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## Multivariable calculus

### Course: Multivariable calculus > Unit 5

Lesson 2: Green's theorem# Green's theorem example 2

Another example applying Green's Theorem. Created by Sal Khan.

## Want to join the conversation?

- How do we know when to use Green's theorem?(23 votes)
- When you have a vector field line integral with no k component and notice that dQ/dx and dP/dy exist.

Those are partial derivatives, btw.(25 votes)

- Why is double integral of dA the area of the circle? It seems like the single integral of dA would be the area. Isn't a single integral usually area? Thanks in advance.(5 votes)
- It's a subtle point. You're right that ∫ f(x) dx will give you the area under a curve, but notice that ∬dA doesn't have a function (or, rather, it does but the function is the constant whose value everywhere is 1).(9 votes)

- If I wanted to do the hard way, what are the bounds of integration?(4 votes)
- -1 <x 1

0 <t<2pi

x =cost

y=sin t(6 votes)

- How did you know that the path (of the unit circle) was going clockwise?(2 votes)
- I think Sal just arbitrarily made that distinction.(3 votes)

- can u show 1 example to solve the above example using parameters?(2 votes)
- I'm not sure about my parametrized integrals but the way I set them up is like this:

our path is defined as x = cost, y = sint. So, dx = -sin(t) dt and dy = cos(t) dt.

Since we have 4 identical regions, in the first quadrant, x goes from 0 to 1 and y goes from 1 to 0 (clockwise). This is the same as t going from pi/2 to 0.

So the integral is 4 * ∫(t = pi/2 to t = 0) 2*sin(t)*(-sin(t))dt - 3*cos(t)*cos(t)dt.

or 4 * ∫(t = 0 to t= pi/2) 2 + (cos(t))^2 dt = 5*pi after integration.(3 votes)

- what would dr be in this case? and when would you not be able to use green's theorem or when would it be best not to use?(2 votes)
- dr = dxi + dyj + dzk where i,j,k are the unit vectors. Always. That way you can take the dot product of the F(x,y) which is also in unit vector notation. Why math teachers cant say just that, i don't know, but I'm pretty sure Sal did in one of the other videos.

Hope I answered that correctly.(3 votes)

- Shouldn't the answer be 10 Pi(3 votes)
- The answer comes out to be 10pi if you compute the double integral using parameterization with theta going from 0 -> 2pi and the radius going from 0 -> 1 of 5 (the integrand). To get the correct answer of 5pi, remember that when parameterizing with d(r)d(theta) you must include an r in the integrand, so redoing the double integral with the same parameterization but with 5r as the integrand yields the correct answer of 5pi.(1 vote)

- How to use Green's theorem for a triangular path? It's posing much difficulty. Can someone please be kind enough to explain this to me?(2 votes)
- How would you use Greens Theorom when the circle's center is at (5,-7) say?(1 vote)
- The center of the circle does not make a difference, looking at the way Sal did it, you just need to know the radius of the circle.(3 votes)

- What if we had a circle with the equation x^2+y^2=3, ie 3 is the radius of the circle.

Would we simply use 5*(r^2)*Pi as the answer or use derivation of the radius to proceed?(1 vote)- I do not quite follow the question, but you can express a circle in polar coordinates instead of cartesian coordinates.

This works by substituting:

x = r * cos(theta)

y = r * sin(theta)

which means that after squaring them in the original equation (the one you wrote) we end up with:

r^2 = 3

which means that the radius of that equation is acutally sqrt(3).

We can solve for the area by integrating the function (whether cartesian or polar) in the following way (with radius 3 this time):

*cartesian*

Int(1)dxdy with the following boundaries (solve for x in order to get boundaries for x, you could do the same with y if you change the integration order).

-sqrt(9-y^2)>=x>=sqrt(9-y^2)

and

-3>=y>=3

The solution of which ends up being 9*pi, which is correct according to the formula we all know for the area; pi*r^2. (see also http://www.wolframalpha.com/input/?i=int%28int1%2Cx%3D-sqrt%289-y^2%29..sqrt%289-y^2%29%29%2Cy%3D-3..3%29).

Now if we want to use polar coordinates it's quite a bit easier, because we know that a full circle is 2pi, and that the r=3.*polar*

boundaries:

0 >= theta >= 2pi

0 >= r >= 3

but because we use polar coordinates we can't use dxdy, we have to use r dr dtheta instead, meaning we get:

int(r)dr dtheta.

If we integrate this we end up with 9pi, just like before (http://www.wolframalpha.com/input/?i=int%28int%28r%2Cr%3D0..3%29%2Ctheta%3D0..2Pi%29).

We can also do this for any r, meaning we take the following boundaries for r:

0 >= r >= r

if we solve this we find the following formula: pi*r^2 which we know already to give us the area of a circle with radius r. (http://www.wolframalpha.com/input/?i=int%28int%28r%2Cr%3D0..r%29%2Ctheta%3D0..2Pi%29)(3 votes)

## Video transcript

Let's say I have a path in the
xy plane that's essentially the unit circle. So this is my y-axis, this is
my x-axis, and our path is going to be the unit circle. And we're going to traverse
it just like that. We're going to traverse
it clockwise. I think you get the idea. And so its equation
is the units circle. So the equation of this is x
squared plus y squared is equal to 1; has a radius
of 1 unit circle. And what we're concerned
with is the line integral over this curve c. It's a closed curve c. It's actually going in that
direction of 2y dx minus 3x dy. So, we are probably
tempted to use Green's theorem and why not? So let's try. So this is our path. So Green's theorem tells us
that the integral of some curve f dot dr over some path where f
is equal to-- let me write it a little nit neater. Where f of x,y is equal to P
of x, y i plus Q of x, y j. That this integral is equal to
the double integral over the region-- this would be the
region under question in this example. Over the region of the partial
of Q with respect to x minus the partial of
P with respect to y. All of that dA, the
differential of area. And of course, the region
is what I just showed you. Now, you may or may not
remember-- well, there's a slight, subtle thing in
this, which would give you the wrong answer. In the last video we said that
Green's theorem applies when we're going counterclockwise. Notice, even on this little
thing on the integral I made it go counterclockwise. In our example, the
curve goes clockwise. The region is to our right. Green's theorem-- this applies
when the region is to our left. So in this situation when the
region is to our right and we're going-- so this
is counterclockwise. So in our example, where we're
going clockwise, the region is to our right, Green's theorem
is going to be the negative of this. So in our example, we're going
to have the integral of c and we're going to go in the
clockwise direction. So maybe I'll draw it like that
of f dot dr. This is going to be equal to the double
integral over the region. You could just swap these two--
the partial of P with respect to y minus the partial of
Q with respect to x da. So let's do that. So this is going to be equal
to, in this example, the integral over the region--
let's just keep it abstract for now. We could start setting the
boundaries, but let's just keep the region abstract. And what is the partial of P
with respect-- let's remember, this right here is our-- I
think we could recognize right now that if we take f dot dr
we're going to get this. The dr contributes
those components. The f contributes
these two components. So this is P of x,y. And then this is Q of x,y. And we've seen it. I don't want to go into the
whole dot dr and take the dot product over and over again. I think you can see that
this is the dot product of two vectors. This is the x component
of f, y component of f. This is the x component of dr,
y component of dr. So let's take the partial of P
with respect to y. You take the derivative of this
with respect to y, you get 2. Derivative of 2y is just 2. So you get 2, and then,
minus the derivative of Q with respect to x. Derivative of this with
respect to x is minus 3. So we're going to get minus
3, and then all of that da. And this is equal to the
integral over the region. What's this, it's
2 minus minus 3? That's the same
thing as 2 plus 3. So it's the integral over
the region of 5 dA. 5 is just a constant, so we can
take it out of the integral. So this is going to turn out
to be quite a simple problem. So this is going to be equal to
5 times the double integral over the region R dA. Now what is this thing? What is this thing right here? It looks very abstract,
but we can solve this. This is just the
area of the region. That's what that double
integral represents. You just sum up all
the little dA's. That's a dA, that's a dA. You sum up the infinite
sums of those little dA's over the region. Well, what's the area
of this unit circle? Here we just break out a little
bit of ninth grade-- actually, even earlier than that--
pre-algebra or middle school geometry. Area is equal to pi r squared. What's our radius? So unit circle,
our radius is 1. Length is 1. So the area here is pi. So this thing over here,
that whole thing is just equal to pi. So the answer to our line
integral is just 5 pi, which is pretty straightforward. I mean, we could have taken the
trouble of setting up a double integral where we take the
antiderivative with respect to y first and write y is equal to
the negative square root of 1 minus x squared y is equal to
the positive square root. x goes from minus 1 to 1. But that would have been
super hairy and a huge pain. And we just have to realize,
no, this is just the area. And the other interesting thing
is I challenge you to solve the same integral without
using Green's theorem. You know, after generating a
parameterization for this curve, going in that direction,
taking the derivatives of x of t and y of t. Multiplying by the appropriate
thing and then taking the antiderivative-- way hairier
than what we just did using Green's theorem to get 5 pi. And remember, the reason why
it wasn't minus 5 pi here is because we're going in
a clockwise direction. If we were going in a
counterclockwise direction we could have applied the straight
up Green's theorem, and we would have gotten minus 5 pi. Anyway, hopefully you
found that useful.