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Section 7.6 Scalar Surface Integrals

Consider again the example in Section 12.1, 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{7.6.1} \end{equation}

where the total area of the surface can be found either as

\begin{equation} \hbox{area} = \Sint \dA\tag{7.6.2} \end{equation}

or from simple geometry. So we need to determine \(\dA\text{.}\) But we already know \(d\AA\) for this surface from (12.1.5)! It is therefore straightforward to compute

\begin{equation} \dA = |d\AA| = |\xhat+\yhat+\zhat|\,dx\,dy = \sqrt{3} \,dx\,dy\tag{7.6.3} \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{7.6.4} \end{equation}