Calculating Centre of area using a method I found but it gives me only approximate answers

Question

As you can see from the title , I am having a problem in calculating centre of areas and specifically the centre of area of a right angle triangle.I know that the centre of area coordinates are $\left( \frac{2a}{3},\frac{b}{3} \right)$ but I get $\left(\frac{a}{\sqrt{2}},\frac{b\sqrt{2}-1}{\sqrt{2}}\right)$ which is approximately equal to the centre coordinates, namely $(0.71,0,29)$. The method I use is a method i found by myself and I didn't find it anywhere on Internet. Simply I thought that the integration of the function from the centre point $x$ to $0$ must be half the whole area of the the shape , unluckily it gave me only approximate answers when I tried it. I understand that my method may be different from the normal method but I don't understand why. Can somebody please explain because it will help me a lot understanding the concept of centre of area.

Answer 1

By your comment, you had a right-angle triangle with vertices $(0,0), (a,0), (a,b)$, took $y=\dfrac{b}{a}x$, integrated vertical strips to get area from $x=0$ to $x=c_x$ as ${\Large\int}_0^{c_x}\dfrac{b}{a}x~dx = \left[\dfrac{b}{2a}x^2\right]_0^{c_x} = \dfrac{bc_x^2}{2a}$, solved for $c_x$ to get half the area of the triangle: $\dfrac{bc_x^2}{2a} = \dfrac{1}{2}\dfrac{ab}{2}, c_x^2=\dfrac{a^2}{2}, c_x=\dfrac{a}{\sqrt{2}}$, and did something similar for $c_y$.

The good news is, there's nothing wrong with your calculations. However, centre of area (or 'centroid') doesn't mean quite what you thought it meant! The Wikipedia article on Centroid has a lot of information, and this page explains how to calculate the centroid using integration: if you can understand example 3 on this page, you can in principle work out how to find any centroid using integration.

For rectangles the centroid is, well, the centre of the rectangle. For triangles, it's where the medians intersect. It turns out that the medians neatly divide the area of the triangle in half, but that's not true for other lines through the centroid, which is why your idea of using a horizontal or vertical line to cut the area of the triangle in half doesn't work.

The 'standard' way to find the centroid involves first calculating the area $A$ of the figure (by integration or geometry). In your example $A=\dfrac{ba}{2}$, you have $f(x)=\dfrac{b}{a}x$ and you would calculate the centroid as follows. Using h(x) to mean the height of a vertical strip:

$\bar{x} = \dfrac{1}{A}{\Large\int} xh(x)~dx = \dfrac{2}{ab}{\Large\int}_0^ax\dfrac{b}{a}x~dx = \dfrac{2}{a^2}{\Large\int}_0^ax^2~dx = \dfrac{2}{a^2}\left[\dfrac{x^3}{3}\right]_0^a = \dfrac{2a^3}{3a^2}=\dfrac{2a}{3}$.

You can find the $y$ coordinate of the centroid in two ways. One is by using horizontal strips. The width $w(y)$ of a horizontal strip at $y$ is $w(y) = a - \dfrac{a}{b}y$, so

$\bar{y}=\dfrac{1}{A}{\Large\int}yw(y)~dy = \dfrac{2}{ab}{\Large\int}_0^by(a-\dfrac{a}{b}y)~dy= \dfrac{2}{b}{\Large\int}_0^by-\dfrac{1}{b}y^2~dy = \dfrac{2}{b}\left[\dfrac{y^2}{2}-\dfrac{y^3}{3b}\right]_0^b = b - \dfrac{2b}{3} = \dfrac{b}{3}$.

Sometimes it's messy using horizontal strips for $y$ (or vertical strips for $x$). Another way to find $\bar{y}$ using vertical strips is to integrate over $x$ using the height $h(x)$ of a vertical strip times the $y$-coordinate of the centre of the strip (in our case, that's $c(x) = \dfrac{f(x)}{2} = \dfrac{b}{2a}x$):

$\bar{y}=\dfrac{1}{A}{\Large\int}h(x)c(x)~dx = \dfrac{2}{ab}{\Large\int}_0^a \dfrac{b}{a}x\dfrac{b}{2a}x~dx = \dfrac{b}{a^3}{\Large\int}_0^ax^2~dx = \dfrac{b}{a^3}\left[\dfrac{x^3}{3}\right]_0^a = \dfrac{b}{3}$.

(Edit - added further explanation:)

Two differences between the methods are the half-area method isn't well-defined and can jump suddenly. For the first picture below, two equal rectangles connected by an infinitely thin line, any point on the connecting line has half the area on either side. If we add a tiny bit of area to the RHS rectangle (second picture), the centroid shifts slightly to the right (which is what happens with a physical object), but the half-area point is immediately forced into the RHS rectangle - it can't sit on the line.

${}~~~~~~~~~{}$

Here's another figure where the half-area method doesn't work. The main thing to get from the figure, though, is that a smaller area can balance a bigger area if it's further away. Think of an adult and a child on a seesaw (= US teeter-totter).

In your example, the half-area method fails because it just balances area - it doesn't take into account that smaller, further areas can outweigh closer, larger areas. (And also you'll get different locations in the triangle with the half-area method if you rotate the triangle).

The nice thing about moments is that you can calculate them from any point and they will tell you where the centroid is. From the centre of the left rectangle, we have 0(100) + 3(50) = 1(150), which says the whole system is the same as 150 units of area centred one unit to the right (at the balance point). From the centre of the right rectangle, we have -3(100) + 0(50) = -2(150), which says the whole system is the same as 150 units of area centred 2 units to the left. From the balance point we have -1(100)+2(50) = 0(150), which says the whole system is the same as 150 unit of area centred right here.