Surface Integrals

Let us take a surface , that is not necessarily co-planar, and divide it up into (scalar) elements $\delta S_i$ . Then

$\displaystyle \int\!\int_S f(x,y,z)\, dS = \lim_{\delta S_i\rightarrow 0}\sum_i f(x,y,z)\, \delta S_i$

(1.84)

is a surface integral. For instance, the volume of water in a lake of depth is

$\displaystyle V= \int\!\int D(x,y)\,dS.$

(1.85)

To evaluate this integral, we must split the calculation into two ordinary integrals. The volume in the strip shown in Figure A.17 is

$\displaystyle \left[\int_{x_1}^{x_2} D(x,y)\,dx\right]dy.$

(1.86)

Note that the limits and depend on . The total volume is the sum over all strips: that is,

$\displaystyle V = \int_{y_1}^{y_2} dy\left[\int_{x_1(y)}^{x_2(y)} D(x,y)\,dx\right] \equiv \int\!\int_S D(x,y)\,dx\,dy.$

(1.87)

Of course, the integral can be evaluated by taking the strips the other way around: that is,

$\displaystyle V = \int_{x_1}^{x_2} dx \int_{y_1(x)}^{y_2(x)} D(x,y)\,dy.$

(1.88)

Interchanging the order of integration is a very powerful and useful trick. But great care must be taken when evaluating the limits.

**Figure A.17:** Decomposition of a surface integral.
$\includegraphics[width=0.6\textwidth]{AppendixA/figA_17a.eps}$

For example, consider

$\displaystyle \int\!\int_S x \,y^{\,2}\,dx\,dy,$

(1.89)

where is shown in Figure A.18. Suppose that we evaluate the integral first:

$\displaystyle dy\left(\int_0^{1-y} x\, y^{\,2}\,dx\right) = y^{\,2}\,dy\left[ \frac{x^{\,2}}{2}\right]^{1-y}_0 = \frac{y^{\,2}}{2}\,(1-y)^2\,dy.$

(1.90)

Let us now evaluate the integral:

$\displaystyle \int_0^1 \left(\frac{y^{\,2}}{2}-y^3+ \frac{y^{\,4}}{2}\right)dy = \frac{1}{60}.$

(1.91)

We can also evaluate the integral by interchanging the order of integration:

$\displaystyle \int_0^1 x\,dx \int_0^{1-x} y^{\,2}\,dy = \int_0^1\frac{x}{3}\, (1-x)^3\,dx = \frac{1}{60}.$

(1.92)

In some cases, a surface integral is just the product of two separate integrals. For instance,

$\displaystyle \int\int_S x^{\,2} \,y\,dx\,dy$

(1.93)

where is a unit square. This integral can be written

$\displaystyle \int_0^1 dx \int_0^1 x^{\,2} \,y\,dy = \left(\int_0^1 x^{\,2}\,dx\right) \left(\int_0^1 y \,dy\right) = \frac{1}{3}\,\frac{1}{2} = \frac{1}{6},$

(1.94)

because the limits are both independent of the other variable.

**Figure A.18:** An example surface integral.
$\includegraphics[width=0.5\textwidth]{AppendixA/figA_18a.eps}$