Choose the learning badge you'll earn today

Current time:0:00Total duration:10:31

0 energy points

# Green's theorem example 1

Using Green's Theorem to solve a line integral of a vector field. Created by Sal Khan.

Video transcript

Let's see if we can use our
knowledge of Green's theorem to solve some actual
line integrals. And actually, before I show an
example, I want to make one clarification on
Green's theorem. All of the examples that I did
is I had a region like this, and the inside of the region
was to the left of what we traversed. So all my examples I went
counterclockwise and so our region was to the left of-- if
you imagined walking along the path in that direction, it
was always to our left. And that's the situation which
Green's theorem would apply. So if you were to take a line
integral along this path, a closed line integral, maybe we
could even specify it like that. You'll see that in some
textbooks-- along the curve c of f dot dr. This is what
equals the double integral over this region r of the partial of
q with respect to x minus the partial of p with
respect to y d area. And just as reminder, now this
q and p are coming from the components of f in
the situation. f would be written as f of xy
is equal to P of xy time the i component plus Q of xy
times the j component. So this is a situation where
the inside of the region is to the left of the direction
that we're taking the path. If it was in the reverse,
then we would put a minus sign right here. If this arrow went the other
way, we'd put a minus sign and we can do that because we know
that when we're taking line integrals through vector
fields, if we were reverse the direction it becomes
the negative of that. We showed that, I think,
4 or 5 videos ago. With that said, it was
convenient to write Green's theorem up here. Let's actually solve a problem. Let's say I have the line
integral and let's say we're over a curve. I'll define the
curve in a second. But let's say that the integral
we're trying to solve is x squared minus y squared
dx plus 2xy dy. And then our curve-- they're
giving us the boundary. The boundary is the region. I'll do it in a
different color. So the curve is boundary of the
region given by all of the points x,y such that x is a
greater than or equal to 0, less than or equal to 1. And then y is greater than
or equal to 2x squared and less than or equal to 2x. So let's draw this region that
we're dealing with right now. So let me draw my x-axis
or my y-axis, sorry. My y-axis and then my
x-axis right there. And let's see. x goes from
0 to 1, so if we make-- that's obviously 0. Let's say that that is x
is equal to 1, so that's all the x values. And y varies, it's above
2x squared and below 2x. Normally, if you get to large
enough numbers, 2x squared is larger, but if you're below 1
this is actually going to be smaller than that. So the upper boundary is
2x, so there's 1 comma 2. This is a line y is equal to
2x, so that is the line y is-- let me draw a straighter
line than that. The line y is equal to 2x
looks something like that. That right there is
y is equal to 2x. Maybe I'll do that in yellow. And then the bottom curve right
here is y is going to be greater than 2x squared. It might look
something like this. And of course, the region that
they're talking about is this, but we're saying that the curve
is the boundary of this region and we're going to go in I
counterclockwise direction. I have to specify that. So our curve, we could start at
any point really, but we're going to go like that. Then get to that point and then
come back down along that top curve just like that. And so this met the condition
that the inside of the region is always going to be our left,
so we can just do the straight up Green's theorem, we don't
have to do the negative of it. And let's define our region. Let's just do our region. This integral right here is
going to go-- I'll just do it the way-- y varies from y is
equal to 2x squared to 2y is equal to 2x. And so maybe we'll integrate
with respect to y first. And then x, I'll
do the outside. The boundary of x
goes from 0 to 1. So they set us up well to
do an indefinite integral. Now we just have to figure
out what goes over here-- Green's theorem. Our f would look like
this in this situation. f is f of xy is going to be
equal to x squared minus y squared i plus 2xy j. We've seen this in
multiple videos. You take the dot product of
this with dr, you're going to get this thing right here. So this expression right
here is our P of x y. And this expression right
here is our Q of xy. So inside of here we're just
going to apply Green's theorem straight up. So the partial of Q with
respect to x-- so take the derivative of this
with respect to x. We're just going to
end up with the 2y. And then from that we're going
to subtract the partial of P with respect to y. So if you take the derivative
of this with respect to y that becomes 0 and then, here you
have-- the derivative with respect to y here is minus 2y. Just like that. And so this simplifies
to 2y minus minus 2y. That's 2y plus plus 2y. I'm just subtracting
a negative. Or this inside, and just
to save space, this inside-- that's just 4y. I don't want to have to
rewrite the boundaries. That right there is
the same thing as 4y. The partial of Q with respect
to y, 2y minus the partial of P with respect to y. Which is minus 2y. You subtract a negative
you get a positive. You have 4y. Let's take the antiderivative
of the inside with respect to y and we're going
to get 2y squared. Let me do it a
little bit lower. We're going to get 2y squared
if you take the partial with respect to y you're
going to get 4y. We're going to evaluate that
from y is equal to 2x squared to y is equal to 2x. And then of course, we
still have the outside integral still there. x goes from 0 to 1 dx. This thing is going to be equal
to the integral from 0 to 1 and then we evaluate
it first at 2x. So you put a 2x in here,
2x squared is 4x squared. 2 squared, x squared, so
4x squared times 2 is going to be 8x squared. Minus-- put this guy in there. 2x squared squared is
2x to the fourth. 4x to the fourth times
2 is 8x to the fourth. Did I do that right? 2x squared-- going to
put it there for y, substitute y with it. That squared is 4
X to the fourth. Times 2 is 8x x to the fourth. Very good. All right. Now dx. Now this is just a
straightforward one-dimensional integral. This is going to be equal
to-- I'll just do it here. This is going to be equal to
the antiderivative of 8x squared is 8/3 x to the third. And then the antiderivative
of 8x to the fourth is minus 8/5 x to the fifth. Then we're going to have to
evaluate that from 0 to 1, or we can put a little line there. When you put 1 in there
you get-- I'll do it in a different color. We get 8/5 times 1 to
the third, which is 8/5 minus 8/5 minus 8/5. And then we're going to have
minus-- when you put 0 in here you're just going
to get a bunch of 0's. Oh sorry, I made a mistake. It would have been a blunder. It's 8/3. 8/3 times 1 to the third minus
8/5 times 1 to the fifth, so that's minus 8/5. And then, when you subtract the
0, you evaluate 0 here. you're just going to get a bunch of
0's so you don't have to do anything else. So now we just have to
subtract these two fractions. So let's get a common
denominator of 15. 8/3 is the same thing if
we multiply the numerator and denominator by 5. That is 40/15. And then if we multiply this
numerator and denominator by 3, that's going to be 24/15. So minus 24/15 and we get
it being equal to 16/15. And so using Green's theorem we
were able to find the answer to this integral up here. It's equal to 16/15. Hopefully you found
that useful. I'll do one more example
in the next video.