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Surface Volume in 4-d graph: Euclidean Geometry Question

  1. Oct 30, 2005 #1
    "Surface Volume" in 4-d graph: Euclidean Geometry Question

    Suppose you have a smooth parametrically defined volume V givin by the following equation.


    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 the angle theta is arccosine((ru*rv)/(|ru||rv|)),
    and the angle phi is arccosine((ru*rs)/(|ru||rs|).

    Does this sound accurate?

    Inquisitively,

    Edwin
     
  2. jcsd
  3. Nov 1, 2005 #2
    One other thought.

    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

    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 "Volume" over E is:


    the tripple integral over E of |ru*(rv X rs)|dV

    This is based on the parallelpiped formed by ru, rv, rs, which represent the change in volume of the solid parametrically defined in 3-space. Here the cross product works because we have three dimensional vectors defining our volume. It is sort of an analogue to the typical method of finding surface area of a smooth 3-d function in three dimensional space when the surface is traversed exactly once as (u,v) vary throughout the domain D, and the vectors partial derivitaves ru, and rv are continuous and exist over the entire domain D. This surface area is the

    double integral over D |ru X rv|dA

    Where

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

    and ru is the partial derivitive of r(u,v) with respect to u, while holding v constant.

    and rv is the partical derivitive of r(u,v) with respect to v, while holding u constant.

    Does this seem accurate?

    Inquisitively,

    Edwin
     
  4. Nov 2, 2005 #3
    Tested it out, the technique in the second post works, and interestingly enough, doesn't require the jacobian correction factor! I'll give specific details on my next post by deriving the equation for the sphere using spherical coordinates and integrating over the E using the method above...

    Until then!

    Good Day!

    Best Regards,

    Edwin

    p.s. The function "r(u,s,v) = x(u,v,s)i + y(u,v,s)j +z(u,v,s)k" should read
    "r(u,v,s) = x(u,v,s)i + y(u,v,s)j +z(u,v,s)k."
     
    Last edited: Nov 2, 2005
  5. Nov 3, 2005 #4
    The triple scalar product of differentials in your volume element is the Jacobian of r.
     
  6. Nov 3, 2005 #5
    Ah-ha!

    That explains why it works!

    So it's really just integrating the function 1, and the |ru*(rv X rs)| is the jacobian correction factor! Got it!

    Thanks Hypermorphism, I appreciate your clarification on that. By the way, do you know whether the first post is correct?

    Inquisitively,

    Edwin
     
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