Question

If Volume = 1/3 pi r^2 h
&
Volume of f(x) revolved around the x- axis from a to b is solved with Vx = Integral from a to b of pi f(x)^2 dx,
where does the "1/3" go? It seems to drop out of the equation without any reason I can see.

Answer 1

f(x)=mx so int(mx)^2dx=1/3m^2x^3 so m=r/h and interval is [0,h] then v=pi/3r^2h