Verify centroid of a function

[Nedcim]

Suppose you determine a point as the centroid for a single variable function is there a method to verify that the point is indeed the centroid?

[Country Boy]

  I have no idea what you mean by "centroid of a single variable function".  "Centroid" is a geometric concept-there must be a region in the plane or an object in three dimensions.   I presume you have some kind of curve that bounds a region (but then it couldn't be a "function").

[Nedcim]

In general a region bounded by the curve y=f(x) of a continuous single variable function,  the x axis with some  finite limits of integration.  Let's work with a specific case. Suppose we have a region bounded by y=x^4, x=0 and x=1 with a centroid located at (5/6, 5/18).  How do you prove that point is the centroid for the defined region?  

[Nedcim]

I found a method using  theorem of Pappus and any of the various volume of solids and checking that the volumes are equal. 

[Country Boy]

I'm not sure what you mean by 'verify' here. Given a region bounded by $y= x^4$, $x= 0$, and $x= 1$, it is easy to find the area, $\int_0^1 x^4 dx= \left[\frac{x^5}{5}\right]_0^1= \frac{1}{5}$.  Also $\int_0^1 x(x^4)dx= \int_0^1 x^5dx= \left[\frac{x^6}{6}\right]_0^1= \frac{1}{6}$ and $\int_0^1 y(x^4)dx= \int_0^1 x^8 dx= \left[\frac{x^9}{9}\right]_0^1= \frac{1}{9}$.  The centroid is the point $\left(\frac{\frac{1}{6}}{\frac{1}{5}}, \frac{\frac{1}{9}}{\frac{1}{5}}\right)= \left(\frac{5}{6}, \frac{5}{9}\right)$. 

You could also physically test a centroid- cut the figure from some thin material with constant density and "glue" a string to the point you believe to be the centroid.  If it actually Is a centroid then the figure will "balance" there, without any side going up or down.  Or- draw two perpendicular lines through the centroid.  If it is the centroid then the figure will balance on a knife edge along both of those lines.

Edited December 24, 2017 by Country Boy