well, it you take the volume of the pipe, before the hole
diameter is 8.4, meaning radius is half that or 4.2, height is 10
then, you take the volume of the "cylindrical hole"
and subtract that volume from the pipe's, you'd end up with their difference,
which is, whatever is not taken up by the hole, namely, just the metal in the pipe
[tex]\bf \textit{volume of a cylinder}\\\\
V=\pi r^2 h\qquad
\begin{cases}
r=radius\\
h=height
\end{cases}\\\\
-----------------------------\\\\[/tex]
[tex]\bf V=\pi r^2 h\quad
\begin{cases}
h=10\\
r=4.2
\end{cases}\implies
\begin{array}{llll}
\textit{pipe without the hole}\\\\
V_c=4.2^2\cdot 10\pi
\end{array}\\\\
-----------------------------\\\\
V=\pi r^2 h\quad
\begin{cases}
h=10\\
r=3
\end{cases}\implies
\begin{array}{llll}
\textit{cylindrical hole}\\\\
V_h=3^2\cdot 10\pi
\end{array}\\\\
-----------------------------\\\\
difference\implies V_c-V_h\implies 1764\pi -90\pi [/tex]
