Section 10.6 Scalar Surface Integrals
Consider again the example in Section 11.1, which involved the part of the plane \(x+y+z=1\) which lies in the first quadrant. Suppose you want to find the average height of this triangular region above the \(xy\)-plane. To do this, chop the surface into small pieces, each at height \(z=1-x-y\text{.}\) In order to compute the average height, we need to find
\begin{equation}
\hbox{avg height} = \frac{1}{\hbox{area}} \Sint z \,\dA\tag{10.6.1}
\end{equation}
where the total area of the surface can be found either as
\begin{equation}
\hbox{area} = \Sint \dA\tag{10.6.2}
\end{equation}
or from simple geometry. So we need to determine \(\dA\text{.}\) But we already know \(d\AA\) for this surface from (11.1.5)! It is therefore straightforward to compute
\begin{equation}
\dA = |d\AA| = |\xhat+\yhat+\zhat|\,dx\,dy = \sqrt{3} \,dx\,dy\tag{10.6.3}
\end{equation}
and therefore
\begin{equation}
\hbox{avg height}
= \frac{1}{\sqrt{3}/2} \int_0^1 \int_0^{1-x} (1-x-y) \sqrt{3}\,dy \,dx
= \frac13 .\tag{10.6.4}
\end{equation}