z by what it actually

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is equal to.

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And then we just compute
the integral.

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And this is just a regular
old double integral.

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And we get the flux was
equal to minus 1/2.

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And again, I want to point out
that if we wanted instead of

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the upward flux the downward
flux, it would be the same

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with the opposite sign.

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OK.

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That is where I think
I'll stop.

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