Will Underwood

2022-09-24

Volume of a Truncated Right Prism with generic base convex polygon

I can find lots of pages online saying the volume of a Truncated Right Triangular Prism is $\frac{A}{3}({h}_{1}+{h}_{2}+{h}_{3})$, where A is the area of the base and ${h}_{i}$ is the height of vertex i of the top face. Or, in other words, to find the volume you "flatten" the top face by finding the average height of its 3 vertices, and then the volume is just that of a regular prism.

My question is: for a generic base convex polygon, does the same argument apply? Is the volume equal to $\frac{A}{n}\sum _{i=1}^{n}{h}_{i}$, where n is the number of vertices of the base/top? Is there a proof for this anywhere?

My intuition says it's true, but what if the base shape was like a really long kite; wouldn't the height of the vertex at the bottom tip of the kite skew the average height of the top face?

I can find lots of pages online saying the volume of a Truncated Right Triangular Prism is $\frac{A}{3}({h}_{1}+{h}_{2}+{h}_{3})$, where A is the area of the base and ${h}_{i}$ is the height of vertex i of the top face. Or, in other words, to find the volume you "flatten" the top face by finding the average height of its 3 vertices, and then the volume is just that of a regular prism.

My question is: for a generic base convex polygon, does the same argument apply? Is the volume equal to $\frac{A}{n}\sum _{i=1}^{n}{h}_{i}$, where n is the number of vertices of the base/top? Is there a proof for this anywhere?

My intuition says it's true, but what if the base shape was like a really long kite; wouldn't the height of the vertex at the bottom tip of the kite skew the average height of the top face?

skarvama

Beginner2022-09-25Added 5 answers

Step 1

Your formula for the volume cannot be true, in general, for $n>3$. Here's a counter-example with $n=4$.

Take a truncated right quadrangular prism with its base on the $z=0$ plane and its top vertices given by:

${V}_{1}=(0,0,0),\phantom{\rule{1em}{0ex}}{V}_{2}=(1,0,a),\phantom{\rule{1em}{0ex}}{V}_{3}=(0,1,b),\phantom{\rule{1em}{0ex}}{V}_{4}=(2,2,2a+2b),$

with a and b positive constants. Note that those points all lie in the same plane, because $\overrightarrow{{V}_{1}{V}_{4}}=2\cdot \overrightarrow{{V}_{1}{V}_{2}}+2\cdot \overrightarrow{{V}_{1}{V}_{3}}$.

Step 2

The volume of this solid can be computed dividing it into two truncated triangular prisms with a plane passing through y-axis and ${V}_{4}$. Both their bases have unit area, hence applying the formula for the triangular case with ${h}_{1}=0$, ${h}_{2}=a$, ${h}_{3}=b$, ${h}_{4}=2a+2b$ we get:

$V=\frac{{A}_{1}}{3}({h}_{1}+{h}_{2}+{h}_{4})+\frac{{A}_{2}}{3}({h}_{1}+{h}_{3}+{h}_{4})=\frac{1}{3}(3a+2b)+\frac{1}{3}(2a+3b)=\frac{5}{3}(a+b).$

On the other hand, if your generalised formula were true, we would have:

$V=\frac{{A}_{1}+{A}_{2}}{4}({h}_{1}+{h}_{2}+{h}_{3}+{h}_{4})=\frac{3}{2}(a+b).$

Hence the generalised formula doesn't work.

Your formula for the volume cannot be true, in general, for $n>3$. Here's a counter-example with $n=4$.

Take a truncated right quadrangular prism with its base on the $z=0$ plane and its top vertices given by:

${V}_{1}=(0,0,0),\phantom{\rule{1em}{0ex}}{V}_{2}=(1,0,a),\phantom{\rule{1em}{0ex}}{V}_{3}=(0,1,b),\phantom{\rule{1em}{0ex}}{V}_{4}=(2,2,2a+2b),$

with a and b positive constants. Note that those points all lie in the same plane, because $\overrightarrow{{V}_{1}{V}_{4}}=2\cdot \overrightarrow{{V}_{1}{V}_{2}}+2\cdot \overrightarrow{{V}_{1}{V}_{3}}$.

Step 2

The volume of this solid can be computed dividing it into two truncated triangular prisms with a plane passing through y-axis and ${V}_{4}$. Both their bases have unit area, hence applying the formula for the triangular case with ${h}_{1}=0$, ${h}_{2}=a$, ${h}_{3}=b$, ${h}_{4}=2a+2b$ we get:

$V=\frac{{A}_{1}}{3}({h}_{1}+{h}_{2}+{h}_{4})+\frac{{A}_{2}}{3}({h}_{1}+{h}_{3}+{h}_{4})=\frac{1}{3}(3a+2b)+\frac{1}{3}(2a+3b)=\frac{5}{3}(a+b).$

On the other hand, if your generalised formula were true, we would have:

$V=\frac{{A}_{1}+{A}_{2}}{4}({h}_{1}+{h}_{2}+{h}_{3}+{h}_{4})=\frac{3}{2}(a+b).$

Hence the generalised formula doesn't work.

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