1 An example of a surface integral

An engineer involved with the construction of a dam to hold back the water in a reservoir needs to be able to calculate the total force the water exerts on the dam so that the dam is built with sufficient strength.

In order to calculate this force, two results are required:

1. The pressure $p$ of the water is proportional to the depth. That is

   $p=kd$ (1)

   where $k$ is a constant.

2. The force on an area subjected to constant pressure is given by

   force $=$ pressure $\times$ area (2)

The diagram shows the face of the dam. The depth of water is $h$ and $\delta A$ is a small area in the face of the dam with coordinates $\left(x,y\right)$ .

Figure 1

Using (1), the pressure at $\delta A\sim k\left(h-y\right)$ . Using (2), the force on an area $\delta A\sim k\left(h-y\right)\delta A$ .

Both of these expressions are approximate as $y$ is slightly different at the top of $\delta A$ to the bottom.

Now

For a better approximation let $\delta A$ become smaller, and for the exact result find the limit as $\delta A\to 0$ . Then

where ${\int \; <!--nolimits\; -->}_{A}k\left(h-y\right)dA$ stands for the surface integral of $k\left(h-y\right)$ over the area $A$ . Surface integrals are evaluated using double integrals . The following Section shows a double integral being developed in the case of the volume under a surface.