Hi, everyone! Welcome back to integralcalc.com. Today we’re going to be talking about how
to use Green’s theorem and in this particular problem, we’ve been given two functions
p of x,y equals x plus y squared and q of x,y which equals y plus x squared.
We’ve been asked to find the line integral that’s defined by the square that has vertices
(+/-1,+/-1). So we have these four points here on the square that’s defined by (1,1),
(-1,1), (-1,-1), and (1,-1). So we have this line integral that’s a square and this is
the formula (Green’s theorem) that allows us to convert the line integral given here
to a double integral which can be a lot easier for us to evaluate. So depending on the kind
of problem, you may be given your problem in terms of p of x,y and q of x,y which is
what we have here. We’ve got these two formulas. You also may be given something that looks
like this. But either way, Green’s theorem tells us that p is what we have here before
dx and q is what we have here before dy. So whether we’re given the problem in terms
of the actual functions p of x,y or we’re given the line integral, we can pull out here,
p and q and we see our two functions. Once we have those two functions, we’re kind
of off to the races because all we need to do is identify the partial derivative of q
with respect to x and the partial derivative of p with respect to y and then put together
some limits of integration and we’ll be able to move forward with our double integral
for Green’s theorem. So again, the first thing we’re going to
want to do is find the partial derivative of q with respect to x because that’s the
first thing that we need here to convert this line integral into a double integral. So we’ll
find the partial derivative of q with respect to x and you always find the partial derivative
of q with respect to x and p with respect to y. So in this case, q of x,y equals y plus
x squared. So to find the partial derivative with respect to x, we’ll get zero plus 2x
which of course would just give us 2x. So now we’ll find the partial derivative of
p with respect to y. And p is x plus y squared so finding the partial derivative with respect
to y gives us zero plus 2y which of course just equals 2y. now that we have here partial
derivative of q with respect to x, we’ll be plugging it in to our double integral here
and we have the partial derivative of p with respect to y and we’ll be plugging that
into our double integral here. So let’s take a look at what that looks
like. We’ll have the double integral. and I’m going to go ahead and leave a little
bit of room for our limits of integration because we’re going to need to identify
limits of integration for both of these upper and lower limits of integration for both integrals
so I’m going to leave a little space. And then we have partial derivative of q with
respect to x which is 2x minus the partial derivative of p with respect to y which we
know to be 2y. and we’re going to be evaluating with respect to a. So we’ll turn this da
into dy and dx in a second. But we always want to, for this double integrals
here, integrate first with respect to y and then with respect to x. So this inner integral
is going to be y and the outer integral will be x. And this da here will become dy,dx.
So because y here is the inner integral, dy has to be on the inside. x is the outer integral
so dx has to be on the outside. So we’re going to talk about our limits of integration
with respect to y first. We’ve got this line integral that defines the square here
and we’ve graphed it. If we’re going to integrate with respect to y, realize that
we’re integrating from the line y equals negative which runs along here to the line
y equals positive 1. So we’re integrating this distance. So from the line, essentially
y equals negative to y equals positive 1. So our lower limit of integration with respect
to y is negative 1. Upper limit of integration is positive 1. Then we’re talking about
limits of integration with respect to x. So now this is something your more familiar with.
We’re integrating from the line x equals negative 1 to the line x equals positive 1.
And we’re integrating this entire distance. So our lower limit of integration with respect
to x is also negative 1 and our upper limit of integration is positive 1.
So now that we have our partial derivatives plugged in to our theorem and our lower and
upper limits of integration, we can go ahead and perform the integration. So we’ll integrate
with respect to y first. I’m going to go ahead and leave this negative 1 and positive
1 out here for the outer integral. For the inner integral, when we integrate with respect
to y, remember that we’ll be treating y as the variable and treating x like a constant.
So for 2x, if x is a constant and 2x is also a constant so we just need to add a y to that
term so we’ll have 2xy. Integrating -2y with respect to y is easy. That should be
familiar to us. We’ll get y squared and we’re going to be evaluating that on our
inner limits of integration here, -1 and 1 are going to translate like this. What remains
are the outer limits of integration and dx because we haven’t integrated yet with respect
to x. So before we do that, we have to go ahead
and evaluate at these limits of integration. So remember we plug in the upper limit of
integration first so we’ll go ahead and leave these outer limits here. And we’ll
be plugging in for the variable y because we integrated with respect to y. So we’ll
have 2x times 1 minus 1 squared and then we will subtract what we get when we plug in
-1. So we’ll get 2x times -1 minus -1 squared and then of course we’ve got dx out here.
Alright, so now we just need to simplify this. We’ll get -1 and positive 1 so here we’ll
get 2x minus 1, which will give us a -2x which is going to turn into a positive 2x. and then
we’ll get a positive 1 but that’s a -1. We’ve got another negative so that ends
up being a positive 1 dx. So as you can see, the -1 and the positive 1 cancel here so we’re
just left with 4x dx. And now, we can integrate with respect to x. So doing so is going to
give us 2x squared and now we’re going to be evaluating our outer limits of integration
so again from negative 1 to positive 1. So when we do that, we’ll plug in our upper
limit of integration first and we get 2 times 1 squared minus 2 and then our lower limit
of integration, negative 1 squared. That’s going to give us 2 minus negative squared
gives us a positive 1 times 2 is 2 so that’s 2. 2 minus 2 gives us zero and that is our
final answer. So that is how you use Green’s theorem to
evaluate the line integral and in this particular case when we have the line integral that defines
the square, our final answer is zero. So that’s it. I hope this video helped you guys and
I will see you in the next one. Bye!