**NOOOOOOOOOOOOOOOOOOOO**

*He is not a phy teacher bro*

Yup 👍

He showed this in another video eight days ago.

Here: The integral formulas for the centroid of a region (center of mass) kzfaq.info/get/bejne/n9Bnarypqp_JnIk.html

Here The integral formulas for the centroid of a region (center of mass) kzfaq.info/get/bejne/n9Bnarypqp_JnIk.html

@@bprpcalculusbasics Makes sense. Thanks!

You don't do that in general. It's for this case specifically, due to the area of the semicircle having a 1/2 factor in it.

@@carultch I see now. I think a better way to show it (especially when explaining to people) would be to say we are integrating y as y goes from 0 to r (instead of -r to +r), so the 1/2 factor is not needed. It makes more sense and matches what is actually happening.

this video explains the formula of the centroids kzfaq.info/get/bejne/n9Bnarypqp_JnIk.html How I think of it: If you were to find the centroid of a shape composed of a finite number of rectangles you essentially take weighted average of the centroid each rectangle. It's fairly easy to find the centroid of a rectangle without any special tools. So centroid of full shape = Σ [ (Area of rectangle k / Area of full shape) ⋅ centroid of rectangle k ] Area of full shape is a constant so we can factor that out centroid of full shape = Σ [ Area of rectangle k ⋅ centroid of rectangle k ] / Area of full shape --> this finite sum is the general structure both the integral expressions follow A key idea of integral calculus is that if you take approximations of the area under the curve with rectangles, and then take the limiting process of those rectangles as the number of rectangles go to infinity you get the exact area. So if we take the weighted average of all of these rectangles we can get the centroid of the region under any curve. for the x coordinate: the x coordinate of the centroid of any given rectangle drawn under the curve is x+width/2. The area of any rectangle is f(x) * width. Multiplying the two gets us xf(x)*width + width²/2. The width of each rectangle is already very small so width²/2 is so small we can ignore it. We can think of dx in an integral as the width of the rectangle. We can use an integral to sum up an infinite number of things so we get x̄ = ∫xf(x)dx / Area of full shape for the y coordinate: the y coordinate of the centroid of any given rectangle is f(x)/2 (this is why there is a 1/2 in the formula for ȳ). Again, the area of any rectangle is f(x) * width. Multiplying the two gets us f(x)²/2 * width. Using simalar logic as before from this we get ȳ = ∫f(x)²/2 dx / Area of full shape and of course if we need to we can find the area of the full shape using another integral if we need to. It wasn't needed in this case since it was just a semi-circle

Average

What you're doing, is taking a "weighted average" of how the area is distributed through space. You calculate the first moment of area (i.e. y*dA) contribution of each area element. You then add up (integrate) to find the total first moment of area, and divide by the area when done. Moment in general means quantity multiplied by distance from a reference point.

Because the 1/2 is being multiplied and you can pull constant factors out of an integral. The r^2 isn't a factor. I mean, if you wanted to, after pulling the 1/2 out front, you could break the integral into 2 integrals: 2/(pi r^2) [∫r^2 dx - ∫x^2 dx], and then factor the r^2 out of the first one to get: 2/(pi r^2) [r^2 ∫dx - ∫x^2 dx] But that doesn't really make things all that much simpler than just the single integral.

@@phiefer3 ah, so Got it Thanks

yes thats correct

yeah same question, ive never done this but the x symmetry seems clear enough - never known about this ybar function though, so is it just true of all centroids ?

this video explains the formula of the centroids kzfaq.info/get/bejne/n9Bnarypqp_JnIk.html How I think of it: If you were to find the centroid of a shape composed of a finite number of rectangles you essentially take weighted average of the centroid each rectangle. It's fairly easy to find the centroid of a rectangle without any special tools. So centroid of full shape = Σ [ (Area of rectangle k / Area of full shape) ⋅ centroid of rectangle k ] Area of full shape is a constant so we can factor that out centroid of full shape = Σ [ Area of rectangle k ⋅ centroid of rectangle k ] / Area of full shape --> this finite sum is the general structure both the integral expressions follow A key idea of integral calculus is that if you take approximations of the area under the curve with rectangles, and then take the limiting process of those rectangles as the number of rectangles go to infinity you get the exact area. So if we take the weighted average of all of these rectangles we can get the centroid of the region under any curve. for the x coordinate: the x coordinate of the centroid of any given rectangle drawn under the curve is x+width/2. The area of any rectangle is f(x) * width. Multiplying the two gets us xf(x)*width + width²/2. The width of each rectangle is already very small so width²/2 is so small we can ignore it. We can think of dx in an integral as the width of the rectangle. We can use an integral to sum up an infinite number of things so we get x̄ = ∫xf(x)dx / Area of full shape for the y coordinate: the y coordinate of the centroid of any given rectangle is f(x)/2 (this is why there is a 1/2 in the formula for ȳ). Again, the area of any rectangle is f(x) * width. Multiplying the two gets us f(x)²/2 * width. Using simalar logic as before from this we get ȳ = ∫f(x)²/2 dx / Area of full shape and of course if we need to we can find the area of the full shape using another integral if we need to. It wasn't needed in this case since it was just a semi-circle