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The answer to my question is far beyond my knowledge (as I believe a basic knowledge of calculus is needed) -- so I'm hoping that I might be able to get someone to give me the answer here... My need for an answer has a practical reason -- I'm trying to calculate the ground-fill needed to level an rectangular, area that is currently sloped. (The gravel/fill company is unable to calculate the volume-value for me.)

This is the basic dimensions and description of the "object".
Starting with the top of the object (which is the resulting level plane of land), the dimensions are 20ft x 36ft (rectangular). The four corners have four varying heights to the existing ground level: h1=2ft, h2=.67ft (2/3), h3=0ft, and h4=1ft (the h-values are given in a clockwise sequence). The h sides are all considered to be perpendicular (90deg) to the top plane (the top of object) I'm hoping that this is enough information to calculate the volume (in cubic feet) between the top plane and the (estimated flat) bottom plane.

I've spent hours online looking for a calculator the would give me the result -- and was surprised that I couldn't find such a practical item. So, the result of this problem would be very much appreciated.

Edit1: the dimensions are actually 20x32 (not 20x36)

EDIT2: the distance between h1 and h2 is 20ft, and consequently the distance btw h2 and h3 is 32ft (not 36ft)

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migrated from mathematica.stackexchange.com Dec 4 '15 at 11:36

This question came from our site for users of Wolfram Mathematica.

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Probably:

Integrate[Interpolation[{{0, 0, 2}, {0, 36, 2/3}, {20, 36, 0}, {20, 0, 1}}][x, y],
          {x, 0, 20}, {y, 0, 36}]
(*660*)

If I got your coordinates right ...

Mathematica graphics

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  • $\begingroup$ I've added a correction and some details, to the question, So when your diagram is adj accordingly, it should be correct. $\endgroup$ – user36090 Dec 3 '15 at 21:22
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Belisarius, I used

ListPlot3D[{{0, 0, 2}, {0, 36, 2/3}, {20, 36, 0}, {20, 0, 1}}] 

enter image description here

to display the plot above. I assume that's what you did. So, the interpolation function represents the surface, and Integrate finds the volume between the plane and a {0,0,0} surface? I assume from this that you integrate between two curved surfaces, too. Would Integrate figure which volume you were trying to get?

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  • $\begingroup$ well, the interpolation degree is 1, so I guess it seems right $\endgroup$ – Dr. belisarius Dec 3 '15 at 23:19
  • $\begingroup$ A possible change may be needed (as in the edits found in question) -- the distance between h1(2) and h2(2/3) should be 20 and the distance between h2(2/3) and h3(0) should be 32 (not 36). And, I'm wanting the volume of the upper portion -- between the slope and the top plane. (in cubic feet) (I noticed after I had submitted the question, that the slope was not a flat plane, but a curved surface -- but thought it would have min effect.) $\endgroup$ – user36090 Dec 4 '15 at 4:56

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