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Studying for a test? Prepare with these 11 lessons on Green's, Stokes', and the divergence theorems.

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# Green's theorem proof (part 1)

Video transcript

Let's say we have a
path in the xy plane. That's my y-axis, that is
my x-axis, in my path will look like this. Let's say it looks like that;
trying to draw a bit of an arbitrary path, and let's say
we go in a counter clockwise direction like that
along our path. And we could call this path--
so we're going in a counter clockwise direction --we
could call that path c. And let's say we also
have a vector field. And our vector field is going
to be a little unusual; I'll call it p. p of xy. It only has an i component, or
all of its vectors are only multiples of the i-unit vector. So it's capital P of xy
times the unit vector i. There is no j component, so if
you have to visualize this vector field, all of the
vectors, they're all multiples of the i-unit vector. Or they could be negative
multiples, so they could also go in that direction. But they don't go diagonal
or they don't go up. They all go left to
right or right to left. That's what this vector
field would look like. Now what I'm interested in
doing is figuring out the line integral over a closed loop--
the closed loop c, or the closed path c --of p dot dr,
which is just our standard kind of way of solving
for a line integral. And we've seen what
dr is in the past. dr is equal to dx times i plus
dy times the j-unit vector. And you might say, isn't
it dx, dt times dt? Let me write that: can't dr
be written as dx, dt times dti plus dy, dt times dtj? And it could, but if you
imagine these differentials could cancel out, and you're
just left with the dx and a dy, and we've seen
that multiple times. And I'm going to leave it in
this form because hopefully, if we're careful, we won't have to
deal with the third parameter, t. So let's just look at it in
this form right here with just the dx's and the dy's. So this integral can be
rewritten as the line integral, the curve c-- actually let
me do it over down here. The line integral over the path
of the curve c of p dot dr. So we take the product of each of
the coefficients, let's say the coefficient of the i component,
so we get p-- I'll do that in green, actually do that purple
color --so we get p of xy times dx plus-- well there's no 0
times j times dy; 0 times dy id just going to be 0 --so this
our line integral simplified to this right here. This is equal to this original
integral up here, so we're literally just taking the line
integral around this path. Now I said that we play our
cards right, we're not going to have to deal with the third
variable, t; that we might be able just solve this integral
only in terms of x. And so let's see if
we can do that. So let's look at our minimum
and maximum x points. That looks like our
minimum x point. Let's call that a. Let's call that our maximum
x point; let's call that b. What we could do is we can
break up this curve into two functions of x.
y is functions of x. So this bottom one right here
we could call as y1 of x. This is just a standard curve;
you know when we were just dealing with standard calculus,
this is just you can imagine this is f of x and
it's a function of x. And this is y2 of x. Just like that. So you can imagine two paths;
one path defined by y1 of x-- let me do that in a different
color; magenta --one path defined by y1 of x as we go
from x is equal to a to x is equal to b, and then another
path defined by y2 of x as we go from x is equal to
b to x is equal to a. That is our curve. So what we could do is, we
could rewrite this integral-- which is the same thing as that
integral --as this is equal to the integral-- we'll first do
this first path --of x going from a to b of p of x. And I could to say p of x and
y, but we know along this path y is a function of x. So we say x and y1 of x. Wherever we see a y we
substitute it with a y1 of x, dx. So that covers that first path;
I'll do it in the same color. We could imagine this is c1. This is kind of the first half
of our curve-- well it's not exactly the half --but that
takes us right from that point to that point. And then we want to
complete the circle. Maybe I'll do that, and
I'll do that in yellow. That's going to be equal to--
sorry we're going to have to add these two --plus the
integral from x is equal to b to x is equal to a of-- do it
in that same color --of p of x. And now y is going
to be y2 of x. Wherever you see a y, you
can substitute with y2 of x along this curve. y2 of x, dx. This is already getting
interesting and you might already see where
I'm going with this. So this is the curve c2. too I think you appreciate if
you take the union of c1 and c2, we've got our whole curve. So let's see if we can simplify
this integral a little bit. Well one thing we want to
do, we might want to make their end points the same. So if you swap a and b
here, it just turns the integral negative. So you make this into a b, that
into an a, and then make that plus sign into a minus sign. And now we can rewrite this
whole thing as being equal to the integral from a to b of
this thing, of p of x and y1 of x minus this thing, minus p of
x and y2 of x, and then all of that times dx. I'll write it in a third color. Now, I'm going to do something
a little bit arbitrary, but I think you'll appreciate why I
did this by the end of this video, and it's just a
very simple operation. What I'm going to do is I'm
going to swap these two. So I'm essentially going to
multiply this whole thing by negative 1, or essentially
multiply and divide by negative 1. So I can multiply this by
negative 1 and then multiply the outside by negative 1, and
I will not have changed the integral; I'm multiplying
by negative 1 twice. So if I swap these two things,
if I multiply the inside times negative 1, so this is going to
be equal to-- do the outside of the integral, a to b. If I multiply the inside--
I'll do a dx out here --if I multiply the inside of the
integral by negative 1, these two guys switch. So it becomes p
of x of y2 of x. And then you're going to have
minus p of x and y1 of x. My handwriting's getting
a little messy. But I can't just multiply
just the inside by minus 1. I don't want to change the
integral, so I multiplied the inside by minus 1, let me
multiply the outside by minus 1. And since I multiplied by
minus 1 twice, these two things are equivalent. Or you could say this is
the negative of that. Either way, I think you
appreciate that I haven't changed the integral
at all, numerically. I multiplied the inside and
the outside by minus 1. And now the next step I'm going
to do, it might look a little bit foreign to you, but I
think you'll appreciate it. It might be obvious to you
if you've recently done some double integrals. So this thing can be rewritten
as minus the integral from a to b of-- and let me do a new
color --of the function p of x, y evaluated at y2 of x minus--
and let me make it very clear; this is y is equal to y2 of x
--minus this function evaluated at y is equal to y1 of x. And of course all
of that times dx. This statement and what we
saw right here-- this statement right here --are
completely identical. And then if we assume that a
partial derivative of capital P with respect to y exists,
hopefully you'll realize-- and I'll focus on this a little
bit because I don't want to confuse you on this step. Let me write the outside
of this integral. So this is going to be equal
to-- and this is kind of a neat outcome, and we're starting
to build up to a very neat outcome, which will probably
have to take the next video to do --so we do the outside dx. If we assume that capital P has
a partial derivative, this right here is the
exact same thing. This right here is the exact
same thing as the partial derivative of P with respect to
y, dy, the antiderivative of that from y1 of x to y2 of x. I want to make you feel
comfortable that these two things are equivalent. And to realize they're
equivalent, you'll probably just have to start here
and then go to that. We're used to seeing this;
we're used to seeing a double integral like this, and then
the very first step we say, OK to solve this double integral
we start on the inside integral right there, and we say, OK
let's take the antiderivative of this with respect to y. So if you take the
antiderivative of the partial of p with respect to y, you're
going to end up with p. And since this is a definite
integral, the boundaries are going to be in terms of x,
you're going to evaluate that from y is equal to y2 of x, and
you're going to subtract from that y is equal to y1 of x. Normally we start with
something like this, and we go to something like this. This is kind of unusual that we
started, we kind of solved, we started with the solution of
the definite integral, and then we slowly built back to
the definite integral. So hopefully you realize that
this is true, that this is just we're kind of going
in a reverse direction than we normally do. And if you do realize that,
then we've just established a pretty neat outcome. Because what is
this right here? Let me go back, let me see
if I can fit everything. I have some function-- and I'm
assuming that the partial of P with respect to y exists
--but I have some function defined over the xy plane. You know, you could
imagine we're dealing in three dimensions now. We'll draw a little bit neater. So that's y, that's x, that's
z, so this, you could imagine, is some surface; it just
happens to be the partial of P with respect to x. So it's some surface on
the xy plane like that. And what are we doing? We're taking the double
integral under that surface, around this region. The region's boundaries in
terms of y are defined by y2 and y1 of x. So you literally
have that curve. That's y2 on top,
y1 on the bottom. And so we're essentially
taking the volume above. So if you imagine with the base
is-- the whole floor of this is going to be the area inside of
this curve, and then the height is going to be the function
partial of P with respect to y. It's going to be a little hard
for me to draw, but this is essentially some type of a
volume, if you want to visualize it that way. But the really neat outcome
here is if you call this region r, we've just simplified
this line integral. And this was a special one. It only had an x-component, the
vector field, but we've just simplified this line integral
to being equivalent to-- maybe I should write this line
integral because that's what's the really neat outcome. We've just established that
this thing right here, which is the same as our original
one, so let me write that. The closed line integral around
the curve c of p of xy, dx, we've just established that
that's the same thing as the double integral over the region
r-- this is the region r --of the partial of
P with respect to y. And we could write dy, dx, or
we could write da, whatever you want to write, but this is
the double integral over that region. The neat thing here is using a
vector field that only had an x-component, we were able to
connect its line integral to the double integral over
region-- oh, and I forgot something very important. We had a negative
sign out here. So this was a minus
sign out here. Or we could even put the minus
in here, but I think you get the general idea. In the next video, I'm going to
do the same exact thing with the vector field that only has
vectors in the y-direction. And then we'll connect
the two and we'll end up with Green's theorem.