area of the duocylinder

Discussion of shapes with curves and holes in various dimensions.

area of the duocylinder

Postby PWrong » Sun Aug 27, 2006 10:55 am

I'm wondering how to compute the area of a duocylinder. Specifically, I'd like to be able to do it using calculus. I know how to calculate areas of a surface in 3D, and volumes of a 4D object, but an area in 4D is difficult, because we don't have a cross product.

I think all we have to do is find the area of a parallelogram, determined by two 4D vectors. Does anyone know how to do this?
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Postby bo198214 » Sun Aug 27, 2006 1:43 pm

Isnt the area of a parallelogram ah<sub>a</sub> where a is the length of a side, h<sub>a</sub> the hight perpendicular to a?

The hight h<sub>a</sub> is sqrt(|b|<sup>2</sup> - (<b,a>/|a|)<sup>2</sup>) so
A=|a||h<sub>a</sub>|=|a|sqrt(|b|<sup>2</sup> - (<b,a>/|a|)<sup>2</sup>) =sqrt(<b,b><a,a>-<b,a><sup>2</sup>)

But this can you already do yourself!

You can find the corresponding integral formula at wolfram surface area (the one with E, F and G).
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Postby PWrong » Sun Aug 27, 2006 2:22 pm

<b,a>

Is that a notation for the dot product?

It's nice that this formula holds even in 4D:
|axb|<sup>2</sup> + (a.b)<sup>2</sup> = |a|<sup>2</sup>|b|<sup>2</sup>

So the area of the duocylinder is 2pi r<sub>1</sub>*2pi r<sub>2</sub>.

Conjecture:
The content of the minimal form of the rotope AB is the product of the contents of the minimal forms of A and B.

To prove this, we'll need a formula for the content of a kD parallelopiped determined by k vectors in nD.
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Postby pat » Sun Aug 27, 2006 4:06 pm

PWrong wrote:
<b,a>

Is that a notation for the dot product?


Yes. It's a notation used for any inner product.

PWrong wrote:It's nice that this formula holds even in 4D:
|axb|<sup>2</sup> + (a.b)<sup>2</sup> = |a|<sup>2</sup>|b|<sup>2</sup>


?? assuming that by "|axb|<sup>2</sup>" you just mean the area rather than actually a cross-product.... or, what sort of cross-product did you have in mind? the quaternion product?
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Postby PWrong » Sun Aug 27, 2006 4:37 pm

I just meant the area. Even though there's no cross product, if we pretend there is we can still take its magnitude.

We could use the same formula to integrate a scalar field over the duocylinder, or any surface in 4D, and I think it would be easy to find formulae to integrate a scalar or vector field over a 3D or 4D region in 4D. But how could we go about integrating a vector field F over a duocylinder? The integrand for a surface in 3D is F.(r<sub>u</sub>xr<sub>v</sub>) dudv.

So in general, do we just integrate the volume of the parallelepiped determined by F, r<sub>u</sub> and r<sub>v</sub>?

If so, then that's all the integrals we can do in 4D :D. The only other thing we need is a 4D version of Stoke's theorem or the Divergence theorem.
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Postby bo198214 » Sun Aug 27, 2006 8:21 pm

Its an old hat. The volume of the parallelepiped is
|det(a<sub>1</sub>,...,a<sub>n</sub>)|
for spanning vectors a<sub>1</sub>,...,a<sub>n</sub>.
For integration you set a<sub>i</sub> = df/dt<sub>i</sub> i think and take the integral over |det(a<sub>1</sub>,...,a<sub>n</sub>)|.
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Postby PWrong » Fri Sep 01, 2006 3:05 pm

The 4D divergence theorem should be pretty simple. It would be:

Integral(div F) dB over B = Integral(F) dV over V
where V is the boundary of B

I don't know about stoke's theorem. Let S be a 2D surface in 4D with a simple closed boundary C. I can't think of a nice example.
We need an integral of F over C to be equal to an integral of something over S. In 3D, we have (Del x F).dS.

But in 4D, (Del ^ F) and dS are both tensors. Can we multiply them to get a scalar? :?
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Postby bo198214 » Fri Sep 01, 2006 9:41 pm

Stoke's theorem already exists in multidimensional form.
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Postby PWrong » Sat Sep 02, 2006 7:15 am

I know, but I only understand the 3D case so far. I'm trying to learn the extention of 3D vector calculus to 4D, preferably without using tensors too much.
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Postby wendy » Sat Sep 02, 2006 7:47 am

One might note that the area of the margin of the duocylinder is:

integral (x= 0-2pi), integral (y= 0-2pi) ×(r dx dy)

where r is a vector from the origin to the point P.

Even though r changes, the dot product is constant, and r is normal to this product. This is then the simple product of this, being 4pi.

One can evaluate the volume of any body by noting the equation,

volume = moment of surface,

This is then volume = moment of the gradient of density (which assigns a vector to the surface).

You can then evaluate the surface by a volume-like argument, and so forth. The idea is the basis of the out-vector and endoanalysis in the polygloss.
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the dream we dream together is reality.
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