Scalar Line Integrals

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Section 10.2 Scalar Line Integrals

What if you want to determine the mass of a wire in the shape of the curve \(C\) if you know the density \(\lambda\text{?}\) The same procedure still works; chop and add. In this case, the length of a small piece of the wire is \(ds=|d\rr|\text{,}\) so its mass is \(\lambda\,ds\text{,}\) and the integral becomes

\begin{equation} m = \Lint \lambda \, ds\tag{10.2.1} \end{equation}

which can also be written as

\begin{equation} m = \Lint \lambda(\rr) \, |d\rr|\tag{10.2.2} \end{equation}

which emphasizes both that \(\lambda\) is not constant, and that \(ds\) is the magnitude of \(d\rr\text{.}\)

Another standard application of this type of line integral is to find the center of mass of a wire. This is done by averaging the values of the coordinates, weighted by the density \(\lambda\) as follows:

\begin{equation} \bar{x} = {1\over m} \Lint x\lambda(\rr) \, ds\tag{10.2.3} \end{equation}

with \(m\) as defined above. Similar formulas hold for \(\bar{y}\) and \(\bar{z}\text{;}\) the center of mass is then the point \((\bar{x},\bar{y},\bar{z})\text{.}\)