I am using Paul's Online Notes to practice Stokes theorem. Here is the link to the exact problem which is giving me trouble: https://tutorial.math.lamar.edu/Solutions/CalcIII/StokesTheorem/Prob4.aspx.
C is given in the problem as having "a counter clockwise rotation if you are above the triangle and looking down towards the xy-plane."
Here is a graph of C: graph of C
Projecting C onto the xy-plane gives this: xy-projection
The arrows indicating orientation aren't in the projection; if they were, they would point from (4,0) up to (0,2) (or from the x-intercept up to the y-intercept).
I understand all aspects of the problem except for the bounds on the final double-integral. My question is why the final double-integrals x bounds are from 0->4 and not from 4->0?
The line is oriented from right to left, so it does not make sense to integrate from left to right. Does projecting down onto the xy-plane change or negate the orientation somehow?
This is my first question on here: I appreciate feedback on how to improve this question or math questions in general.
I understand now! When computing this surface integral, the bounds of integration on the surface integral describe some 2D region, D.
That is, for some surface S and some 2D region D, the double integral over S DS is computed by converting it into the double integral over D DA (dxdy or dydx). This has nothing to do with stokes theorem! You can see this on step six and seven of the solution to the problem here: https://tutorial.math.lamar.edu/Solutions/CalcIII/StokesTheorem/Prob4.aspx.