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Integration over Surfaces

Integration of a function over a surface is a straightforward generalization of $\int \int f(x,y) dx dy = \int f(x,y) dA$ . The set of all little rectangles $dx dy$ defines a planar surface. A non-planar surface $\vec{x}(u,v)$ is composed of a set of little parallelogram patches with sides given by the infinitesimal vectors

$$\begin{split}\vec{r_u} du = & \ensuremath{\frac{\partial{\vec... ...nsuremath{\frac{\partial{\vec{x}}}{\partial{v}}} dv \end{split}$$ (15-2)

Because the two vectors $\vec{r_u}$ and $\vec{r_v}$ are not necessarily perpendicular, their cross-product is needed to determine the magnitude of the area in the parallelogram:

$$dA = \norm {\vec{r_u} \times \vec{r_v}} du dv$$ (15-3)

and the integral of some scalar function, $g(u,v) = g(x(u,v),y(u,v)) = g(\vec{x}(u,v))$ , on the surface is

$$\int g(u,v) dA = \int \int g(u,v) \norm {\vec{r_u} \times \vec{r_v}} du dv$$ (15-4)

However, the operation of taking the norm in the definition of the surface patch $dA$ indicates that some information is getting lost--this is the local normal orientation of the surface. There are two choices for a normal (inward or outward).

When calculating some quantity that does not have vector nature, only the magnitude of the function over the area matters (as in Eq. 15-4). However, when calculating a vector quantity, such as the flow through a surface, or the total force applied to a surface, the surface orientation matters and it makes sense to consider the surface patch as a vector quantity:

$$\begin{split}\vec{A}(u,v) & = \norm {\vec{A}} \hat{n}(u,v) = ... ...{n}(u,v)\ d\vec{A} & =\vec{r_u} \times \vec{r_v}\ \end{split}$$ (15-5)

where $\hat{n}(u,v)$ is the local surface unit normal at $\vec{x}(u,v)$ .

MATHEMATICA$^{\text{\scriptsize {\textregistered }}}$ Example

(notebook Lecture-15)

(html Lecture-15)

(xml+mathml Lecture-15)

Integrals of Anisotropic Surface Energy The surface energy of single crystals often depends on the surface orientation. This is especially the case for materials that have covalent and/or ionic bonds.

To find the total surface energy of such a single crystal, one has to integrate an orientation-dependent surface energy over the surface of a body.

This example compares the total energy of such an anisotropic surface energy integrated over a sphere and a cube that enclose the same volume.

1. Enter the parametric equation of the sphere in terms of longitude $v \in (0, 2 \pi)$ and latitude $u \in (-\pi/2, \pi/2)$ .
2. Calculate the tangent plane vectors $\vec{r_u}$ and $\vec{r_v}$
3. Calculate their cross product for subsequent use in the surface integral:
   $$\int \int g(\hat{n}(u,v)) \norm {\vec{r_u} \times \vec{r_v}} du dv$$

4. Use the vectors to calculate the local unit normal.
5. Integrate a model function:
   $$\gamma_{anis}(\hat{n}) = 1 + \gamma_{111} n_1^2 n_2^2 n_3^3$$
   over a sphere
6. Compare this result to the integration over a cube with the same volume.

The above calculation compares two fixed shapes--to find the surface which has the least energy for enclosing a given volume, one would employ a construction known as the Wulff Theorem.