Let's see if we can apply some of our new tools to solve

some line integrals.

So let's say we have a line integral along a closed curve

-- I'm going to define the path in a second -- of x squared

plus y squared times dx plus 2xy times dy.

And then our curve c is going to be defined by

the parameterization.

x is equal to cosine of t, and y is equal to sine of t.

And this is valid for t between 0 and 2 pi.

So this is essentially a circle, a unit circle,

in the xy plane, and we know how to solve these.

Let's see if we can use some of our discoveries in the last

couple of videos to maybe simplify this process.

So the first thing you might say, hey, this looks like a

line integral, but you have a dx and dy, I don't

see a dot dr here.

It's not clear to me that this is some type of even

a vector line integral.

I don't see any of vectors.

What I want to do first, and the reason why I wanted to show

you this example, is just to show you that this is just

another form of writing really a vector line integral.

To show you that you just have to realize if I have some are

r of t -- this is our curve.

I don't even write these functions in there.

I'm just going to write it's x of t times i

plus y of t times j.

We've seen several videos now that we can write dr dt as

being equal to dx dt times i plus dy dt times j.

We've seen this multiple times.

And we've seen multiple times we want to get the differential

dr, we can just multiply everything times dt.

And normally I just put a dt here and a dt there

and get rid of this dt.

But if you multiply everything times dt, if you view the

differentials as actual numbers, you can multiply and

normally you can treat them like that.

Then you just get rid of all of the dt's.

So dr you could imagine is equal to dx times the unit

vector i plus dy times the unit vector j.

So put that aside, and you might already

see a pattern here.

So if we define our vector field f, f of xy, as being

equal to x squared plus y squared, i plus 2xy j,

what is this thing?

What is this thing over here?

Well, what is f dot dr going to be?

Dot products, you just multiply the corresponding components

of our vectors and then add them up.

So it's going to be if you take this f and dot it with that dr

you're going to get the i component, x squared plus y

squared times that dx plus -- I'll do it in the pink again --

plus the y component, the j component 2xy times that dy.

That's the dot product.

And notice, this thing right here is identical to

that thing right there.

So our line integral, just to put it in a form that we're

familiar with, this is the same exact thing as the line

integral over this curve c, this closed curve c, of this f

-- maybe I'll write it in that magenta color, or actually it's

more of a purple or pink color -- f dot this dr.

That's what this line integral is, it's just a different

way of writing it.

Now that you see it, in the future if you see in kind of

this differential form, you'll immediately know OK, there's

one vector field that this is its x component, this is its y

component, dotting with the dr. This is the x component of dr

or the i component, and this is the y component or the j

component of the dr. So you immediately know what the

vector field is that we're taking a line integral of.

This is the x, that's the y.

Now, let's ask ourselves a question.

Is f conservative?

So is f equal to the gradient of some scalar field, we'll

call it capital F -- is this the case?

So let's assume it is and see if we can solve for a scalar

field whose grade it really is f.

Then we know that f is conservative.

And then if f is conservative, and this is the whole reason we

want to do it, that means that any closed loop, any line

integral over a closed curve of f is going to be equal

to 0 and we'd be done.

So if we can show this then the answer to this question or this

question is going to be 0.

We don't even have to mess with the cosine of t's and the

sign of t's and all that.

Actually, we don't even have to take antiderivatives.

So let's see if we can find an f whose gradient is

equal to that right there.

So in order for f's gradient to be that, that means that the

partial derivative of our capital F with respect to x has

got to be equal to that right there.

It's got to be equal to x squared plus y squared.

And it also tells us that the partial derivative of capital

F with respect to y has got to be equal to 2xy.

And just as a review, if I have the gradient of any function,

of any scalar field is equal to the partial of f with respect

to x times i plus the partial of capital F with

respect to y times j.

So that's why I'm just pattern matching.

I'm just saying well, gee, if this is the gradient of that,

then this must be that, which I wrote down right here, and this

must be that, which I wrote down here.

So let's see if I can find an f that satisfies both

of these constraints.

So we could just take the antiderivative with respect to

x on both sides -- remember, you just treat y like a

constant or y squared like a constant -- it's just a number.

So then we could say that f is equal to the antiderivative of

x squared is x to the third over 3.

And then the antiderivative of y squared -- remember,

this is with respect to x.

So you just treat it like a number.

That could just be the number k, or this

could be the number 5.

So this is just going to be that times x.

So plus x times y squared.

And then there could be some function of y here.

So plus some, I don't know, I'll call it g of y.

Because there could have been some function of y here.

If it's a pure function of y, when you take the derivative or

the partial with respect to x, this would have disappeared.

So it would reappear when we take the antiderivative.

And just to be clear, let me make it clear that f is going

to be a function of x and y.

So we just have the, I guess you could say

the antiderivative with respect to x.

Let's see if we take the antiderivative with respect

to y and then we can reconcile the two.

So based on this, f of xy, f of xy is going to have to look

like -- so let's take the antiderivative with

respect to y here.

So remember, you just treat x like it's just some number --

it could be a k, it could be an m, it could be a 5.

It's just some number.

So if x is just some -- the antiderivative

of 2y is y squared.

And if x is just a number there, the antiderivative of

this with respect to y is just going to be xy squared.

Don't believe me?

Take the partial of this with respect to y.

Treat x like a constant you'll get 2 times xy

with no exponent there.

And, of course, if you took the antiderivative with respect to

x, there might be some function of x here.

We were just basing it off of that information.

Now given that, this information says f of xy

is going to have to look something like this.

This information tells us f of xy's going to have to

look something like that.

Let's see if there is an f of xy that looks like

both of them essentially.

So let's see.

On this one we have xy squared here, we have

an xy squared there.

So good.

That looks good.

And it over here we have an f of x -- we have something

that's a pure function of x.

And here we have something that is a pure function of x.

So these two things could be the same thing.

Then here we have a pure function of y that might be

there, but it didn't really show up anywhere over here.

So we could just say hey, that's going to be 0.

0 is a pure function of y.

You could have something called g of y is equal to 0.

And then we get that capital F of xy is equal to x to the

third over 3 plus xy squared.

And the gradient of this is going to be equal to f.

And we've already established that.

But just to hit the point home, let's take the gradient of it.

Just if you don't believe this little stuff that I did right

there, let's take the gradient.

The gradient of f is equal to, and sometimes people put a

little vector there because you're getting

a vector out of it.

You could put a little vector on top of that gradient sign.

The gradient of f is going to be what?

The partial of this with respect to x times i.

So the partial of this with respect to x.

The derivative here is 3 divided by 3 is 1.

So it's just x squared plus the derivative of this with respect

to x is y squared times i plus the partial with respect to y.

Well, the partial with respect to y of this 0, partial with

respect to y of this is 2xy or 2xy to the first.

So it's 2xy times j.

And this is exactly equal to f, our f that we wrote up there.

So we've established that f can definitely be written -- f is

definitely the gradient of some potential scalar

function there.

So f is conservative, and that tells us that this closed loop

integral, line integral, of f is going to be equal to 0.

And we are done.

We could even ignore the actual parameterization of the path.