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Area element, volume element and matrix

  1. Mar 6, 2014 #1
  2. jcsd
  3. Mar 6, 2014 #2


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    Suppose you have a surface parametrized as
    Then the surface-element normal vectors are given by
    [tex]\mathrm{d}^2 \vec{F}=\frac{\partial \vec{x}}{\partial u} \times \frac{\partial \vec{x}}{\partial v} \mathrm{d} u \mathrm{d} v.[/tex]
    This is clear from the geometric meaning of the vector product as a (axial) vector with magnitude of the area of the parallelogram spanned by the two vectors and directed perpendicular to the surface with the orientation given by the right-hand rule.

    In the same way, from the triple product giving the volume of a parallelipiped spanned by three vectors
    [tex]\mathrm{d}^3 \vec{x}=\mathrm{d} u \mathrm{d} v \mathrm{d} w \left (\frac{\partial \vec{x}}{\partial u} \times \frac{\partial \vec{x}}{\partial v} \right ) \cdot \frac{\partial \vec{x}}{\partial w},[/tex]
    where [itex](u,v,w)[/itex] are arbitrary "generalized coordinates". This is of course identical with the Jacobian determinant of the transformation law from generalized to Cartesian coordinates,
    [tex]\mathrm{d}^3 \vec{x} =\mathrm{d} u \mathrm{d} v \mathrm{d} w \det \left (\frac{\partial(x,y,z)}{\partial(u,v,w)} \right ).[/tex]
  4. Mar 6, 2014 #3


    Staff: Mentor

    Actually, the formula shows a determinant.
    The formula gives dA, not d2A. The reason it is not shown as dxdy is that the area is for a region that is approximately a sector of a circle, rather than for a rectangular area element. In the drawing in the wiki article, the coordinates of point M' are (x + dx, y + dy) and the coordinates of M are (x, y). If you connect point M with a segment perpendicular to OM', you get something that is nearly a right triangle. The area of this triangle would be approximately (1/2)dx dy, so it's clear that the area of the sector can't be dx dy.
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