Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We got this fuzzy course in the university, there was a problem which it's result led to multiple overlapping fuzzy values. in order to conclude a value out of that there was different approaches which like Centroid of area and Bisector of area.

these to functions give us approximately the same value. I'm aware that the formula to each function is different,

but logically thinking centroid of area is point which holds two equal weights by it's side. on the other hand bisector of area divides the area into two equal values. Supposing the weight (in COA) as the area (in BOA) we must get the same value, but it isn't so.

So philosophically speaking, what is the reason?

share|improve this question
add comment

2 Answers 2

The two concepts are quite different. Think of a thin rod, with weights of 1kg and 2kg at its ends.

The centroid of the apparatus (the balancing point) is one-third of the way along the rod, nearer the heavier weight.

A bisector, on the other hand, is any plane that divides the apparatus into two parts of equal weight. For instance, any plane passing through the heavier weight (but not the lighter weight), dividing it into two parts of weights 1/2 and 3/2 (with the 1/2 portion on the same side of the plane as the lighter weight).

share|improve this answer
add comment

If you have some background in probability, the following family of examples may be useful. Let $X$ be a random variable with a continuous distribution, and probability density function $f(x)$. Let $\mathcal{A}$ be the plane region below $y=f(x)$ and above the $x$-axis. Recall that $\mathcal{A}$ has area ("weight") $1$.

A midpoint of $\mathcal{A}$ is a point $m$ (commonly uniquely determined) such that $$\int_{-\infty}^m f(x)\,dx=\int_m^\infty f(x)\,dx=\frac{1}{2}.$$ Such a point $m$ is called a median of $X$.

By way of contrast, the centroid of $\mathcal{A}$ is the point $\mu$ (if such a point exists) with the property that if we think of the whole weight of $mathcal{A}$ as concentrated at $\mu$, then this has the same moment about the $y$-axis as the actual region $\mathcal{A}$. It turns out that $$\mu=\int_{-\infty}^\infty xf(x)\,dx,$$ so the centroid of $\mathcal{A}$ is the mean of $X$.

For symmetric regions, the concepts more or less coincide, apart from some technical issues. But in general, the midpoint of the area is not equal to the centroid. For example, consider the exponential distribution with parameter $\lambda$. So $f(x)=\lambda e^{-\lambda x}$ for $x\ge 0$, and $f(x)=0$ elsewhere. Then $$m=\frac{\ln 2}{\lambda}\qquad\text{and}\qquad \mu=\frac{1}{\lambda}.$$ So in this case $m<\mu$. Informally, $\mu$ is more sensitive to area in the tail than $m$ is.

There are some interesting extreme examples. For instance, let $X$ have density function $$f(x)=\frac{2}{\pi}\frac{1}{1+x^2}$$ (for $x \ge 0$). Then the midpoint $m$ of the area turns out to be $1$, while the mean $\mu$ does not exist, or more informally is infinite. The "tail" goes down so slowly that though the area is $1$, the centroid is "at infinity."

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.