# Area element, volume element and matrix

1. Mar 6, 2014

2. Mar 6, 2014

### vanhees71

Suppose you have a surface parametrized as
$$\vec{x}=\vec{x}(u,v).$$
Then the surface-element normal vectors are given by
$$\mathrm{d}^2 \vec{F}=\frac{\partial \vec{x}}{\partial u} \times \frac{\partial \vec{x}}{\partial v} \mathrm{d} u \mathrm{d} v.$$
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
$$\mathrm{d}^3 \vec{x}=\mathrm{d} u \mathrm{d} v \mathrm{d} w \left (\frac{\partial \vec{x}}{\partial u} \times \frac{\partial \vec{x}}{\partial v} \right ) \cdot \frac{\partial \vec{x}}{\partial w},$$
where $(u,v,w)$ are arbitrary "generalized coordinates". This is of course identical with the Jacobian determinant of the transformation law from generalized to Cartesian coordinates,
$$\mathrm{d}^3 \vec{x} =\mathrm{d} u \mathrm{d} v \mathrm{d} w \det \left (\frac{\partial(x,y,z)}{\partial(u,v,w)} \right ).$$

3. Mar 6, 2014

### 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.