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

Part 2 of the proof of Green's Theorem

Missed the previous lesson?

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!

[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.

[MisterTutor]

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.

[SerdceDikarya]

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

[malem_]

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. 🙏

[GoogleUser-eero]

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

[panotsampas]

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

[abdelrahmangamalmahdy]

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

[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.

[spiderkent]

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

[nathanbenson]

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

[austinkubiniec]

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

[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?

[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!

[Ferrus]

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

[hikmatullahpakhtoon]

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).

[tantalides]

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

[aparnasadhukhan]

Writing Green's Theorem in Green!!!

[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)