Answer
$15\pi \text{ cubic inches}
$
Work Step by Step
The variation model described by the problem is $
V=khr^2
,$ where $V$ is the volume, $h$ is the height, and $r$ is the radius.
Substituting the known values of the variables results to
\begin{array}{l}\require{cancel}
32\pi=k(6)(4)^2
\\
32\pi=k(6)(16)
\\
32\pi=96k
\\
32\pi=96k
\\
\dfrac{32\pi}{96}=k
\\\\
k=\dfrac{\cancel{32}\pi}{\cancel{32}\cdot3}
\\\\
k=\dfrac{\pi}{3}
.\end{array}
Hence, the equation of variation is
\begin{array}{l}\require{cancel}
V=\dfrac{\pi}{3}hr^2
.\end{array}
Using the variation equation above, then
\begin{array}{l}\require{cancel}
V=\dfrac{\pi}{3}hr^2
\\\\
V=\dfrac{\pi}{3}(5)(3)^2
\\\\
V=\dfrac{\pi}{3}(5)(9)
\\\\
V=\dfrac{\pi}{\cancel3}(5)(\cancel3\cdot3)
\\\\
V=15\pi
.\end{array}
Hence, the volume of the cone, $V,$ is$
15\pi \text{ cubic inches}
.$
