4.12 Oriented Surfaces

If it is possible to choose a unit normal vector \(\mathbf n\) at every point \((x,y,z)\) of a surface \(S\), varying continuously over \(S\), then \(S\) is called an oriented surface, and that choice of \(\mathbf n\) gives \(S\) an orientation. Every orientable surface has two possible orientations. (Not every surface is orientable: the Möbius strip has only one side.)

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For a surface \(S\) given by \(z = g(x,y)\), the natural orientation is given by \[\mathbf n = \frac {-\frac {\partial g}{\partial x} \,\mathbf i - \frac {\partial g}{\partial y}\,\mathbf j + \mathbf k}{\sqrt {1 + \left (\frac {\partial g}{\partial x}\right )^2 + \left (\frac {\partial g}{\partial y}\right )^2}} .\] Since the \(\mathbf k\)-component is positive, this gives the upward orientation of the surface.

If \(S\) is a smooth orientable surface given by a vector function \(\mathbf r(u,v)\), it is automatically supplied with the orientation of the unit normal vector \[\mathbf n=\frac {\mathbf r_u \times \mathbf r_v}{|\mathbf r_u \times \mathbf r_v|} ,\] and the opposite orientation is given by \(-\mathbf n\), or by interchanging \(u\) and \(v\). For a closed surface, the boundary of a solid region, the convention is that the positive orientation is the one for which the normal vectors point outward from the solid; inward-pointing normals give the negative orientation. The spherical parametrisation of the sphere has the positive orientation.

Example 4.12.1. Find the outward flux of \(\mathbf F = x\,\mathbf i + 2y\,\mathbf j + 3z\,\mathbf k\) through the surface of the cube \([0,1]^3\).

Solution. On the face \(x = 1\) the outward normal is \(\mathbf i\) and \(\mathbf F\cdot \mathbf n = x = 1\), giving flux \(1\); on \(x = 0\) the normal is \(-\mathbf i\) and \(\mathbf F\cdot \mathbf n = -x = 0\). Likewise the faces \(y = 1\) and \(z = 1\) contribute \(2\) and \(3\), and the other three faces nothing. The total outward flux is \(6\).

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