Volume of a curved cone

Question

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.

Answer 1

\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.