Surface Integrals

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

$$\int\!\int_S f(x,y,z)\, dS = \lim_{\delta S_i\rightarrow 0}\sum_i f(x,y,z)\, \delta S_i$$ (A.84)

is a surface integral. For instance, the volume of water in a lake of depth $D(x,y)$ is

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

$$\left[\int_{x_1}^{x_2} D(x,y)\,dx\right]dy.$$ (A.86)

Note that the limits $x_1$ and $x_2$ depend on $y$ . The total volume is the sum over all strips: that is,

$$V = \int_{y_1}^{y_2} dy\left[\int_{x_1(y)}^{x_2(y)} D(x,y)\,dx\right] \equiv \int\!\int_S D(x,y)\,dx\,dy.$$ (A.87)

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

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

For example, consider

$$\int\!\int_S x \,y^{\,2}\,dx\,dy,$$ (A.89)

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

$$dy\left(\int_0^{1-y} x\, y^{\,2}\,dx\right) = y^{\,2}\,dy\left[ \frac{x^{\,2}}{2}\right]^{1-y}_0 = \frac{y^{\,2}}{2}\,(1-y)^2\,dy.$$ (A.90)

Let us now evaluate the $y$ integral:

$$\int_0^1 \left(\frac{y^{\,2}}{2}-y^3+ \frac{y^{\,4}}{2}\right)dy = \frac{1}{60}.$$ (A.91)

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

$$\int_0^1 x\,dx \int_0^{1-x} y^{\,2}\,dy = \int_0^1\frac{x}{3}\, (1-x)^3\,dx = \frac{1}{60}.$$ (A.92)

Figure A.18: An example surface integral.

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

$$\int\int_S x^{\,2} \,y\,dx\,dy$$ (A.93)

where $S$ is a unit square. This integral can be written

$$\int_0^1 dx \int_0^1 x^{\,2} \,y\,dy = \left(\int_0^1 x^{\,2}\,dx\right) \left(\int_0^1 y \,dy\right) = \frac{1}{3}\,\frac{1}{2} = \frac{1}{6},$$ (A.94)

because the limits are both independent of the other variable.