Green's theorem proof (part 2) | Multivariable Calculus | Khan Academy

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Part 2 of the proof of Green's Theorem
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Multivariable Calculus on Khan Academy: Think calculus. Then think algebra II and working with two variables in a single equation. Now generalize and combine these two mathematical concepts, and you begin to see some of what Multivariable calculus entails, only now include multi dimensional thinking. Typical concepts or operations may include: limits and continuity, partial differentiation, multiple integration, scalar functions, and fundamental theorem of calculus in multiple dimensions.
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[@Valentinort]

This was beautiful. I just wanted to say thank you Khan for never being condescending towards your students, and for all the math you teach to the world!

[@sandrogiongo]

I was gonna say the same thing! This is beautiful indeed!

[@TanmayMukim_dhab]

17.25"May be i should write it in green color" lol :)

[@APaleDot]

To anyone confused as to why he flipped the previous integral to be negative but not the one in this video, it's not arbitrary, here's an explanation: The negative sign comes directly from the counter-clockwise direction of the curve. If the curve were travelling in the opposite direction, the other function would have picked up the negative sign instead. To see this, notice that he always makes the first half of the curve go from a to b, and the second half goes from b to a. This is consistent in both videos. The difference is because of the direction of the path: In the first video the first half of the curve ends up on the bottom (closer to the x-axis) whereas in this video the first half of the curve ends up on the top (further from the y-axis). So, when converting to a double integral, he needs the curve on the bottom to have a negative sign because an integral is (upper bound - lower bound). In this video, the lower bound is already negative because it's the second half of the curve which goes from b to a. But in the previous video the lower bound was positive because it was the first half of the curve, so he had to flip it. Hopefully it's clear why the path must go away from the y-axis when starting at it's lowest point due to the counter-clockwise direction of the path. This is directly related to the Curl and all kinds of rotations actually. Ever notice that if you rotate the x and y basis vectors by 90 degrees, the x points in the same direction as the y basis was pointing, but the y basis now points in the negative x direction? This is the same effect we are seeing here.

[@pietergeerkens6324]

Khan also covers this at start of the next video.

[@fredesch3158]

Your lower bound, upper bound did it for me, I knew it had to do with the orientation of the curve, but I didn't know how to interpret it, thank you so much.

[@MisterTutor2010]

I had read that conservative vector field have a curl of zero. I was wondering why that was true until I saw the end of this video. The Curl P(x,y)i+Q(x,y)j works out to (dQ/dx-dP/dy)k This video showed that for a conservative vector field dQ/dx=dP/dy so dQ/dx-dP/dy=0 and therefore the Curl of P(x,y)i+Q(x,y)j if conservative must equal zero. These video are amazing informative.

[@SerdceDikarya199]

and even after 10 years, u save end-semester exams!!

[@gonzo_o_o]

IT MAKES SENSE!!!

[@GoogleUser-ee8ro]

it's a very clear and beautiful proof of Green's theorem

[@malem_1346]

Truly my words won’t be able to tell how beautiful this series of videos is.! I’m really really thankful to KHAN ACADEMY for making me understand Green’s theorem. 🙏

[@austinkubiniec9868]

You integrated from y1(x) to y2(x) last time, which gave you a minus sign before the P. This time you integrated from x2(y) to x1(y) which gave you no minus sign. This is why the final result has a minus sign, instead of a sum, but I see absolutely no reason why you switched the orders. Many comments bringing this up, none have satisfying answers :P

[@divykala169]

I was wondering about the same thing. Found any reason why?

[@Jonrulesistaken]

The confusion is that he arbitrarily changed the direction of the x field integral without explaining why. If you express the region in terms of dx and dy in the standard direction, you have two options: dx:[x_a, x_b], dy:[y_1(x), y_2(x)] or dy:[y_a, y_b], dx:[x_2(y), x_1(y)] The x field integral is backwards, but the y field integral is already in the standard direction.

[@divykala169]

I think this makes some sense

[@aparnasadhukhan5567]

Writing Green's Theorem in Green!!!

[@Liaomiao]

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?

[@abdelrahmangamalmahdy]

I just wanted to say THANKS! I finally understood this..

[@nathanbenson0288]

You explained Green's theorem very well. Thank you very much for this video.

[@spiderkent]

The detailed explanation without skipping any little steps is very helpful in understandind.

[@Ferrus91]

Would've been nice to have mentioned the intuition linked to the curl.

[@panotsampas5080]

Im confused! I think you made an error. How can you place both partial function under the same integral when the bounds you found for Q(x,y) are from x2(y) to x1(y) and NOT x1(y) to x2(y) 8:20

[@chessaddict5942]

if you look at the diagram you can actually see that the x2(y) is bottom line since you are looking with respective to y. And you can see that the boundaries a and b are closer to C2 then they are to C1, so C2 isbasically lower boundary of integral.

[@chessaddict5942]

if you look at the diagram you can actually see that the x2(y) is bottom line since you are looking with respective to y. And you can see that the boundaries a and b are closer to C2 then they are to C1, so C2 isbasically lower boundary of integral.

[@tantalides]

I am "Stoke"-d to see that proof :)

[@taurus50592]

deja vu for the first part

[@kunai3269]

the best teacher alive

[@michaelempeigne3519]

you are a great teacher.

[@Freddbigghead]

Does the theorem only apply to curves that can be split into two funktions per coordinate (like x_1 and x_2) or can you generalize the proof for more complex curves by splitting it into more functions? (Maybe I missed some assumptions)

[@GhostyOcean]

I presume that it can be between an arbitrary amount of functions as long that they form a closed loop. I don't know for sure since I haven't gotten to this part of my math journey yet, so I'm unfamiliar with this. I assume it can because the top and bottom functions are arbitrary so long as they form a closed loop (without self intersection?) and could be sets of piece wise functions.

[@grlCycling101]

fuckin boss im drunk but still entertained

[@TheTeladras]

Yep Grate job! It was a bit hard finding the video, for Greens equation proof, but was helpful, in fact i got a grate grade and the most important i understood it! Some people can explain things just they way others can understand it, from the basics!

[@BeGhappy]

@Liaomiao 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.

[@batrarohit1]

Mr. Khan, that was an amazing video, really beautiful. Thank you so much.

[@Ferrus91]

Can't wait to see the proof of Stoke's theorem

[@speedbird7587]

nice and neat explanation, thanks

[@ugursoydan8187]

THANK YOU VERY MUCH!! IT WAS AN EXCELLENT PROOF

[@hikmatullahpakhtoon3694]

0:49 if you consider the field along j then you shouldn't write q(x,y) instead you should write it as q(y).

[@samin3997]

10 years ago, I feared math. Now look whrere it got me.

[@GHOSTrex1324]

I learner this condition for jee. Now I understand it.

[@walterfrederick]

Please mister, I want to remember the demonstration of Stokes' theorem, because I forgot it, i.e. how to proof it and I know the Green's theorem is only a corollary from it!

[@mariomuysensual]

beautiful proof

[@Prometeur]

Absolutely beautiful. Thank you so much

[@lindsaysunhendra402]

Wow... this is so amazing.. You explain it very clearly. Thank you very much :)

[@deeparnoldgomes8238]

thanks i needed this!

[@abdelstar3999]

perfect,thank you

[@shibsankarbera9218]

u r my inspiration

[@kpranav513]

Can the limits be same in both the parts of the proof? x=a and then x=b in part1 then, in this part y=a and y=b

[@mayankmahajan4617]

actually your proof is incomplete since it need not be a function. I mean is you writing x=f(y) is wrong since there might exist a curve where for one particular y there might be 2 x.

[@laputahayom]

he has videos on that too.

[@aimuhire]

blackboard :) 12:47

[@johnr1875]

why is the integrand at the very end of the video in the double integral equal to zero? i thought the double integral itself is equal to zero.

[@julianlastname5730]

The integral itself being zero is only guaranteed if the integrand is equal to zero, or if the area in c is equal to zero. Since the latter is false because of the givens at the beginning, the former must be true

[@swarnalihait7201]

It was the most colorful video ever..:D But really really nice usage of colors..:) I like it..:)

[@yli5531]

Maybe you should mention that the left hand side of Green's theorem is the double integration of the 2d curl?

[@dannyboy12357]

thanks bud!

[@tag_of_frank]

its not clear why you made the integral negative in part 1 to get P(x,y2(x))-P(x,y1(x)) but here you just left it positive... why why why

[@whitneycoyle6785]

Check out which integral he switched the limits on for P vs. Q (b/t video 1 and 2). In Video 1 (with P), it was the second. In video 2 (with Q w.r.t y) it is the first.... Perhaps that helps?

[@LaureanoLuna]

8:17 Taking the definite integral not from a particular value of the variable x to another but from one function x_2(y) to another x_1(y), where y has no determinate value makes little sense.

[@TheBoredNerd]

he must be getting quite a few more hits, considering finals are here

[@Ferrus91]

Indeed, good work. :)

[@nikhilsingh6946]

Nice

[@luvley5323]

can anyone explain why the result of the previous video is negative like why didnt he just keep it positive like he did for the result of this video

[@APaleDot]

Sure. The negative sign comes directly from the counter-clockwise direction of the curve. If the curve were travelling in the opposite direction, the other function would have picked up the negative sign instead. To see this, notice that he always makes the first half of the curve go from a to b, and the second half goes from b to a. This is consistent in both videos. The difference is because of the direction of the path: In the first video the first half of the curve ends up on the bottom (closer to the x-axis) whereas in this video the first half of the curve ends up on the top (further from the y-axis). But when converting to a double integral, he needs the curve on the bottom to have a negative sign because an integral is (upper bound - lower bound). In this video, the lower bound is already negative because it's the second half of the curve which goes from b to a. But in the previous video the lower bound was positive because it was the first half of the curve, so he had to flip it. Hopefully it's clear why the path must go away from the y-axis when starting at it's lowest point due to the counter-clockwise direction of the path. This is directly related to the Curl and all kinds of rotations actually. Ever notice that if you rotate the x and y basis vectors by 90 degrees, the x points in the same direction as the y basis was pointing, but the y basis now points in the negative x direction? This is the same effect we are seeing here.

[@africanprincess7896]

Does it mean the volume for a conservative vector field is zero?

[@Nikifuj908]

MAGENTA

[@saketjoshi3125]

Do not understand the negative sign. Everything is symmetric between x and y. why should the coefficient for partial derivative wrt y and x have opposite signs?

[@effortless35]

Saket Joshi The orientation of the curve breaks the symmetry between x and y. This particular version is true for positively oriented (counterclockwise) curves. If it was a negatively oriented curve it would be the exact opposite: the partial derivative wrt x would have the negative sign and the partial wrt y the positive. You can go through the mechanics of the derivation in the two videos side by side and you'll see how and why the signs of the corresponding terms are the exact opposite.

[@austinkubiniec9868]

Going through side by side makes the discrepancy even more obvious. For seemingly no reason, he switches from doing y1(x) to y2(x) in the first video to x2(y) to x1(y) in the second. No explanation.

[@MJforever127i]

why did i do this degree!! lol but you made it very simple. Thanks :D

[@HAWXLEADER]

but you have no whiteboard.... you have an old classic blackboard!(well not so classic)

[@kdmdlo]

What the heck software are you using to write your notes? And on what platform?

[@martinelenkov2113]

satz von gauss. anyone?

[@lerneninverschiedenenforme7513]

aaaaaaaaaaaaaaaaawwwwwweeeeeeeeeessoooooooooooomeeeeeeeeeeeee!! Thank you!!!

[@user-qm6up7kz4n]

a and b from 1st video are not same as a and b from second. should've called them differently

[@KRISHNU10]

You play with colours a lot.

[@dannyboy12357]

sal loves magenta loooooool

[@a86ziz]

getting sick of these theorem! who will relate it to practical application?