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## Multivariable calculus

### Course: Multivariable calculus>Unit 5

Lesson 2: Green's theorem

# Green's theorem proof (part 2)

Part 2 of the proof of Green's Theorem. Created by Sal Khan.

## Want to join the conversation?

• Loved this proof but this negative sign is bugging me, in the previous video, he used integral over y2(x) to y1(x), hence the negative sign but in this one he does x1(y) to x2(y), that gives us no negative signs, hence we get *(dq/dx-dp/dy)dA*. I ask why? why evaluate the second case from 1 to 2? Is it a preconceived notation to make this case simpler? An elaborate answer would be much appreciated.
• The main idea is that for the double integral, he want to integrate from a LOWER x-boundary to an GREATER x-boundary, and in the second integral, from a LOWER y-boundary, to a GREATER y-boundary. In other words, integrate from left to right (with respect to x), and from bottom to top (with respect to y). (You could also have gone from right to left AND top to bottom to get the same answer, but Sal didn't do that in the first video.)

The names of the curves (y1, y2, x1, x2) don't matter too much, so long you integrate in the right direction. If you look at part 1, you'll see Sal also integrates from left to right, and from bottom to top.
• Hi I have a confusion about this video. He proved the Green's Theorem by assuming that F is conservative. Because of that, wouldn't Green's Theorem only true if and only if F is conservative? So Green's Theorem cannot be used as a general formula then. And since it's conservative, wouldn't the answer just going to be 0 anyway?

Thank you
• No, it's built under the assumption that the vector field F can have a partial derivative with respect to x and y. If f was conservative, the line integral would be zero, because of the gradient theorem.
• Why does it matter that the path goes counterclockwise? I don't see how that affects turning the line integral into a double integral.
I know that changing the direction changes the sign of the line integral, but is it not kind of arbitrary which direction you denote as positive in the first place?
• If you do a line integral for a circle and use the parameterization x=cos(t) and y=sin(t), the counterclockwise direction yields the positive result and vice versa for the clockwise direction. Perhaps using this parameterization is due to convention, but I don't know of an alternative. Since Green's Theorem is built off Line Integrals of circles (which are the most basic closed loops), the fact that counterclockwise is positive remains true. If you work through a few problems and try switching the direction of integration you'll find the sign does indeed switch.
• At the end of the video, Sal says:
If f is conservative then the integrand is equal to zero. (dQ/dx=dP/dy)
Is it an if and only if? Because if it is, this would be another method to show that f is conservative, right? Would it also be a faster method than the previous one (where you try to equal f to the gradient of a scalar field?
• Thats false... Take this counterexample:
F(x,y)=(P(x,y), Q(x,y)) with (x,y) in R2 /(0,0)
with P(x,y) = -y / (x^2+y^2) and Q(x,y) = -y / (x^2+y^2)
This F is not conservative, you can prove it by taking a circle with radius r, if you take the circular integral you'll get 2pi instead of 0.
Also F is not conservative and dQ/dx = dP/dy.
So in conclution it is not an "if and only if" just an "if".
• When Sal worked on the part 1, he found the double integral of dP/dy negative, because he had reversed the second integral (from y2-y1 to y1-y2). But in this video, he finds dQ/dx positive, KEEPING the second integral as it is (from x2 to x1). My question is, why can we reverse the second integral in the first video (thus getting a negative result) and not in the second video too? It seemed a bit arbitrary to me and I am confused.
• because in the first we had (y2>y1) now we have (x1>x2) so that we will not do the reverse!
• Isn't the argument of that double integral the 2D curl of the vector field?
• Why is the closed line integral in the i dimension negative, while the one in the j dimension is positive?
• is because of the inverse use of the fundamental theorem of calculus for properties of integrals that is generally has the limits of integration from left to right; and from the bottom to the top.
• im kind of confused did you say that the line integral is equal to the area of the bounded region R
• The result from the line integral is the volume encompassed by the surfaced form by dQ/dx and the region formed by the paths C1 and C2