Sign in to follow this  
Nedcim

Verify centroid of a function

Recommended Posts

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?

Share this post


Link to post
Share on other sites

  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").

Share this post


Link to post
Share on other sites

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?  

Share this post


Link to post
Share on other sites

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

Share this post


Link to post
Share on other sites

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 by HallsofIvy

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this