I want to find the volume of the cut-cone in the picture with respect to $R, r$ and $h$. Can you check my work
The volume of the cut-cone is is the volume of the full cone minus the volume of the small cone which gone so
$V = \dfrac{\pi}{3} R^2 x – \dfrac{\pi}{3} r^2 (x-h) $
But from the similarity of the triangles we get $\dfrac{x}{R} = \dfrac{x-h}{r} \rightarrow x = \dfrac{hR}{R-r}$
$V = \dfrac{\pi}{3} \left(x(R^2 – r^2) + rh \right)=\dfrac{\pi}{3}\left(\dfrac{Rh(R^2-r^2)}{R-r} +r^2h \right)= \dfrac{\pi h}{3}(R(R+r)+ r^2) $
Best Answer
Man what's the question the cone is more like a frustum and volume of small cone =volume of original cone - volume of frustum $= 1/3πR^2x-1/3π(R^2+r^2+rR)(x-h)$. Here $h$ is height of frustum.