Suppose you have a smooth parametrically defined volume V givin by the following equation.(adsbygoogle = window.adsbygoogle || []).push({});

f(x,y,z,w)= r(u,s,v) = x(u,v,s)i + y(u,v,s)j +z(u,v,s)k + w(u,v,s)l

Consider the vectors ru=dr/du, where dr/du is the partial derivitive of r with respect to the parameter u. Similarly, rv = dr/dv, rs=dr/ds are the partial derivitives of r with respect to the parameters v and s, respectively.

I presume that ru(u0,v0,s0), rv(u0, v0, s0), and rs(u0,v0,s0) form a three dimensional parallelpiped that represents the rate of change of the three dimensional surface volume in four dimensional Euclidean space.

The angle theta between ru and rv is arccosine((ru*rv)/(|ru||rv|))

Similarly, the angle phi between ru and rs is arccosine((ru*rs)/(|ru||rs|).

The height h1 of the parallelogram p1 formed by ru and rv is the magnitude of the projection of rv onto the perpindicular of p1 and is equal to

|rv|sin(theta). The area A1 of this parallelogram is

|ru|*h1 = |ru||rv|sin(theta).

The height h2 of the parallelogram p2 formed by ru and rs is the magnitude of the projection of rs onto the perpindicular of p2, and is equal to |rs|sin(phi)

The area of A2 of this parallelogram is |ru|*h2 = |ru||rs|sin(phi).

The volume of the parallelpiped V = the area of either of the parallelograms times the height of the other parallelgram.

A2*(h1) = A1*(h2) =|ru||rv||rs||sin(phi)*sin(theta)|

Based on this, I conjecture the following:

If a smooth parametrically defined volume V is givin by the following equation:

r(u,s,v) = x(u,v,s)i + y(u,v,s)j +z(u,v,s)k + w(u,v,s)l

Where (u,s,v) are elements of E, and

V is covered just once as (u,v,s) varies throughout the parameter domain E, then the "Surface Volume" is

the tripple integral over E of =|ru||rv||rs||sin(phi)*sin(theta)|dV

Where ||sin(phi)*sin(theta)|is a constant.

Does this sound accurate?

Inquisitively,

Edwin

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# Euclidean Geometry Question

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