Use integration to write aformula that gives the volume of cone , the radius of whose base is r and whose height is h
Alternatively,
$\displaystyle R=radius,\;H=height,\;r=varying\;cross-sectional\;radius,$
$\displaystyle h=height\;of\;internal\;cone\;of\;radius\;r$
$\displaystyle \displaystyle\frac{R}{H}=\frac{r}{h}=constant\Righ tarrow\ R=Hk,\;\;r=hk$
The constant is $\displaystyle tan\theta$ but this form is unnecessary.
$\displaystyle \displaystyle\int_{0}^H{{\pi}r^2}dh={\pi}k^2\int_{ 0}^H{h^2}dh$
$\displaystyle \displaystyle\ ={\pi}k^2\frac{H^3}{3}=\frac{{\pi}R^2}{H^2}\;\frac {H^3}{3}=\frac{{\pi}R^2H}{3}$