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# Stokes' theorem proof part 3

Video transcript

We've now laid the groundwork
so we can express this surface integral, which is
the right-hand side of the way we've
written Stokes' theorem. We can now express this
as a double integral over the domain of the
parameters that we care about. And we're going to do
that in this video. And then in the next
series of videos, we'll do the same thing
for this expression. We're actually going to do
that using Green's theorem. What we're going to
do is we're going to see we're going to get
the same expressions, which will show us that
Stokes' theorem is true, at least for this
special class of surfaces that we are studying right here. But they're pretty general. Now let's now try to do that. So our surface
integral-- I'm just going to rewrite it down here. It's the surface integral,
so over our surface of the curl of f. Actually, let me go
a little bit lower. So we have our surface integral
of the curl of F dot ds. Well, we've already figured
out what our curl of F is here two videos ago. And we've almost
figured out what ds is. ds is the cross product of
these two vectors times dA. The cross product
of these two vectors is this right over here. So we could just
write that ds is going to be equal to
this thing times dA. This is the cross
product of the partial of r with respect to x and the
partial of r with respect to y. And then we have to multiply
that times dA right over there. So this expression is
just going to be the dot product of the curl of F, which
is this stuff up here dotted with this stuff down here. And essentially, we're
going to take the dot product of this
vector and that vector and then multiply it times
this-- we can actually consider this to
be a scalar value. So let's do that. So this is going to
be equal to-- and when we do this, all
of these we are now going to start operating in
the domain of our parameters. So it's going to turn
from a surface integral into a double integral
over that domain, over that region
that we care about. This is the domain of our
parameters, this region R. And this is how we've
manipulated any of the surface integrals that we've
come across so far. We've turned them
into double integrals over the domain
of the parameters. So this is going to turn into a
double integral over the domain of our parameters, which is
the region R. It's the region R in the xy plane right up here. And now we can take the dot
product of the curl of F with dotted ds, which is all of
this business right over here. And let me see if I
can show both of them on the screen at the same time. There you go. So first let's think
about the x components. So you have that
right over there. And then you have
this right over here. You multiply the two. The negative we can swap
the order right over there. You have the partial of z
with respect to x times-- and we're going to swap
this order right over here. The partial of Q with respect
to z minus the partial of R with respect to y. Now let's think about
the j component. You have negative z sub y
times all of this up here, at least the stuff that's
multiplied times the j component. This negative can cancel out
with that negative, so you have plus z sub y, the partial
of z with respect to y, times the partial
of R with respect to x minus the partial
of P with respect to z. Let me make that clear, that
is an R right over here. And then we have
the k component. And the k component is actually
the easiest because it's just 1 here. So it's just going
to be 1 times-- and I'll just do it in
that same color-- 1 times the partial of Q with respect
to x minus the partial of P with respect to y. And then finally, we just
have this dA over here. And this dA is
multiplied by everything. So we'll put some parentheses,
and we'll write dA. So we're done. We've expressed our surface
integral as a double integral over the domain
of our parameters. And what we're going to
do in the next few videos is do the same thing with
this using Green's theorem. And we're going to see that
we get the exact same value.