Green's Theorem: area under an arch | MIT 18.02SC Multivariable Calculus, Fall 2010

[@henriquenunes7196]

This is a great video, very demanding when one is starting to trying to understand GT, but the fact is it makes you think a lot, and hopefully one will profit significantly from this video. Thank you Prof. Breiner.

[@justpaulo]

4:58 I believe the integral of x.dy is also OK because in the curve C1 dy = 0. So you'll be integrating x.dy only in C2.

[@anuragbose4283]

Thank you so much for the nice explanation.

[@isaacmandell-seaver7223]

4:58 couldn't you still ignore c1 using xdy by seeing that F would be perpendicular to the x-axis (perpendicular to the unit tangent vector)? Or am I seeing this wrong? This gave the correct answer but that may have been by chance.

[@vh73sy]

There's no F here. You can discard C1 upon taking the closed integral of x dy, because this time dy is zero due to y being zero all the way, though x is changing from 0 to 2 a pi. There is no escape from a difficult integration, if you take x dy, then you will face ( t sint - (sint)² ) for integration over C2 path.

[@wasa1997]

Is the basic (unsaid) assumption here that the vector field is just equal to ?

[@KeanuHolz-kb9gy]

By Green's thm you have that the "double integral (over R) of [partial(N)/partial(x) - partial(M)/partial(y)]dA" = "closed line integral (along C) of [Mdx + Ndy]". Since we want to find the Area, we should start with an integral of the form "double integral of [1] dA". => "[partial(N)/partial(x) - partial(M)/partial(y)] = [1]" so we're basically free to choose the M,N components, as long as curl(F) will be 1

[@setarehsiralek6924]

:( still i cant get the principles

[@ziye4710]

This is different from our teacher's equation which is area = 0.5*integral(xdy-ydx)

[@bigmantm2532]

its the same

[@mooseonshrooms]

@@bigmantm2532 Definitely. as the -y part= the x part then 1/2 of both summed is equal to one on their own. like if 1=1 then 1/2 (1+1)=1=1.

[@JoyErnst]

Por atrás.

[@usnxx3755]

Can somebody tell me why dA = -ydx = xdy at 4:23?

[@henriquenunes7196]

To find the area you need the Integral to be S 1dA . So -ydx and xdy are the components of the vector field , that will make you have an integrand function of 1 (so S 1dA gives you the area). But obviously here you will use the other side of the GT equation , which is what Prof. Breiner did. Hope this helps

[@soccertort24]

I was also confused by this, but this is discussed in another video here: kzbin.info/www/bejne/qaa5lIeueZ6fj9k