Volume of Frustum of Right Circular Cone

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Theorem

Let $F$ be a frustum of a right circular cone.

The volume $\VV$ of $F$ is given as:

$\VV = \dfrac {\pi h \paren {a^2 + a b + b^2} } 3$

where:

$a$ and $b$ are the radii of the bases of $F$
$h$ is the altitude of $F$.


Proof

From Volume of Frustum of Cone or Pyramid:

$\VV = \dfrac {h \paren {A_1 + A_2 + \sqrt {A_1 A_2} } } 3$

where:

$A_1$ and $A_2$ are the areas of the bases of $F$
$h$ is the altitude of $F$.


Here we have that $F$ be a frustum of a right circular cone.

Hence the bases of $F$ are circles.

From Area of Circle, the areas of the bases of $F$ are therefore:

$A_1 = \pi a^2$
$A_2 = \pi b^2$

Hence:

\(\ds \VV\) \(=\) \(\ds \dfrac {h \paren {A_1 + A_2 + \sqrt {A_1 A_2} } } 3\) Volume of Frustum of Cone or Pyramid: see above
\(\ds \) \(=\) \(\ds \dfrac {h \paren {\pi a^2 + \pi b^2 + \sqrt {\pi a^2 \pi b^2} } } 3\)
\(\ds \) \(=\) \(\ds \dfrac {\pi h \paren {a^2 + b^2 + a b} } 3\) after simplification

Hence the result.

$\blacksquare$


Sources

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