Nedcim Posted October 18, 2017 Posted October 18, 2017 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 Posted October 19, 2017 Posted October 19, 2017 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 Posted October 19, 2017 Author Posted October 19, 2017 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 Posted October 31, 2017 Author Posted October 31, 2017 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 Posted December 24, 2017 Posted December 24, 2017 (edited) 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
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