What if you want to determine the mass of a wire in the shape of the curve $C$ if you know the density $\lambda$? The same procedure still works; chop and add. In this case, the length of a small piece of the wire is $ds=|d\rr|$, so its mass is $\lambda\,ds$, and the integral becomes \begin{equation} m = \Lint \lambda \, ds \end{equation} which can also be written as \begin{equation} m = \Lint \lambda(\rr) \, |d\rr| \end{equation} which emphasizes both that $\lambda$ is not constant, and that $ds$ is the magnitude of $d\rr$.

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