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## Stokes' theorem (videos)

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# Stokes example part 3

## Video udskrift

We're now ready to get into the
meat of evaluating this surface integral. And we really need
to re-express it in terms of a double integral
in the domain of the parameters. And the first thing
I'm going to do is rewrite this part right
over here using our parameters. And we already know that n, our
normal vector times our surface differential, can
also be written as kind of a vector version
of our surface differential that points in the same
direction as our normal vector. And this is going to
be the same thing. And we need to make
sure that we get the order on the
cross product right as the partial
derivative-- and I'm going to confirm
this in a second. The partial derivative
of the parametrization with respect to one
of the parameters crossed with the
partial derivative of the parametrization with
respect to the other parameter. And then that whole
thing-- I'm not going to take the absolute
value because I need a vector right over here--
times differentials of the parameters-- d theta
dr. And we can swap these two things around depending
on what will make our eventual double
integral easier. But we can't swap
these two things around because this would
actually change the direction of the vector. So we need to make sure
that this is popping us out in the right direction. So let's think
about the direction that the partial with
respect to r will take us. So as r increases, we're
going to be moving radially outward from the
center of our surface. Let me do this in
a different color. As r increases, we'll be
moving radially outwards. So this quantity
will be a vector that looks something like that. And then as theta
increases, we'll be going roughly
in that direction. And so if we take the cross
product of those two things-- and we could use
the right-hand rule. Take your right hand,
point your index finger in the direction of
that yellow vector. Let me make it clear-- this
is the orange vector right over here. Take your index finger
in the direction of that yellow vector. So that's my index finger,
my shakily drawn yellow index finger. Put your middle finger in the
direction of the orange vector. So my middle, you
bend it and put it in the direction of
the orange vector. And we don't care what
the other two fingers do. And then, your thumb
will be in the direction of the cross product. So your thumb will
point outward like that. That's my best
attempt at drawing it. Which is exactly the direction
we needed to point it. We need it to point
upward here in order to be oriented properly
with the direction that we are actually
traversing the path. So this is actually
the right order. If when we did this we
got the thumb pointing into it or below
the plane, then we would actually have
to swap these orders. So with that out
of the way, let's actually evaluate
this cross product. So the cross product of the
partial of our parametrization with respect to r
crossed with the partial of our parametrization
with respect to theta. I like to set up the matrix
to take the cross product. So we'll put i, j, and k
components, < like that. And then, first, I will write
the partial with respect to r. So the i-component, if you
take the derivative of this with respect to r, is just
going to be cosine theta. The derivative of
this with respect to r is just sine theta. And the derivative of
this with respect to are r is just going to be a
negative sine theta. And then we're going
to cross it with this, so the derivative of this
with respect to theta is going to be negative
r sine of theta. The derivative of
our j-component with respect to theta
will be r cosine theta. And the derivative
or our k-component-- or our z-component
with respect to theta-- is going to be-- let's see--
negative r cosine theta. Is that right? Yeah. Derivative of sine
theta's cosine theta. Yeah. So it's negative r cosine theta. And now we just evaluate
this determinant over here. Our i-component is
going to be-- so ignore that row and this column. And we get sine times
negative r cosine theta. I'll just do this
in a new color. So we're going to get negative
r-- that wasn't a new color. I'll do it in purple. We will get negative r cosine
theta sine theta minus-- well, this is going to be
a negative number. So when you subtract a negative
number, it's going to be plus. So it's going to be plus
r cosine theta sine theta. It's always nice when
things cancel out like this. This plus this is just 0. Negative of it plus
the positive of it. So that all canceled out to 0. We don't have an i-component. Now, let's go to
the j-component. And remember, we need to do our
little checkerboard pattern. So it's going to be minus j. And it's going to be-- ignore
this column, that row-- cosine theta times negative r. Cosine theta is negative
r cosine squared theta. I just multiplied those two. And then, from that I'm going
to subtract this times that. And so this times that,
the negatives cancel out. We get r sine squared theta. So this is minus--
let me make sure. Yeah, I'm going to subtract
the product of these two. The product is positive. It's going to be r
sine squared theta. This is always a hard part. You can make a lot of
careless mistakes here. And this looks like we
might be able to simplify this in a second, but I'll wait. Actually, I'll just distribute
this negative sign just for fun. So if we distribute
the negative sign, these all become positive. Helps simplify
things a little bit. And now let's worry
about the k-component. The k-component I will be
doing in purple-- well, I'll do it in blue. The k-component. Ignore this row,
ignore this column. So plus k times cosine
theta times r cosine theta is r cosine squared of theta. And then from that, I'm going
to subtract negative r sine theta times sine theta. So that's going to be
negative r sine squared theta, but I'm subtracting it. So it's going to be plus
r sine squared theta. And this looks like we're
going to simplify it as well. And so this piece
right over here, we can factor out--
let me just rewrite it. This can be rewritten as r
times cosine squared theta plus sine squared theta. Basic trig identity. That just evaluates to 1. So this is just r times j. And this over here simplifies
for the exact same reason. This is r times cosine squared
theta plus sine squared theta. This also is just 1. So this just simplifies
to r times k. This whole cross product--
all of this business right over here-- simplified
to, quite luckily, is equal to r times
our j unit vector plus r times our k unit vector. And so now we can write
our surface integral. Our original surface
integral we can write as the double integral. Or we might want to change the
order in which we integrate, but we'll give ourselves that
option a little bit later. Double integral. Now, it's going to be over our
parameter domain or the r theta domain. So it's the double integral of--
we still have the curl of f. And we're going to have
to evaluate the curl of f. So I'll just write "curl
of our vector field f" dotted with this
business, rj plus rk. And then we have
our two parameters. And we might want
to switch the order. So maybe we could
write d theta dr. And then if we do
it in this order, theta goes from 0 to 2 pi,
and r is going from 0 to 1. But if we swap these
two, then obviously, we're going to have to
swap these two as well. I'll leave you there. In the next video, we will
evaluate the curl of f. And maybe in that
video, if we have time, we will get to the finish line.