The line integral of the scalar $f\left(x\,y\,z\right)$ along a path described parametrically by $x\=x\left(t\right)\,y\=y\left(t\right)\,z\=z\left(t\right)$, $t\in \left[a\,b\right]$, is given by

${\int }_a^{b}f\left(x\left(t\right)\,y\left(t\right)\,z\left(t\right)\right) \frac{\mathrm{ds}}{\mathrm{dt}} \mathrm{dt}$

where $s$ is arc length, so $\frac{\mathrm{ds}}{\mathrm{dt}}\=\sqrt{{\stackrel{\.}{x}}^2\+{\stackrel{\.}{y}}^2\+{\stackrel{\.}{z}}^2}\=\mathrm{\ρ}$ = $∥\stackrel{\.}{\mathbf{R}}∥$, with $\mathbf{R}\left(t\right)\=x\left(t\right) \mathbf{i}\+y\left(t\right) \mathbf{j}\+z\left(t\right) \mathbf{k}$ being the vector form of the parametric representation of the path. However, there are two linear segments for the polygonal line.

A parametric representation of the first line segment is

$\mathbf{R}\=\left[\begin{array}{c}x\(t\)\\ y\(t\)\\ z\(t\)\end{array}\right]\=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\+t \(\left[\begin{array}{c}5\\ 3\\ 2\end{array}\right]-\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\)$= $\left[\begin{array}{c}1\+4 t\\ 2\+ t\\ 3-t\end{array}\right]\,0\le t\le 1$ 

so that

 $\mathrm{\ρ}\=\frac{\mathrm{ds}}{\mathrm{dt}}\=\sqrt{{\left(\frac{d}{\mathrm{dt}}\left(1\+4 t\right)\right)}^2\+{\left(\frac{d}{\mathrm{dt}}\left(2\+ t\right)\right)}^2\+{\left(\frac{d}{\mathrm{dt}}\left(3-t\right)\right)}^2}$ = $\sqrt{4^2\+1^2\+1^2}\=3\sqrt{2}$ 

and the line integral along this segment is given by

${\int }_0^1\left(1\+4 t\right) \left(2\+ t\right) \left(3-t\right) \left(3 \sqrt{2}\right) \mathrm{dt}\=3\sqrt{2} {\int }_0^1\left(-4t^3\+3t^2\+25t\+6\right) \mathrm{dt}\=111\/\sqrt{2}$ 

A parametric representation of the second line segment is

$\mathbf{R}\=\left[\begin{array}{c}x\(t\)\\ y\(t\)\\ z\(t\)\end{array}\right]\=\left[\begin{array}{c}5\\ 3\\ 2\end{array}\right]\+t \(\left[\begin{array}{c}0\\ 1\\ 4\end{array}\right]-\left[\begin{array}{c}5\\ 3\\ 2\end{array}\right]\)$= $\left[\begin{array}{c}5-5 t\\ 3-2 t\\ 2\+2 t\end{array}\right]\,0\le t\le 1$ 

so that

 $\mathrm{\ρ}\=\frac{\mathrm{ds}}{\mathrm{dt}}\=\sqrt{{\left(\frac{d}{\mathrm{dt}}\left(5-5 t\right)\right)}^2\+{\left(\frac{d}{\mathrm{dt}}\left(3-2 t\right)\right)}^2\+{\left(\frac{d}{\mathrm{dt}}\left(2\+2 t\right)\right)}^2}$ = $\sqrt{5^2\+2^2\+2^2}\=\sqrt{33}$ 

and the line integral along this segment is given by

${\int }_0^1\left(5-5 t\right) \left(3-2 t\right) \left(2\+2 t\right) \sqrt{33} \mathrm{dt}\=\sqrt{33} {\int }_0^{1}10\left(2 t^3-3 t^2-2 t\+3\right) \mathrm{dt}\=15\sqrt{33}$ 

Thus, the line integral along the polygonal path is the sum

$\frac{111}{\sqrt{2}}\+15\sqrt{33}$ ≐ $164.66$ 

Clearly, integrating along a polygonal line is equivalent to integrating along each segment of the path, and adding the resulting integrals.

A solution from first principles is also possible, but the bulk of the work consists in obtaining parametric representations of each line segment.

Now write and evaluate two integrals of the form ${\int }_0^{1}f\left(x\left(t\right)\,y\left(t\right)\,z\left(t\right)\right) \mathrm{\ρ} \mathrm{dt}$.