91Ó°ÊÓ

Consider the vector field below:

The goal is to calculate the line integral${∫}_C\mathbf{F}(x,y,z)·d\mathbf{r}\mathbf{}$for the above vector field in which $C$the circle is in the $xy$-plane and $x=2\cos t,y=2\sin t$and $n$points upward.

The following is how the curve$C$in $R^3$is parametrized:

$\mathbf{r}\left(t\right)=⟨2\cos t,2\sin t,0âŸ\copyright \text{for}0≤t≤2\mathrm{Ï€}\text{.}$

The derivative will then be,

${\mathbf{r}}^{\text{'}}\left(t\right)=\frac{d}{dt}\mathbf{r}\left(t\right)$

$$\begin{gathered} \mathit{r}\mathbf{\text{'}}\left(t\right)=\frac{d}{dt}\mathit{r}\left(t\right) \\ =\frac{d}{dt}(2\cos t,2\sin t,0) \\ =\left(\frac{d}{dt}\right(2\cos t),\frac{d}{dt}(2\sin t),\frac{d}{dt}(0\left)\right) \\ =(-2\sin t,2\cos t,0) \end{gathered}$$

In order to parametrize: $\mathbf{r}\left(t\right)=⟨2\cos t,2\sin t,0âŸ\copyright$

Replace the vector field with the following:

$=(8{\sin }^{3}t,0,16{\cos }^{2}t)$

The required line integral then becomes,

$$\begin{gathered} {∫}_C\mathit{F}(x,y,z)·d\mathit{r} \\ ={∫}_c\mathit{F}\left(x\right(t),y(t\left)\right),\mathit{r}\mathbf{\text{'}}\left(\mathrm{t}\right)\mathrm{dt} \\ ={∫}_0^{2\mathrm{π}}(8{\sin }^3\mathrm{t},0,16{\cos }^2\mathrm{t})·(-2\mathrm{sint},2\mathrm{cost},0)\mathrm{dt}\mathrm{Substitute}\mathrm{values} \\ ={∫}_0^{2\mathrm{π}}(8{\sin }^3\mathrm{t}·(-2\mathrm{sint})+0·2\mathrm{cost}+16{\cos }^2\mathrm{t})\mathrm{dt}\mathrm{Apply}\mathrm{Dot}\mathrm{Product} \end{gathered}$$

$$\begin{gathered} {∫}_0^{2\mathrm{π}}-16{\sin }^{4}tdt \\ =-16{{\left[\frac{3t}{4}-\frac{1}{4}\sin 2t+\frac{1}{32}\sin 4t\right]}_0}^{2\mathrm{π}} \\ =-16\left[(\frac{3·2\mathrm{π}}{8}-\frac{1}{4}\sin 4\mathrm{π}+\frac{1}{32}\sin 8\mathrm{π})-(\frac{3·0}{8}-\frac{1}{4}\sin 0+\frac{1}{32}\sin 0)\right] \\ =-16\left[(\frac{3}{4}\mathrm{π}-\frac{1}{4}·0+\frac{1}{32}·0)-(0-\frac{1}{4}·0+\frac{1}{32}·0\right] \\ =-16\left(\frac{3}{4}\mathrm{π}\right) \\ =-12\mathrm{π} \end{gathered}$$

As a result, the necessary line integral is,

${∫}_C\mathbf{F}(x,y,z)·d\mathbf{r}=-12\mathrm{π}\text{.}$