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# Green's theorem proof (part 1)

Part 1 of the proof of Green's Theorem. Created by Sal Khan.

## Want to join the conversation?

• why does sal use the integral with the circle in it sometimes, and the normal one with a sub c other times? •  when it has th circle it means we want to integrate around a CLOSED loop :)
other than that , if there is no circle there will be another path to be followed which is not a closed path .
• î·î=1 ?
Is this expression true?

That is, when one has the x-unit-vector and the dot-product with another x-unit-vector (as at ), does that always equal one? Because it always seems to disappear when this happens. • So at around Sal decides to swap the inside of the integral by multiplying with negative one. This isn't really an obvious thing to do but it is essential to get the right result for Green's Theorem. Does anybody know why he took that step? • I think you need to do this because of the direction of the curve. The general form given in both these proof videos, that Green's theorem is dQ/dX- dP/dY assumes that your are moving in a counter-clockwise direction. If you were to reverse the direction and go clockwise, you would switch the formula so that it would be dP/dY- dQ/dX.

It might help to think about it like this, let's say you are looking at the unit circle, but only look at the part in the first quadrant. If you move counter-clockwise, you are going from larger to smaller x values, so when you integrate, it is going to be the smaller x minus the larger x (it's the inverse of how we generally view integrals, going from smaller to larger values). The y values on the other hand, are going from smaller values to larger. If you integrate all the same way, you have to subtract how you integrated along the x axis because it is going in a decreasing direction.

Of course, you could argue that this argument really only holds if you look at the first quadrant, but nevertheless I think it is a somewhat useful way to visualize it.

In short, he did this because of the direction of the curve.
• At around , you start talking about how the integral we've taken is the volume under the surface. But that's only true if P is a scalar being treated as a "z" component. Didn't we say earlier that P is a vector with a component only in the X direction? How is it going up? • he did say that p (lowercase) was a vector field with a component only in the x direction. However, he also said that the x component of this vector field p (lowercase) was a scalar function P(x,y) (capital). When he turned it into a double integral, the partial with respect to y of P (capital) became the inside piece, and one way to visualize double integrals is volume. p is a vector field. P is a scalar field. Neither of them really go up, but that is a way to visualize integrals of scalar functions, whether they go up or not. You can think of a scalar field in 2 dimensions as a surface in 3 dimensions with the value of the scalar being the z-value.
• But why does it end up being a volume? The integral in the beginning is supposed to be just an area right? I get the math and intuition behind it, but isn't something wrong considering we started out with something in 2D and ended up with something in 3D? • At he says that he considers the minimum and maximum values of x and states that he can split this into two function of y on x. Why is this legal to do? The way he draws it I think it's legal on this example, but in general those half loops may not be functions of x at all (such as when the curve takes multiple y values for a given x value). • Why is partial of P with respect to y the height of curve?? • Partial P with respect to y is the function inside the integral. When you have an integral of some function f(x), the f(x) is usually represented graphically as the height above the x-axis. In this case dP/dy is f(y) and is the "height" above the y-axis. Later on, you will have f(x,y) being the "height" on the z-axis above the x,y-plane. Etc. this continues into any number of dimensions. (e.g. f(x,y,z) will be the height in some 4th-dimension above the x,y,z-"hyperplane")
• I'm currently working my way through Khan academy in the hopes of pursuing a Physics degree soon. Currently on the Algebra 2 course and I just skipped ahead to see that knowledge jump I would need to obtain. I understand almost none of this but I can't wait until all the hard work pays off and I can!   