What is the volume of this thing?

If $z$ is $\dfrac{1}{2}$ of the vertical axis, then we integrate via slices parallel to the base

$\dfrac { 9 }{ 16 } \left( \dfrac { 1 }{ 3 } 40\cdot \pi \cdot { 20 }^{ 2 } \right) +2 \displaystyle \int _{ 0 }^{ 10 }{ \left( 10\sqrt { { \left( 20-z \right) }^{ 2 }-{ 10 }^{ 2 } } +{ \left( 20-z \right) }^{ 2 }\arcsin\left( \dfrac { 10 }{ 20-z } \right) \right) dz } =14910.9374...$

The volume of the top half of the cone is $\dfrac{1}{8}$ of the whole cone, while the volume of half of the conical frustum is $\dfrac{7}{16}$ of the whole cone. Add both to get $\dfrac{9}{16}$ , and then add that integrated volume.

The volume removed from the original right cone of vertical height $H$ & base radius $R$ cut by a plane parallel to the symmetrical axis at a distance $x$ is given by the most Generalized Formula

$\color{#D61F06}{\Large V_{removed}=\frac{H}{3R}\left(R^3\cos^{-1}\left(\frac{x}{R}\right)-2Rx\sqrt{R^2-x^2}+x^3\ln\left(\frac{R+\sqrt{R^2-x^2}}{x}\right)\right)}$

substituting the corresponding values, $H=40, R=20, x=10$ , one should get

$V_{removed}=\frac{40}{3\cdot 20}\left(20^3\cos^{-1}\left(\frac{10}{20}\right)-2\cdot 20\cdot 10x\sqrt{20^2-10^2}+10^3\ln\left(\frac{20+\sqrt{20^2-10^2}}{10}\right)\right)\approx 1844.223384$

hence, the volume of cut cone

$V_{left}=\frac13\pi R^2H-V_{removed}=\frac13\pi (20)^210-1844.223384\approx \color{#3D99F6}{14910.93744\ cm^3}$