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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

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and we want S to
be the graph of z

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above the unit square
in the xy-plane.

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And what we'd like to do is
for F equal to z*i plus x*k,

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find the upward
flux of F through S.

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So S is our surface
that's a graph,

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over the xy-plane
and the unit square,

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of z 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,

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

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bring 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

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to find the upward flux of
F through this surface S.

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And let's think about 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--

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we can think of it as a function
of x and y over the unit

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square.

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

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is equal to x squared plus y.

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And then we know how to
compute the normal-- well,

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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-- or sorry.

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

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

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minus f sub y comma 1 dx dy.

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So that's exactly what this
n*dS-- so n is the vector,

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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?

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f sub x-- because f is
equal to z-- f sub x is 2x

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and f sub y is 1.

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

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minus 1, comma 1, dx dy.

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And now to compute the surface
integral what we do-- or sorry,

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to compute the flux along
the surface, what we do

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is we integrate over
the surface-- which

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I guess we should remember
that's a double integral,

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because it's over a surface--
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

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below that defines the
surface in the xy-coordinates.

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So it's integrating over
the region of F dotted

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with this vector here.

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Because n*dS-- dS is n*dS,
and in the x-y components

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it's 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--
we were integrating F dot

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dS on the surface,
to now taking F

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dotted with this
vector on the region

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in the xy-plane over
which we can define S.

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So the region we're
interested in, remember,

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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

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looking at F as a
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-- I'm going
to remind myself--

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F was equal to z*i plus x*k.

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Which, if I write that
in the component form,

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it's z comma, 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-- we see we get--

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minus 2x dotted with z-- we get
minus 2x*z and then we get 0

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and then we get x.

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So we get minus 2x*z 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 plus y plus x

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again.

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So I'm going to just expand that
so it's easier to deal with.

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So we get negative 2x
cubed minus 2x*y plus x.

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And now we have exactly
what-- if we look over here--

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we have exactly this
entire part here written

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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 bounds

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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,

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we know 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

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depends on another function.

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And then we're integrating
exactly this function.

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Negative 2 x to the third
minus 2x*y plus x dy dx.

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So when we integrate in
y, we should be careful,

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what do we get here?

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

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and then we're going to have--
this we get a negative 2 x

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cubed times y, and then
evaluated at 0 and 1,

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so we just get a
negative 2 x cubed again.

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We integrate this we have a
negative 2x y 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 y

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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 any mistakes there.

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So this one, I'm
integrating it in y

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and so I get a negative 2 x
cubed y, evaluated at 0 and 1.

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

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In this one, I have
a negative 2x*y.

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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 x y squared.

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

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and at 1 I get 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 x*y 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.

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So those subtract off and I'm
left with minus the integral

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from 0 to 1 of 2 x cubed dx.

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

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it's going to be x
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

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is equal to 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,

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that 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

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that 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 n*dS 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

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over some region
in the xy-plane.

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And we wanted to compute
the flux of a certain vector

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field-- 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 ultimately

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understand what n*dS was--
which you actually did in class.

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You saw what n*dS is, this
is the upward normal through

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the surface.

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And then recognize
that the flux-- again,

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we saw this from class-- that
the flux is equal to the double

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integral over the
surface of F dot dS,

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which 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 n*dS, now, I'm referring
to as-- n is the vector and dS

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is-- this whole component is
n*dS-- 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

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a 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

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of the upward flux
the downward flux,

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it would be the same
with the opposite sign.

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OK.

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That is where I think I'll stop.