Green's and Stokes' Theorems are actually the same thing (Stokes' is more general). These sections will be easier to understand if you understand dot products, curl, and circulation.
Green's theorem states that the amount of circulation around a boundary is equal to the total amount of circulation of all the area inside. Remember that curl is circulation per unit area, so our theorem becomes:
The total amount of circulation around a boundary = curl * area.
Another way to look at the theorem is to think of each point inside the area. Each point is twisting a certain amount. If you add up the total of all these twists, then you have the amount that the outside is twisting (which is the net twist of the entire area).
The reason this works is as follows. Although each point is turning, some of its motion is canceled out by other points twisting nearby in the opposite direction. This cancels all the twisting except for that on the boundary. Thus, the total amount of twisting in the inside is equal to the amount of twisting on the boundary.
Stokes' theorem is a more general form of Green's theorem. Stokes' theorem states that the total amount of twisting along a surface is equal to the amount of twisting on its boundary. Suppose we have a hemisphere and say that it is bounded by its lower circle. (picture)
Now, suppose we have a cone, which is bounded by the same circle. Stokes' theorem states that both of these regions would have the same circulation, because they have the same boundary. This works for the same reason as Green's theorem. All the twisting of the peices gets cancelled except for the twisting along the boundary. Because the two regions have the same boundary, they must have the same circulation.
Green's theorem is simply Stokes' theorem in the plane. Green's theorem deals with 2-dimensional regions, and Stokes' theorem deals with 3-dimensional regions. Thus, Stokes' is more general, but it is easier to learn Green's theorem first, then expand it into Stokes'.https://betterexplained.com/~kazad/resources/math/Stokes_and_Green/green.htm
$$ \int_D d\omega = \int_{\partial D} \omega $$