 Posers and Puzzles

1. 11 Mar '05 12:34
many 3d forms of can be expressed as a 2d shapes elevated a given distance. their volumes are expressed as the base area times the hight.
for instance, cubes, cilinders et al. their volumes, expressed as functions of their base shapes are all V(A)=Ah

othere 3d forms can be experessed as a 2d shape rotated about an axis of symitry. for example sphears, cones, cilinders et al.
is there a universal function for the volume of such forms in terms of the rotated shape?
2. 11 Mar '05 13:494 edits
I don't know if this would be a "universal function" but here's the general solution:

Assuming the shape is bilaterally symmetrical about the line of revolution, you begin by cutting it in half along that line. Call that line y=0, and plot it on a 2 dimensional cartesian coordinate system such that the half shape rises above y=0. The whole shape would be found by mirroring the part with y>0 across the y=0 line.

Then, find the equation for the half shape.

For example, if the original shape were a 90-45-45 triangle revolved about the line that bisects the 90 degree angle, then the half shape would be another 90-45-45 triangle, with the 90 degree angle at the origin. The equation would be y=1-x (or you could do y=x, which would be easier). You take as a representative rectangle the area y times dx. Then you revolve this representative rectangle about the y=0 line to make a thin disk with thickness dx and circular area pi times y^2. The volume of this representative disk would be pi times y^2 times dx.

Then you integrate this volume from x=0 (where the shape touches the y axis) to x=1 (where the shape touches the y axis again).

In this case, the total volume would be the definite integral evaluated from 0 to 1 of pi times (1-x)^2 dx (or integral from 0 to 1 of pi times x^2 dx).

If the shape is not bilaterally symmetrical about the line of revolution, you need to superimpose the two halves over one another. For every x, use the y for whichever side has the greater y, as it will cover the other side at that point.

Summarized - find the area of a representative disk with radius from the line of rotation out to the edge of the shape. Integrate this area dx down the line. From the top of the shape to the bottom.

Does that make sense?
3. 11 Mar '05 16:08
is there a universal function for the volume of such forms in terms of the rotated shape?
V = 2pi*A*x

where x is the distance of the shape's centroid from the axis of revolution.
4. 11 Mar '05 20:29
both seem right, but Thunder i think has a bit more what i was looking for. I just dont know what a centroid is.
5. 11 Mar '05 21:35
I just dont know what a centroid is.
Non-mathematical definition: the geometric centroid of a 2D shape is the point on which it would balance if placed on a needle.
6. 12 Mar '05 04:39
Originally posted by THUDandBLUNDER
Non-mathematical definition: the geometric centroid of a 2D shape is the point on which it would balance if placed on a needle.
That doesn't work.

If you take a circle and rotate it about a line that passes through a diameter of the circle, then the distance from centroid to axis of revolution is zero. Therefore, you get a volume of zero for any sphere.
7. 12 Mar '05 12:31
Originally posted by AThousandYoung
That doesn't work.

If you take a circle and rotate it about a line that passes through a diameter of the circle, then the distance from centroid to axis of revolution is zero. Therefore, you get a volume of zero for any sphere.
Good point. OK, the axis should be 'external'.

In fact, the second theorem of Pappus states that the volume V of a solid of revolution generated by the revolution of a lamina about an external axis is equal to the product of the area A of the lamina and the distance d traveled by the lamina's geometric centroid.
http://mathworld.wolfram.com/PappussCentroidTheorem.html

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8. 12 Mar '05 21:06
Originally posted by THUDandBLUNDER
the second theorem of Pappus http://mathworld.wolfram.com/PappussCentroidTheorem.html
exelent!
9. 13 Mar '05 03:06
Originally posted by THUDandBLUNDER
Good point. OK, the axis should be 'external'.

In fact, the second theorem of Pappus states that the volume V of a solid of revolution generated by the revolution of a lamina about an external axis is equal to the product of the area A of the lamina and the distance d traveled by the lamina's geometric centroid.
http://mathworld.wolfram.com/PappussCentroidTheorem.html

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This is nice, but it doesn't answer the original question, which wants formulas for spheres, cones, etc.

othere 3d forms can be experessed as a 2d shape rotated about an axis of symitry. for example sphears, cones, cilinders et al.
is there a universal function for the volume of such forms in terms of the rotated shape?
10. 13 Mar '05 05:351 edit
is there a universal function for the volume of such forms in terms of the rotated shape

Originally posted by AThousandYoung
This is nice, but it doesn't answer the original question, which wants formulas for spheres, cones, etc.
I don't think his original question is well er...formed.
'Shape' is a function of what?
11. 13 Mar '05 11:21
His question is - if you rotate a two dimensional shape about an axis of symmetry (like the diameter of a circle) you get a three dimensional shape (like a sphere). Is there a general formula with which one can determine the volume of the 3d object knowing only what the 2d shape is and what the axis of rotation/symmetry is?
12. 13 Mar '05 21:552 edits
Originally posted by THUDandBLUNDER
I don't think his original question is well er...formed.
'Shape' is a function of what?
Well, a 2-manifold embedded in 3-space has a shape operator S, which is the differential of the normal to the surface. The shape operator is a 2x2 matrix which tells you exactly how the surface is curved at any point. For example, a sphere has shape operator I/r, where I is the identity and r is the radius. It'd be interesting to see a concise formula linking the shape operator with the volume enclosed by a compact manifold (if such a formula exists).

The shape operator of a surface of revolution is of a nice form (assuming I'm not mistaken): if we assume the curve f(x) to be rotated lives in the x-y plane, and we rotate about the x-axis, S = diag(k,-1/f(x)) where k is the intrinsic curvature of f at x, and S is given in the basis {t,b} (t = tangent of f, b = binormal of f).
13. 14 Mar '05 15:54
a single unified function based on Pappus's therom would be difficult, simpily because it isn't always easy to find the centroid. in practice, the functon in terms of the distance between the axis and the centroid as stated above will do.

it may not always be simple to find the centroid, but it also isn't alwas easy to find the area of a 2d shape.

it dosn't answer my question as stated, but it's just as good.