The integral formulas for the centroid of a region (center of mass)

[@ranjithkumararunachalam3844]

Calculus Teacher ~ transform ~ physics teacher.

[@qav_cnzo_]

first time seeing him teaching phisics😅

[@ridesafealways4929]

​@@qav_cnzo_Because he is a mathematician. He focuses more on much much harder maths than those we use in engineering

[@MrUtah1]

ℒ{calculus teacher} = physics teacher

[@Patrik6920]

Well .. ∂F(Math)dX = Practical application aka Physics ..Usually all math was and are invented to solve real world problems...

[@kingoreo7050]

A lot of the number theory that early mathematicians like euclid did never found usefulness until thousands of years later. Most high level maths done now is in that same boat of just discovering interesting things in whatever mathematical object they happen to be playing with. Maths is discovered almost always for the purpose of discovery itself and there is no obligation that it becomes useful, though it often does end up being.

[@jeffeloso]

That takes me back to the 1980s when 0:06 I was playing with my Sinclair ZX81. I wrote a neat little program to find the centroid of an I beam. I then extended it to do the same for any shape as long as it was made up of rectangles. I think the initial data entry was first how many rectangles, and then for each rectangle, the location of the bottom left of each rectangle from any convenient origin, and it's width and height. The result was the coordinates from the previously defined origin. It was a nice little problem to code as I was learning the principles of simple coding .

[@DAMN__________]

nothing beats the feeling of finally understanding a topic, thanks a bunch

[@tobybartels8426]

4:00 : Since you're doing a centroid rather than a centre of mass, a more direct 1-dimensional analogy is where you (arbitrarily) cut the 10-metre bar somewhere (not in the middle), find the centroid of each piece, and compare those to the (obvious) centroid of the entire bar. Then you'll see that you need to weight each piece by its length.

[@RamPatel-r5f]

Great explanation 👌

[@cdkw2]

bprp physics basics?

[@HeManGNichtDualismus]

thanks for the Y centre of mass explanation with Y (underscore) i element

[@AlbertTheGamer-gk7sn]

Real centroid formulas: m = ∬ρ(x, y)dxdy Mx = ∬yρ(x, y)dxdy My = ∬xρ(x, y)dxdy Centroid: (My/m, Mx/m) Also, in 3D, m = ∭ρ(x, y, z)dxdydz, and the centroid is equal to (Myz/m, Mzx/m, Mxy/m). Using this formula, we can derive the centroid of a given function z = f(x, y) under the curve is equal to: x- = x∬(f(x0, y0) - g(x0, y0))dxdy y- = y∬(f(x0, y0) - g(x0, y0))dxdy z- = ∬(f(x0, y0))^2dxdy

[@joeythreeclubs]

What do you mean "real centroid formulas"?

[@joeythreeclubs]

Also what are m, Mx, and My?

[@AlbertTheGamer-gk7sn]

@@joeythreeclubs Real centroid formulas are the formulas they are derived from. The formulas used in the video are only used for EXPLICIT R -> R functions (y = f(x)), etc. However, if you have an implicit f(x, y) function in the 2D plane, you use the real centroid formulas.

[@AlbertTheGamer-gk7sn]

@@joeythreeclubs m = mass M = first moment of inertia (Mx = moment of inertia of x-axis, My = moment of inertia of y-axis)

[@emanuellandeholm5657]

The x coordinate of a rectangle to be integrated would be (x + 1/2 dx). The area of that same rectangle is (x + 1/2 dx) f(x) = x f(x) + x/2 f(x) dx. Integrating this we get Int (x f(x) + x/2 f(x) dx) dx = Int x f(x) dx + Int [ x/2 f(x) dx ] dx. The second integral vanishes as dx approaches 0.

[@headshotgaming6808]

I did the same thing some months ago but I used inverse function to find the y coordinate

[@DEYGAMEDU]

balancing the torque to find centroid in a line.

[@tensu7260]

Perfect explanation.

[@janda1258]

”Just hold up something heavy like the two markers here”

[@mhm6421]

He's really strong

[@quandarkumtanglehairs4743]

haha @4:26: 'So what, exactly, does d1*m1 do, though? This, right here, is called the 'moment'... at the moment, we are doing moments in Calculus. heh." Love this dude lol

[@ThAlEdison]

You could instead do x̅=∫xdA/∫dA and y̅=∫ydA and setup double integrals or integrals in terms of inverse functions as appropriate. And if an area is bounded by piecewise functions or other complexities, you can still break it apart into components and sum them.

[@adrified9352]

Please do AP Calculus AB 2024 FRQs whenever you can, those are the ones I took. Great video 👍🏾

[@SanjayBhowmick-uj8tc]

Sir please make a video on how to find standard deviation

[@sinekavi]

Were you able to slove that integral BPRP?

[@niom9446]

this is very cool

[@sonyaraman]

Cool! Would you made a video with the same calculations but for unevenly distributed mass/density?

[@afernandesrp]

What if the density wasn’t uniformly distributed?

[@jamescollier3]

it gets more complicated lol.

[@yplayergames7934]

Then, integrate

[@CptnWolFox]

For a one-dimensional object like a rod, if you know how the density ρ varies as a function of position, you can use this: ( ∫ xρ(x) dx ) / (total mass).

[@headshotgaming6808]

Double integrals

[@danielhinrichsen6927]

You'd probably have to deal with line integrals

[@shahkhalid4384]

(2,4)

[@DEYGAMEDU]

Moment of force

[@mausamthapa3007]

Hello there, can you help me with my integration question? The question is Integrate e^-x . secx

[@richardgratton7557]

In the first example, does the centroid of the whole shape necessarily lie on the line joining the 2 centroids of the rectangles?

[@ACheateryearsago]

If the mass is distributed uniformly within the body

[@HenryBriskin]

An equation common for structural engineering

[@theunkown2982]

Thank u❤

[@zzzluvsk]

Sometimes i be feelin like the person with mass m2 lately…

[@khizarzulfi8517]

Shouldn't the X coordinate of bigger rectangle be 3 because 1/2 of 4 + 1/2 of 2 = 2 + 1 = 3?

[@Ninja20704]

No its 4 because u have to add 1/2 of 4 (which is 2. Starting from the left border of the big rectangle and not the centroid of the smaller rectangle.

[@Gattrecity]

Isn’t that barycenter?

[@mastfamastfa1256]

7:21 hahaha