Green's Theorem Proof (part 2)
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This proof does not cover the general case.
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For any grand conclusion... it's gotta be magenta. Thanks Sal!
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awesome
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he must be getting quite a few more hits, considering finals are here
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@Ferrus91 why has it not been put up here yet?
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It is indeed confusing, though he is not wrong. If you look at the start of the video he took the left part of the curve and called it X=X2(y) and the right part he called X=X1(y). It is here where you're confusion is. If you want to define the region R, you take y to be going from a to b and then automatically x will go from X2 to X1 (look at the picture!). So R will in the end be defined as the double integral with Y going from a to b, and X going from X2 to X1.
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Hmm I dont get it again. The last video you took the boundary from y1 to y2 and made the entire integral negative. Here, you left the boundary as from x2 to x1 making the integral positive. But you said that they both denote the are R, which kind of makes the - or + a bit arbitrary. Yet the ultimate formula requires the i component of the partial derivative to be negative. Can anyone in the know explain this to me please?
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deja vu for the first part
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Can't wait to see the proof of Stoke's theorem
Ferrus91 10 months ago 3
Salman Khan -- Thank you so much.
I remember using your Algebra videos in preparation for my GED (I have never been to High School; by the way, I passed), then using your videos in preparation for my ACT (got a high enough score for a free 2 year tuition at the college I'm going to now). Then, I used your precalculus videos to prepare for Calculus I. Now, I'm using your Calculus videos to get ahead of everyone in my Calculus II class. :)
Thanks from,
A 17 year old with a 3.8 GPA in college.
MyOverflow 1 year ago