Centroid using integration?

Look at problems 2, 3, and 4 in Week 10 on http://www.itcanbeshown.com/forstudents-itcanbesho... . The author goes through considerable detail (e.g. http://www.itcanbeshown.com/For_Students/Solutions... ) in developing why the formulation should be constructed the way it is, does it for 3D shapes using a single integral and triple integral approach, 2D shapes for single and double integration. I also encourage you to look around at problems near Week 10, maybe it will help elucidate any confusion you have.

A centroid is just an average value, the equation you see in discrete over integral form are the same thing (the integral form is a limit of the discrete formulation, it happens when your shapes get infinitesimally small). It is the same as how you compute any average, say a test score average among a class, as an example consider a test that was 100 points total with total number of test takers to be 1 + 5 + 7 + 9 = 22 students

Average = { (1)(25) + 5(50) + 7(89) + 9(92) } / {1 + 5 + 7 + 9 } = 78.45 points

The numerator here is a sum of products = (how many people got this score)(the score itself), the denominator is the total number of students. This kind of average you should be familiar with. This is how you calculate averages though, you take the averaging quantity (the score) and you "weight" it by the appropriate factor (here, the total number of students).

If you want to find the center of area in the x-direction, you do the same thing: here the averaging quantity is the x-coordinate, and the "weights" are the area, so you have

x-centroid = {sum of products (x-coordinate)(area of each piece) } / {total area}

that is if you have a discrete areas you can work with. Suppose you have a shape that does not have easy areas you can break it up into, that is still no problem but you have to make the areas so small so that they can fill up the entire odd shape you have. When you take a formal limit on the above expression for the area approaching zero (i.e. really small), you get the integral representation

x-centroid = integral{ x dA} / integral {dA}

where dA = differential area. You can do it with anything:

x-center-of-anything = integral { x dQ} / {integral dQ}

where Q = anything, it could be volume, pressure, force, mass, weight, etc. All the definitions of center-of-mass and so forth are the same formula, just an average, except we weight them differently.

Centroid Integral

Finding Centroids

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RE:

Centroid using integration?

I'm doing Further Maths in A-Level and I can't find a single resource about centroids of areas and volumes using integration. In one of the examiners' reports to a past paper I was solving I saw a reference to some formulae. I searched high and low to no avail. Please do help by either...

Im not sure but I'm pretty sure u can say y=50 and y=5x right? if you take the sqaure root of the second problem. Idk if that will help but i hope it does. Im not sure how to estimate curves though. If it were lines, the centroid would be (10,1)

If its rotated about x-axis , for parametric equations ;

x-coordinate = integration( xy dx/dt * dt)/integration (y dx/dt *dt)

y-coordinate = integration(y^2 dx/dt *dt)/integration (y dx/dt *dt )

If rotated about y-axis just replace dx/dt by dy/dt and y by x , this may help you all .....