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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.

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• How do we know when to use Green's theorem?
• 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.
• 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.
• 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).
• If I wanted to do the hard way, what are the bounds of integration?
• -1 <x 1
0 <t<2pi
x =cost
y=sin t
• How did you know that the path (of the unit circle) was going clockwise?
• I think Sal just arbitrarily made that distinction.
• can u show 1 example to solve the above example using parameters?
• 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.
• 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?
• 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.

• Shouldn't the answer be 10 Pi
• 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?
• 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.
• 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)