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### Course: Multivariable calculus>Unit 5

Lesson 9: Proof of Stokes' theorem

# Stokes' theorem proof part 6

More manipulating the integrals... Created by Sal Khan.

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• I had an hard time understanding the anology between N and M and (p+RZx) and (Q+RZy) . I don't understand how we can replace those terms by M and N?
• M and N are functions of x and y -- (p+RZx) and (Q+RZy) are also functions of x and y. This is only possible because Z can be written as a function of x and y, which is exactly the assumption that Sal made at the very beginning of this video. When sal found the original line integral involving M and N, he did so with M and N being arbitrary -- meaning that we can actually pretend that M and N are (p+RZx) and (Q+RZy) respectively. Now, here's the key -- Were you to plug in (p+RZx) and (Q+RZy) as M and N into the line integral involving C_1, you would get an identical value to the line integral involving C. This is because each line integral actually traverses a different field. the C_1 integral traverses through the field defined by (p+RZx), and (Q+RZy), whereas the C integral traversed the field defined by P, Q, R. Sal managed to simplify the P, Q, R traversal into a scalar field defined in terms of (p+RZx) and (Q+RZy), and, this scalar field being a function only of x and y, is identical for all paths that look identical along the x and y axes -- C_1 and C are identical along said axes, they differ only in their Z components. This means that you could change the C in the original (p+RZx) and (Q+RZy) line integral to a C_1, and get the exact same result. What Sal noticed is that pretending (p+RZx) = M and (Q+RZy) = N, then substituting those into the C_1 integral is exactly the same as swapping the C with a C_1, since (p+RZx) and (Q+RZy) appear everywhere you would expect to see an M and an N. Hence why he is able to pretend they are the same, and interchange them freely! While it appears he is replacing M's and N's everywhere, what he's really doing is swapping a C for a C_1, but going through the algebra required to prove that that is possible, as opposed to making an intuitive argument.
• At ish, I am a little confused about how having the result of F dot dr in the same form as the line integral over c1 (i.e. some M times dx + some N times dy) leads to said product of F dot dr necessarily being equal to the line integral over c1 specifically. I see how it represents the same form as the general line integral, but I am not sure why that necessarily makes it equal to the line integral specifically over c1. I hope that question makes sense; any clarification would be appreciated.