Centroid of volume of revolution

Question

Consider a solid generated by the curve $y^2 =ax^2+2bx+c$,rotated about the $x$-axis, and two plane surfaces at right angles to the latter, distance $h$ apart, and with areas $A$ and $B$. To prove that the centroid of the solid is at a distance $h/2 + (B-A)h^2/12V$ from $A$, where $V$ is the volume of the solid.

I have calculated $$V=\frac{(B+A)h}2 -\frac{\pi a h^3}6$$ by integration, but cannot see how to proceed further. Any help would be appreciated.

Answer 1

I suggest that you use Pappus's ($2^{nd}$) Centroid Theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $2\pi R$. The bottom line is that the volume is given simply by $V=2\pi RA$. I assume that that you can calculate the area, hence you can get centroid quite easily, i.e., $R=\frac{V}{2\pi A}$. (Be sure to calculate the area between the curve and the axis of revolution.)

Answer 2

I have now proved the result, by expressing the stated result in terms of integrals and comparing terms in the usual formula for the centroid. This method was somewhat laborious, and did not give any hint of a more elegant method!