Question #bb3d8

Jul 5, 2016

$V = \frac{7}{9} \pi {r}^{2} H$

Explanation:

Volume of this solid is a sum of two volumes - the one of a cone and that of a cylinder.

Since we know the radius of both, all we need is the height of each - ${h}_{1}$ (height of a cone) and ${h}_{2}$ (height of a cylinder).
We do not know these heights but we know two important equations they participate in:
(1) ${h}_{2} = 2 {h}_{1}$
(2) ${h}_{1} + {h}_{2} = H$
where $H$ is a known height of an entire solid.

From the two equations above we can easily find ${h}_{1}$ and ${h}_{2}$ in terms of $H$:
substituting (1) into (2), we get
${h}_{1} + 2 {h}_{1} = H$
$\Rightarrow$ ${h}_{1} = \frac{H}{3}$
$\Rightarrow$ ${h}_{2} = \frac{2 H}{3}$

Knowing heights and radiuses of a cone and a cylinder, we can calculate each volume.
The volume of a cone is
${V}_{1} = \frac{1}{3} \pi {r}^{2} {h}_{1} = \frac{1}{3} \pi {r}^{2} \frac{H}{3} = \frac{\pi {r}^{2} H}{9}$
The volume of a cylinder is
${V}_{2} = \pi {r}^{2} {h}_{2} = \pi {r}^{2} \frac{2 H}{3}$

Total volume of a solid is
$V = {V}_{1} + {V}_{2} =$
$= \pi {r}^{2} H \left(\frac{1}{9} + \frac{2}{3}\right) =$
$= \frac{7}{9} \pi {r}^{2} H$