The volume for truncated pyramid with irregular base












1












$begingroup$


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?






geometry volume







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    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?






    geometry volume







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      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?






      geometry volume







      share|cite|improve this question











      $endgroup$




      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?






      geometry volume




      geometry volume






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 24 '18 at 9:34









      Glorfindel

      3,41381930




      3,41381930










      asked Oct 13 '16 at 8:06









      GravitonGraviton

      1,09622044




      1,09622044






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          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






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I know how to find the top and bottom surface area, but I don't know how to find the volume
            $endgroup$
            – Graviton
            Oct 13 '16 at 14:33










          • $begingroup$
            @Graviton, mathworld.wolfram.com/PyramidalFrustum.html
            $endgroup$
            – Seyed
            Oct 13 '16 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
            Oct 13 '16 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
            Oct 13 '16 at 14:44



















          0












          $begingroup$

          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 $a(h)=kh^2implies k=frac{a(h)}{h^2}$ 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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I can understand your argument, but your $k$ is not determined; can you show how $k$ can be determined here?
            $endgroup$
            – Graviton
            Oct 14 '16 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
            Oct 14 '16 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
            Oct 14 '16 at 5:50












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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          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






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I know how to find the top and bottom surface area, but I don't know how to find the volume
            $endgroup$
            – Graviton
            Oct 13 '16 at 14:33










          • $begingroup$
            @Graviton, mathworld.wolfram.com/PyramidalFrustum.html
            $endgroup$
            – Seyed
            Oct 13 '16 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
            Oct 13 '16 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
            Oct 13 '16 at 14:44
















          1












          $begingroup$

          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






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I know how to find the top and bottom surface area, but I don't know how to find the volume
            $endgroup$
            – Graviton
            Oct 13 '16 at 14:33










          • $begingroup$
            @Graviton, mathworld.wolfram.com/PyramidalFrustum.html
            $endgroup$
            – Seyed
            Oct 13 '16 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
            Oct 13 '16 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
            Oct 13 '16 at 14:44














          1












          1








          1





          $begingroup$

          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






          share|cite|improve this answer











          $endgroup$



          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







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Oct 14 '16 at 0:33

























          answered Oct 13 '16 at 13:09









          SeyedSeyed

          7,18641526




          7,18641526












          • $begingroup$
            I know how to find the top and bottom surface area, but I don't know how to find the volume
            $endgroup$
            – Graviton
            Oct 13 '16 at 14:33










          • $begingroup$
            @Graviton, mathworld.wolfram.com/PyramidalFrustum.html
            $endgroup$
            – Seyed
            Oct 13 '16 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
            Oct 13 '16 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
            Oct 13 '16 at 14:44


















          • $begingroup$
            I know how to find the top and bottom surface area, but I don't know how to find the volume
            $endgroup$
            – Graviton
            Oct 13 '16 at 14:33










          • $begingroup$
            @Graviton, mathworld.wolfram.com/PyramidalFrustum.html
            $endgroup$
            – Seyed
            Oct 13 '16 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
            Oct 13 '16 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
            Oct 13 '16 at 14:44
















          $begingroup$
          I know how to find the top and bottom surface area, but I don't know how to find the volume
          $endgroup$
          – Graviton
          Oct 13 '16 at 14:33




          $begingroup$
          I know how to find the top and bottom surface area, but I don't know how to find the volume
          $endgroup$
          – Graviton
          Oct 13 '16 at 14:33












          $begingroup$
          @Graviton, mathworld.wolfram.com/PyramidalFrustum.html
          $endgroup$
          – Seyed
          Oct 13 '16 at 14:38




          $begingroup$
          @Graviton, mathworld.wolfram.com/PyramidalFrustum.html
          $endgroup$
          – Seyed
          Oct 13 '16 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
          Oct 13 '16 at 14:40




          $begingroup$
          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
          Oct 13 '16 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
          Oct 13 '16 at 14:44




          $begingroup$
          I would also appreciate if you can (re)derive the volume formula in the wolfram page link; I can quite see it.
          $endgroup$
          – Graviton
          Oct 13 '16 at 14:44











          0












          $begingroup$

          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 $a(h)=kh^2implies k=frac{a(h)}{h^2}$ 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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I can understand your argument, but your $k$ is not determined; can you show how $k$ can be determined here?
            $endgroup$
            – Graviton
            Oct 14 '16 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
            Oct 14 '16 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
            Oct 14 '16 at 5:50
















          0












          $begingroup$

          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 $a(h)=kh^2implies k=frac{a(h)}{h^2}$ 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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I can understand your argument, but your $k$ is not determined; can you show how $k$ can be determined here?
            $endgroup$
            – Graviton
            Oct 14 '16 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
            Oct 14 '16 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
            Oct 14 '16 at 5:50














          0












          0








          0





          $begingroup$

          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 $a(h)=kh^2implies k=frac{a(h)}{h^2}$ 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.






          share|cite|improve this answer









          $endgroup$



          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 $a(h)=kh^2implies k=frac{a(h)}{h^2}$ 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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 13 '16 at 15:38









          SophieSophie

          1,6021526




          1,6021526












          • $begingroup$
            I can understand your argument, but your $k$ is not determined; can you show how $k$ can be determined here?
            $endgroup$
            – Graviton
            Oct 14 '16 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
            Oct 14 '16 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
            Oct 14 '16 at 5:50


















          • $begingroup$
            I can understand your argument, but your $k$ is not determined; can you show how $k$ can be determined here?
            $endgroup$
            – Graviton
            Oct 14 '16 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
            Oct 14 '16 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
            Oct 14 '16 at 5:50
















          $begingroup$
          I can understand your argument, but your $k$ is not determined; can you show how $k$ can be determined here?
          $endgroup$
          – Graviton
          Oct 14 '16 at 0:10




          $begingroup$
          I can understand your argument, but your $k$ is not determined; can you show how $k$ can be determined here?
          $endgroup$
          – Graviton
          Oct 14 '16 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
          Oct 14 '16 at 0:23




          $begingroup$
          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
          Oct 14 '16 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
          Oct 14 '16 at 5:50




          $begingroup$
          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
          Oct 14 '16 at 5:50


















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          Musa

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          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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