91Ó°ÊÓ

Assume a vector field $\mathbf{F}(x,y,z)={\mathrm{F}}_1\mathbf{i}+{\mathrm{F}}_2\mathbf{j}+{\mathrm{F}}_3\mathbf{k}$with component functions that are each continuous on $C$which is defined in $R^3$with a smooth parametrization $\mathit{r}\left(t\right)$for$t∈[a,b]$and a domain that is open, connected, and simply connected.

Then. the line integral of $\mathbf{F}(x,y,z)$along $C$is as follows:

${∫}_{C}F(x,y,z)d\mathit{r}={∫}_a^b\left(F_1\right(x,y,z)x\text{'}(t)+F_2(x,y,z)y\text{'}(t)+F_3(x,y,z)z\text{'}(t\left)\right)dt$

Example: Find the line integral ${∫}_{C}F.dr$for the vector field role="math" localid="1650948342018" along the given curve $C$which is the straight segment fromrole="math" localid="1650948318632" $(2,1,0)$to role="math" localid="1650948328716" $(5,2,1)$.

First, find $\mathit{r}\left(t\right)$from the point $(a,b,c)$to the point $(d,e,f)$, which can be parametrized as follows:

$\mathit{r}\left(t\right)=(a+t(d-a),b+t(e−b),c+t(f−c\left)\right)$, for .

Thus, $\mathit{r}\left(t\right)$is as follows:

role="math" localid="1650948391029" $\begin{array}{rcl}\mathit{r}\left(t\right)& =& (2+t(5-2),1+t(2-1),0+t(1-0\left)\right)\\ & =& (2+3t,1+2t,t)\end{array}$

Find $\mathit{r}\mathbf{\text{'}}\left(t\right)$.

role="math" localid="1650948404728" $\mathit{r}\mathbf{\text{'}}\left(t\right)=(3,2,1)$

The curve $\mathbf{F}(x,y,z)$is as follows:

Substitute the values of $\mathbf{F}(x,y,z)$and $\mathit{r}\text{'}\left(t\right)$into the formula mentioned in Step 1.

Simplify further.

$\begin{array}{rcl}{∫}_C\mathbf{F}(x,y,z)& =& \left[-8\left(1\right)-\frac{14{\left(1\right)}^2}{2}-0\right]\\ & =& -8-\frac{14}{2}\\ & =& -8-7\\ & =& -15\end{array}$