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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}
