Question #bb3d8

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Question #bb3d8

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Answer 1 · 2016-07-05T03:24:23

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

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 - $h_{1}$ $h_1$ (height of a cone) and $h_{2}$ $h_2$ (height of a cylinder).
We do not know these heights but we know two important equations they participate in:
(1) $h_{2} = 2 h_{1}$ $h_2 = 2h_1$
(2) $h_{1} + h_{2} = H$ $h_1 + h_2 = H$
where $H$ $H$ is a known height of an entire solid.

From the two equations above we can easily find $h_{1}$ $h_1$ and $h_{2}$ $h_2$ in terms of $H$ $H$ :
substituting (1) into (2), we get
$h_{1} + 2 h_{1} = H$ $h_1+2h_1 = H$
$\Rightarrow$ $rArr$ $h_{1} = \frac{H}{3}$ $h_1 = H/3$
$\Rightarrow$ $rArr$ $h_{2} = \frac{2 H}{3}$ $h_2 = (2H)/3$

Knowing heights and radiuses of a cone and a cylinder, we can calculate each volume.
The volume of a cone is
$V_{1} = \frac{1}{3} π r^{2} h_{1} = \frac{1}{3} π r^{2} \frac{H}{3} = \frac{π r^{2} H}{9}$ $V_1 = 1/3 pi r^2h_1 = 1/3 pi r^2 H/3 = (pi r^2H)/9$
The volume of a cylinder is
$V_{2} = π r^{2} h_{2} = π r^{2} \frac{2 H}{3}$ $V_2=pir^2h_2 = pi r^2 (2H)/3$

Total volume of a solid is
$V = V_{1} + V_{2} =$ $V=V_1+V_2 =$
$= π r^{2} H (\frac{1}{9} + \frac{2}{3}) =$ $= pir^2H(1/9+2/3) =$
$= \frac{7}{9} π r^{2} H$ $= 7/9pi r^2 H$

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Question #bb3d8

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