Surface Area/Volume Calc Help Needed

[MKSheppard]

Okay, is this a better one?

Height 0.6 meters
Base Length (lo) 2.4 meters
Base Width (wo) 2.4 meters
Top Length (l1) 1.54 meters
Top Width (w1) 1.62 meters

Volume 2 m3

[MKSheppard]

Am I correct in thinking as long as the Lengths and Widths are parallel with each other; slope differences, like 55 deg front, and 30 deg back are not needed when you are doing volume only?

[Gil Hamilton]

Am I correct in thinking as long as the Lengths and Widths are parallel with each other; slope differences, like 55 deg front, and 30 deg back are not needed when you are doing volume only?

Yes and that's the beauty of calculus.

Mike gave you a general equation for that shape, so long as the lengths and widths are parallel. It makes sense if you look at what actually happened in the integration. In plain English terms, he noted that the cross-section of the object was a rectangle and then added up all the cross-sections of the object. Kind of like stacking up infinitely thin sheets to build the object. As long as the top sheet is the same height from the bottom most sheet and the edges are parallel, no matter where it is, you're using the same sheets. So, same volume.

That's integration in a nutshell. You are adding up an infinite number of infinitely small bits.

[Darth Holbytlan]

Height 0.6 meters
Base Length (lo) 2.4 meters
Base Width (wo) 2.4 meters
Top Length (l1) 1.54 meters
Top Width (w1) 1.62 meters

Volume 2 m3

Should be about 2.4 m^3 (2.40936 to be exact).

[sketerpot]

Am I correct in thinking as long as the Lengths and Widths are parallel with each other; slope differences, like 55 deg front, and 30 deg back are not needed when you are doing volume only?

Yes and that's the beauty of calculus.

Mike gave you a general equation for that shape, so long as the lengths and widths are parallel. It makes sense if you look at what actually happened in the integration. In plain English terms, he noted that the cross-section of the object was a rectangle and then added up all the cross-sections of the object. Kind of like stacking up infinitely thin sheets to build the object. As long as the top sheet is the same height from the bottom most sheet and the edges are parallel, no matter where it is, you're using the same sheets. So, same volume.

To expand on this, the length and width of the rectangles depend on the slopes of two lines added together. Since this is all linear, Mike was able to use simple linear interpolation to get length and width for any value of z between 0 and h.

I mention this because it's the only clever part of the derivation, and my brother was having trouble with this exact same concept on some of his calculus homework.

[Kuroneko]

Addendum: I do agree that if it isn't [the length is non-uniform and/or sides not perpendicular] then one has to use Integration however, especially if it isn't a symmetry shape.

That's not quite true, although calculus is still a good idea. The formula for a prismatoid, V = (h/6)[A0 + 4M + A1], where M is the area of the middle cross-section (z=h/2) and A0,A1 of the top and bottom faces, respectively, can be derived geometrically, and is in fact true regardless of whether the top and bottom are rectangular or not.

A rather surprising result that calculus adds, however, is that the exact same relation holds also if the cross-sections vary as a cubic polynomial of the height (instead of quadratic as in the OP and in other prismatoids), because it simply Simpson's rule, which has an error term of order d^4f (=0 for cubics).

[MKSheppard]

Should be about 2.4 m^3 (2.40936 to be exact).

Are you sure on this?

Checking my formula in Excel, and making the result cell show 2 place decimals gives me:

Height 0.6 m
Base Length (lo) 2.4 m
Base Width (wo) 2.4 m
Top Length (l1) 1.54 m
Top Width (w1) 1.62 m

Volume 1.86 m3

[Kuroneko]

MATLAB:

Code: Select all

» syms l0 w0 l1 w1 h; V = ((l1*w0+l0*w1)/2 + (l0-l1)*(w0-w1)/3)*h;
» subs(V,{h,l0,w0,l1,w1},{0.6,2.4,2.4,1.54,1.62})

ans =

    2.4094

If you're having trouble using that the formula, you might want to try V = (h/6)*[ l0w0 + lw + l1w1 ], where l = (l0+l1) and w = (w0+w1) are the combined length and width. It's the same thing, just rearranged.

[Darth Wong]

Should be about 2.4 m^3 (2.40936 to be exact).

Are you sure on this?

Checking my formula in Excel, and making the result cell show 2 place decimals gives me:

Height 0.6 m
Base Length (lo) 2.4 m
Base Width (wo) 2.4 m
Top Length (l1) 1.54 m
Top Width (w1) 1.62 m

Volume 1.86 m3

Maybe your Excel formula is mal-formed.

[MKSheppard]

Maybe your Excel formula is mal-formed.

This is the formula:

(((Top Length*Base Width)+(Top Length*Top Width))/2)*Height

[Kuroneko]

That formula is valid iff l0 = l1 or w0 = w1, which is not the case for your parameters. (Thanks to Mr. Wong for the correction.)

[MKSheppard]

aw crap

[Darth Wong]

Shep, the correct formula to use is the one just above the line "This is a rather ugly formula" in my calculus proof. The other two are special cases.

[Darth Holbytlan]

That formula is valid iff l0 = l1 or w0 = w1, which is not the case for your parameters. (Thanks to Mr. Wong for the correction.)

No, that formula is only valid when l0 = l1. w0 = w1 requires a different, but similar formula: (((Base Length*Top Width)+(Top Length*Top Width))/2)*Height

[Kuroneko]

Oh, yes. I accidentally read that as
(((Top Length*Base Width)+(Base Length*Top Width))/2)*Height,
which is correct in either of those two cases.

[MKSheppard]

Right fixed it; thanks