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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. enter image description here

Joe
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    this has nothing to do with physics. – AccidentalFourierTransform Mar 20 '16 at 18:56
  • What shape would $\pi r^2h$ be the volume of? – lemon Mar 20 '16 at 19:17
  • @lemon 1/3(pi)hr^2 is the volume of a regular cone. But i cant find anything on curved cones. – Joe Mar 20 '16 at 19:29
  • That's not what I asked... – lemon Mar 20 '16 at 19:36
  • @lemon 3 cones? – Joe Mar 20 '16 at 19:38
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    A cylinder of radius $r$ and length $h$. So the geometrical rationale behind the (1/16) is you can fit 16 of them inside a cylinder. Now, why 1/16 and not some other number? Well it depends entirely on the sort of curvature - whether it's parabolic, cubic, etc... There's nothing significant about it. – lemon Mar 20 '16 at 19:40
  • @lemon you're right, but what about in a general case. Isn't there a general equation to solve the volume of these types of cones like regular cones? – Joe Mar 20 '16 at 19:47
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    @Joe yes, the general equation is called Calculus. – AccidentalFourierTransform Mar 20 '16 at 19:59
  • As pointed out by lemon, you need to know the functional form of the curved side in order to be able to use your calculus knowledge in calculating the volume due to revolution of a curve around some axis. –  Mar 20 '16 at 19:59
  • @Benjamin thanks, i understood it now. i forgot about the equation for volume of revolution – Joe Mar 20 '16 at 20:05

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\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.

Frobenius
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