${\int }_C\mathit{F}(x,y,z)·d\mathit{r}$, where$\mathit{F}(x,y,z)=(y^3,-4z,4x^2),$$C$is the

circle in the $xy$-plane parametrized by $x=2\cos t,$$y=2\sin t$

and $n$points upwards.

Consider the vector field below:

The goal is to calculate the line integral${\int }_C\mathbf{F}(x,y,z)·d\mathbf{r}\mathbf{}$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:

$\mathbf{r}\left(t\right)=⟨2\cos t,2\sin t,0⟩\text{for}0\le t\le 2\pi \text{.}$

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: $\mathbf{r}\left(t\right)=⟨2\cos t,2\sin t,0⟩$

Replace the vector field with the following:

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

The required line integral then becomes,

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

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

As a result, the necessary line integral is,

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