Volume of a curved cone Apparently, the volume of this cone is (1/16)π(r^2)h. My question is why this is the case, can someone please geometrically explain the reason behind the 1/16 bit. The radius is supposed to be proportional to the square of its height. Thanks.


\begin{equation} 
r=a \cdot h^2
\tag{01}
\end{equation}
\begin{equation}   
dV=\pi r^{2} dh = \pi a^{2}h^{4}dh
\tag{02}
\end{equation}
\begin{equation}  
V=\int_{h=0}^{h=H}\pi a^{2}h^{4}dh=\dfrac{\pi a^{2}H^{5}}{5}=\dfrac{\pi R^{2}H}{5}
\tag{03}
\end{equation}
If instead of eq.(01) your curve was
\begin{equation} 
r=a \cdot \sqrt{h^{15}}
\tag{04}
\end{equation}
then you would have 16 in the denominator.
