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Definition and properties.

Consider the curve $\mathcal{C}$ defined by the parametrization

\begin{displaymath}\overrightarrow{r} (t)= x(t)\overrightarrow{i} +y(t)\overrightarrow{j} +z(t)\overrightarrow{k} , \; a \leq t \leq b
\end{displaymath}

If the curve $\mathcal{C}$ passes through the domain $\mathcal{D}$ of a function w=f(x,y,z), the values of f on the curve $\mathcal{C}$ are given by w=f(x(t),y(t),z(t)). If we integrate this function of the variable t with respect to the arc length for t in the interval [a,b], we get the line integral of f on $\mathcal{C}$; it is denoted

\begin{displaymath}\int_{\mathcal{C}} f(x,y,z) \; ds
\end{displaymath}

We suppose that the curve $\mathcal{C}$ is smooth, i.e. for every value of t, the functions x, y and z are differentiable and nowhere all equal to 0. Denote $\overrightarrow{V} (t)= x'(t)\overrightarrow{i} +y'(t)\overrightarrow{j} +z'(t)\overrightarrow{k} $.

Theorem 8.1  

\begin{displaymath}\int_{\mathcal{C}} f(x,y,z) \; ds
= \int_{t=a}^{t=b} f(x(t),y...
...t)) \begin{vmatrix}
\overrightarrow{V} (t) \end{vmatrix} \; dt
\end{displaymath}

Example 8.2   Let f(x,y,z)=xy+yz2+x and let $\mathcal{C}$ be the segment joining the origin to the point A(1,2,1).

A parametrization of $\mathcal{C}$ is given by $\overrightarrow{r} (t)= t \overrightarrow{i} +
2t \overrightarrow{j} + t \overrightarrow{k} , \; 0 \leq t \leq 1$. Thus $ \overrightarrow{v} (t)= \overrightarrow{i} + 2 \overrightarrow{j} + \overrightarrow{k} $ and $\vert \overrightarrow{v} (t) \vert = \sqrt{6}$.

We have:
\begin{align*}\int_{\mathcal{C}} f(x,y,z) \; ds & =
\int_0^1 (t \cdot 2t + 2t \...
... \frac 12 t^4 + \frac 23 t^3 + \frac 12 t^2 \right]_0^1 = \frac 53.
\end{align*}

Example 8.3   We define a function by f(x,y,z)=z+y-z and consider the curve $\mathcal{C}$, dispalyed on Fig.  1, with the following parametrization:

\begin{displaymath}\overrightarrow{r} (t)= \cos t\overrightarrow{i} +\sin t\overrightarrow{j} +t\overrightarrow{k} , 0 \leq t \leq 2\pi
\end{displaymath}


  
Figure 1: A loop of an helix.
\begin{figure}
\mbox{\epsfig{file=boucle.eps,height=8cm} }
\end{figure}

Then:

\begin{displaymath}\overrightarrow{V} (t)= -\sin t\overrightarrow{i} + \cos t \overrightarrow{j} + \overrightarrow{k}\end{displaymath}

and
\begin{align*}\int_{\mathcal{C}} f(x,y,z) \; ds
& = \int_{t=0}^{t=2 \pi}
(\cos ...
...\cos t -\frac 12 t^2 \right]_0^{2 \pi}\sqrt{2}
=-2 \pi ^2 \sqrt{2}.
\end{align*}

Remark 8.4   After replacing ds by $\left\vert \overrightarrow{V} (t) \right\vert dt$, the integral becomes a regular integral of a function of one real variable on an interval.

Proposition 8.5       
1.
$\int_{\mathcal{C}} (f(x,y,z) + g(x,y,z) ) \; ds = \int_{\mathcal{C}} f(x,y,z)\;
ds + \int_{\mathcal{C}} g(x,y,z) \; ds$.
2.
$ \int_{\mathcal{C}} \alpha f(x,y,z) \; ds = \alpha \int_{\mathcal{C}} f(x,y,z)\;
ds$.
3.
If $\mathcal{C}= \mathcal{C_1} \cup \mathcal{C_2} \cup \dots \cup \mathcal{C_r}$, where each $\mathcal{C_i}$ is a smooth curve and the endpoint of $\mathcal{C_i}$ coincides with the startpoint of $\mathcal{C_{i+1}}$, then

\begin{displaymath}\int_{\mathcal{C}} f(x,y,z) \; ds = \underset{i=1}{\overset{r}{\sum}}\int_{\mathcal{C_i}} f(x,y,z) \; ds.
\end{displaymath}

Example 8.6   Let f(x,y,z)=x+y+2z; we compute the integral of f on the broken line $\mathcal{C}$ displayed on Fig.  2. This curve is the union of two line segments:
  
Figure 2: A ``broken'' path.
\begin{figure}
\mbox{\epsfig{file=path1.eps,height=4cm} }
\end{figure}

We have:

Finally, we have:

\begin{displaymath}\int_{\mathcal{C}} f(x,y,z) \; ds =
\int_{OA} f(x,y,z) \; ds + \int_{AB} f(x,y,z) \; ds
=\frac {5 \sqrt{13}}{2} +28.
\end{displaymath}


next up previous contents
Next: Mass of a wire, Up: Line integrals. Previous: Line integrals.
Noah Dana-Picard
2001-05-30
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