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# Stokes' theorem proof part 5

Video transcript

Now that we have
a parameterization for the boundary of our
surface right up here, let's think a little bit
about what the line integral-- and this was the left side of
our original Stokes' theorem statement-- what the line
integral over the path C of F, our original vector
field F, dot dr is going to be. And now, once
again, this R we're talking about the path of
this boundary right over here, C. Not C1 that we
did to kind of get some of our foundational
material going. So F dot dr-- well,
we remember what F is. F was all the way up here. Its i, j, and k components
were just P, Q, and R. So that's kind of
easy to remember. But let's think about
what dr is equal to. And we're going to
have to break out a little bit of our
three-dimensional or our multivariable chain
rule right over here. So dr is the same thing
as dr dt times dt. So we really just
have to figure out what the derivative of
R with respect to t. And this one we're going
to have to break out a little bit of the chain rule. So let me write this down. So dr dt is going to be
equal to the derivative of x with respect to t times i
plus the derivative of y with respect to
t-- I'm just taking the derivative with
respect to t-- j. Because z is a function of
x, which is a function of t, and z is also a function of
y, which is a function of t, we're going to have to break out
our multivariable chain rule. So if we want to take
the derivative of z with respect to t-- I'll
do it separately here, and then I'll write
it down down here. The derivative of z with respect
to t-- the way I conceptualize is, what's all the
different ways that z can change from a change in t? Well, it could
change because x is changing due to a change in t. So z could change due
to x, the partial of z with respect to x, when x
changes with respect to t. But then that's not the
only way that z can change. We have to add to that how z
can change with respect to y, partial of z with respect
to y times how fast or how y is changing
with respect to t. And this is just our
multivariable chain rule. And so this is dz dt. So I'll just write
it right over here. And I'll use slightly
different notation that's consistent with
what we were doing before, and it'll help make things
a little bit clearer. So it's going to be the partial
of z with respect to x, dx dt, plus-- actually, let
me write it this way-- plus the partial of z
with respect to y dy dt. And then we're going to
multiply everything times our k. So with this out of the way,
if we wanted what dr is, dr is just going to be
this whole thing times dt. So let's do that. So now we can rewrite our
line integral right over here. And we're going to now
go into the t domain. And so t is going to
go between a and b. And F dot dr--
remember, F's components were just the functions P,
Q, and R. And each of those were functions of x, y, and z. And z is a function
of x and y, so we'll have to think about all
of that in a little bit. Use a little more of our
multivariable chain rule. But when we take
the dot products, we're just going to take
the corresponding components and multiply them. So it is going to be-- actually,
let me just copy and paste it. Let me rewrite it down here. Our vector field
F-- and I'm going to write it a
little bit shorter. Our vector field F is P times i
plus Q times j plus R times k. So when we take our dot
product of F dot dr, we're essentially taking the
dot product of this and this. And we have to throw
a dt at the end. So we're going to get P
times dx dt plus Q times dy dt plus R times all of
this business over here, which is the partial
of z with respect to x dx dt plus the partial
of z with respect to y dy dt. And then we have to multiply
all of this times dt. We can't forget that
part right over there. So we're going to multiply all
of this right here times dt. Now, I'm going to leave
you there in this video just because I'm afraid of
making careless mistakes. What we're going
to do now is we're going to rearrange
this whole thing, recognize that this is the same
thing as this right over here. Then that gives it a form that
we can apply Green's theorem using this boundary
right over here. And then when we do a little
bit more algebraic manipulation, we're going to see that this
thing simplifies to this thing right over here
and proves Stokes' theorem for our special case.