Does everything cancel out?

Volume of cube $V = \dfrac{1}{3}\left(\pi r^2 h\right) = \dfrac{\pi d^2 h}{3\cdot 4}\phantom{ccccc}{\color{#3D99F6}r =\dfrac{d}{2}}$ (say it is the initial volume). Now , new vertical height $h_n = 2h$ and new diameter is $d_n = \dfrac{2r}{2}=\dfrac{d}{2}$ . Also $r_n = \dfrac{d_n}{2}\implies r_n^2 ={\color{#D61F06} \dfrac{d_n^2}{4}}$ .

New volume of cone is $V_n = \dfrac{\pi {\color{#D61F06}d_n^2 }h_n}{3\cdot {\color{#D61F06}4}} =\left(\dfrac{\pi}{3\cdot 4} \right)\left(\dfrac{d^2}{4}\cdot 2h\right) =\dfrac{1}{2}\left(\dfrac{\pi d^2 h}{3.4}\right)= \dfrac{V}{2}$ New volume become half of the original volume.

$x = \color{#D61F06}{\dfrac{\pi r^2 h}{3}}$

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\[\begin{array} \begin{align}\text{Shape}_A &= \dfrac{\pi (2r)^2 h}{3} \\ \\ &= 4 \color{Red}{\dfrac{\pi r^2 h}{3}} \\ \\&= 4x \end{align}

& \quad \quad \quad \quad

\begin{align}\text{Shape}_A &= \dfrac{\pi r^2 (2h)}{3} \\ \\ &= 2 \color{Red}{\dfrac{\pi r^2 h}{3}} \\ \\&= 2x \end{align}

\end{array}\]

The volume of a cone is $\frac{\pi r^2h}{3}$ , where ‘r’ is the radius and ‘h’ is the height. We can see that (in the volume) r is squared whereas h is to the power 1. Therefore if we decrease the diameter (which decreases the radius proportionally), and increase the height by the same ratio, the volume of the cone halves .