If someone doesn't understand why the hell are the circle chips ("rectangles") are treated like straight ones , it is because the length in the middle of the rec (where dony picked the point*) is equal to the length of the rectangle if it was bent straight , easy to prove with calculus and geometry , so you can treat it's area as a normal rectangle and multiply by height for the summation.
(I'm on 4:42 so I might be wrong in here if dony outright subtracted two circle sections)
@NMLP1 Well, I didn't watch it, I saw the introduction and moved on.
jiramate999 1 week ago
2x speed!!!!!!!!!!!!!!
browno26 1 month ago in playlist Multiple Integrals
captions are hilarious
ebemc 4 months ago
@NMLP1 Well, seriously. This is pretty easy stuff. Its almost hilarious.
jiramate99 1 year ago
@jiramate99 then why u watching video
Garykelleher123 1 week ago
def helped me out. thanks man.
NMLP1 1 year ago
This has been flagged as spam show
ut'll be great if your accent can improve!
jiramate99 1 year ago
it'll be great if your accent can improve!
jiramate99 1 year ago
@jiramate99 dont be a noob.
NMLP1 1 year ago
thanks a lot donny you're a life saver!
sodown 2 years ago 2
An easier way of considering the geometry of "r.dr.d(theta)" is as follows.
By definition, arc length = r.(theta) .
Hence, d(arc) = r.d(theta) .
The area of the small rectangle equals to the arc length times the small change in r. In the limit, this becomes accurate.
Hence, dA = r.d(theta).dr .
bigstas503 3 years ago 5
Another method would be to use the Jacobian, I believe students have already learned partial derivation if they are solving double integrals.
x = r cos(theta) , y = r sin(theta)
J = |del(x)/del(r) del(x)/del(theta)| |del(y)/del(r) del(y)/del(theta)|
This is the determinant of the partial derivatives.
|cos(theta) -r sin(theta)|
|sin(theta) r cos(theta) |
r cos^2(theta) + r sin^2(theta) = r = Jacobian.
So dA becomes r dr d(theta)
illusion422 3 years ago
I think there is a theorem of advanced calculus about the Jacobian in coordinants transforms.
Double integral of f(x,ydx.dy)in R=
Double integral of f(g(u,v),h(u,v)*Jacobian[(x,y)/(u,v)]du.dv in G
2112dim 2 years ago
oh sh* he just did just that , maybe I'm missing something and it's for the proof....
LongShlong125 3 years ago
If someone doesn't understand why the hell are the circle chips ("rectangles") are treated like straight ones , it is because the length in the middle of the rec (where dony picked the point*) is equal to the length of the rectangle if it was bent straight , easy to prove with calculus and geometry , so you can treat it's area as a normal rectangle and multiply by height for the summation.
(I'm on 4:42 so I might be wrong in here if dony outright subtracted two circle sections)
LongShlong125 3 years ago