Green's Theorem in normal form | MIT 18.02SC Multivariable Calculus, Fall 2010

[@cookiethief]

Please see Erik's comment below. Thanks a lot for your hard work! Previous misconception by me: "When he wrote down ∫∫ div(F)dA, I believe he may have meant ∫∫ curl(F)dA. Since the former is Gauss's divergence theorem. However, since the vector field is symmetrical, the answer happened to be right."

[@raymondblake5765]

I thought the same thing.

[@raymondblake5765]

How do you type those integral squiggly lines?

[@erikumble]

Did you watch the lecture corresponding with this? kzbin.info/www/bejne/lXTHoIWfg7ilp6s It is correct to use divergence of F, rather than curl(F) since this is Green's theorem in normal form (applied to flux), rather than the tangential form (applied to work). Since the normal differential is given by [dy, -dx] then Green's theorem uses the curl of that, we can skip the step and just use divergence.

[@cookiethief]

​@@erikumble You're right! It's been almost a decade since I did this stuff, but thanks for clearing this up for the future viewers.

[@roger72715]

The vector field is conservative hence the line integral for work done should be zero. Greens theorem is a vector identify which is equivalent to curl theorem in a plane

[@DoctorFastest]

@faithx92 ∫∫ div(F)dA is correct. Green's theorem is the analog of the divergence theorem for two dimensional vector fields. Perhaps you are confused because Green's theorem can also be generalized to Stokes theorem. Namely, if you go from the plane the three dimensional space. In that case it is the *other* side of the equation you reference that will be replaced by the curl.

[@roger72715]

Greens theorem contains Curl of F

[@vocdex]

@@roger72715 Yes but it can also be expressed as div of F when for the line integral P dy − Qdx is considered instead of the integral P dx + Qdy.

[@herrGaman]

These clips are great help, thanks!

[@raynoldcsya8317]

for the semi circle part of the line integral, isnt it true that that F and n vectors are parallel? It would then become the integral of magnitude of F over the path which is coincidentally the length of the semi circle arc. Beats having to go through the hassle of parameterization

[@erikumble]

yes

[@frade001]

[5:15] Did he just fart?

[@alitaskran9177]

i think he is wrong, since on the semicircle, vector field is perpendicular to path and on the bottom line from -1 to 0 path is in the opposite direction with the field which gives - signed value (say -a), from 0 to 1 the line integral gives a. finally the result must be 0 which means no work is done. Am I wrong?

[@arunsinghmehra7728]

yeh u r right

[@pipertripp]

that's exactly what I got. The curl of that function is 0, so the result that Green's Theorem should give you is 0. But if this video is wrong, why on earth would they leave it up.

[@brodyattwater6190]

It's the flux across the curve. Along the the path [-1,1] (He calls it C1), the vector field is perpendicular to the path and therefore the flux is 0. Along C2, the flux is radially outwards with magnitude 1. Therefore the flux is 1 multiplied by arclength. Equal to 1*pi = pi. There's no error in this video.

[@roger72715]

@@brodyattwater6190 the vector field is a radial one..so along the path C1 it's not perpendicular but parallel

[@y_l__c_h7540]

Nothing else. I love u !!!!!

[@vennilasubash3283]

Thanks a bunch!

[@Thegema101]

where did they get this guy ? lol

[@AlphaBeta-xt3wn]

From your mother's bed you ungrateful fuck.

[@imjrich19]

@frade001 thats what i thought. lol

[@BkSenko24]

lol

[@arunsinghmehra7728]

This guy knows nothing Lmao from where did they get this guy....

[@loganwishartcraig]

He's published a lot of papers for a guy who "knows nothing" arxiv.org/a/lewis_j_1.html

[@RM-gc8lx]

This was a perfectly executed video. The topic was clearly explained in a highly pedagogical 11 minutes. The presenter was very likely a doctoral student, also doing research, and likely quite busy with other matters. You sir, are an abject n00b who does not recognize excellence when seen, and has likely barely understood this material, if at all, and promptly now forgotten it all as you dwell in other, simpler, subject matters that are more suiting to your tastes because they threaten your inattentive and lukewarm conscious ego less. 6 years since you posted this, which is roughly today, you've likely hit wall upon wall in your studies, and opted for some easy way out and convinced yourself that this was better for you, as you chugged down yet another alcoholic beverage, despite your expanding midsection, and nestled comfortably into your increasingly mediocre life. But, oh yes -- do remind us again why you thought this freely posted video of excellent quality from one of the finest learning institutions the world has ever seen regarding one of the greatest gems of intellectual achievement humanity has ever devised from an individual who delivered a superlative presentation as a complete side-project was insufficient for your refined taste. Or just realize that this shitposting of yours, 6 years ago, was just another data point in that straight line of yours from making fun of others' hard work into abject mediocrity, thankless grind, and the inevitable mid-life breakdown where you find some external motive to convince yourself to keep going in a plot reminiscent of the typical B-movie that keeps playing in your head and reminds you that while you're not great, you're at least 'good enough'. P. S. You might have a chance at redemption yet. Don't give up.