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According to here, for truncated pyramid with rectangular base:

enter image description here

The volume is given by:

$volume= h/3*(a*b+c*d+(a*d+b*c)/2)$

What if the base is an irregular surface area, defined by $n$ sets of coordinates?

How to extend the above volume formula to cater for irregular surface area?

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2 Answers 2

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We need to find the area of top and bottom surfaces. enter image description here

You can also visit: http://mathworld.wolfram.com/PyramidalFrustum.html

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  • $\begingroup$ I know how to find the top and bottom surface area, but I don't know how to find the volume $\endgroup$
    – Graviton
    Commented Oct 13, 2016 at 14:33
  • $\begingroup$ @Graviton, mathworld.wolfram.com/PyramidalFrustum.html $\endgroup$
    – Seyed
    Commented Oct 13, 2016 at 14:38
  • $\begingroup$ that seems to be good, I would want to revoke my down-vote, with the condition that you expand your answer to include your wolframalpha link and do some explanation on that. $\endgroup$
    – Graviton
    Commented Oct 13, 2016 at 14:40
  • $\begingroup$ I would also appreciate if you can (re)derive the volume formula in the wolfram page link; I can quite see it. $\endgroup$
    – Graviton
    Commented Oct 13, 2016 at 14:44
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Prove this holds for any pyramid with a triangular base then triangulate your polygon and sum the volumes.

More generally, if you have a closed curve in a plane and a vertex not in that plane the volume of the solid obtained by joining the vertex to the curve is going to be the integral of the sections of area parallel to the curve from it to the vertex: $\int_0^h a(t)dt$ where $a(t)$ is the area of the intersection of the solid and the plane parallel to the one which contains the curve and whose distance to the vertex is t and h is the distance from the vertex to the plane which contains the curve. By a similarity argument we obtain that $a(t)$ is proportional to $t^2$, so let $a(t)=kt^2$ then $a(h)=kh^2\implies k=\frac{a(h)}{h^2}$ so $\int_0^h a(t)dt=\int_0^h kt^2dt=\frac{kh^3}{3}=\frac{a(h)h}{3}$, the familiar formula.

The formula in the question can then be obtained by subtracting a small pyramid from a big pyramid.

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  • $\begingroup$ I can understand your argument, but your $k$ is not determined; can you show how $k$ can be determined here? $\endgroup$
    – Graviton
    Commented Oct 14, 2016 at 0:10
  • $\begingroup$ We want to find the volume as a function of $h$ and $a(h)$ but $a(t)$ is some constant times $t^2$ for all t, I just called it $k$ for the sake of convenience so letting $t=h$, $a(h)=kh^2$. $\endgroup$
    – Sophie
    Commented Oct 14, 2016 at 0:23
  • $\begingroup$ How to determine the $k$? I think this is the key to my question. Because I know that the area is somewhat proportional to $h$ $\endgroup$
    – Graviton
    Commented Oct 14, 2016 at 5:50

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