Find the work done by the given vector field $F$along specified curve $C$.

$\mathbf{F}(x,y,z)=yz\mathbf{i}+xz\mathbf{j}+xy\mathbf{k}$where$C$is curve parametrized by

The given vector field is $\mathbf{F}(x,y,z)=yz\mathbf{i}+xz\mathbf{j}+xy\mathbf{k}$

and curve is parametrized by

The curve is parametrized as

Its derivative

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

$=⟨-\sin t,\cos t,-4\sin t\cos t⟩\text{.}$

It can be written as

or $x=\cos t,y=\sin t,z={\cos }^{2}t-{\sin }^{2}t$as below:

$\mathbf{F}(x,y,z)=yz\mathbf{i}+xz\mathbf{j}+xy\mathbf{k}$

$\mathbf{F}\left(x\right(t),y(t),z(t\left)\right)=\sin t\left({\cos }^{2}t-{\sin }^{2}t\right)\mathbf{i}+\cos t\left({\cos }^{2}t-{\sin }^{2}t\right)\mathbf{j}+\cos t\sin t\mathbf{k}$

$=\left(\sin t{\cos }^{2}t-{\sin }^{3}t\right)\mathbf{i}+\left({\cos }^{3}t-{\sin }^{2}t\cos t\right)\mathbf{j}+\cos t\sin t\mathbf{k}$

The line integral is evaluated as ${\int }_C\mathbf{F}(x,y,z)·d\mathbf{r}$

$={\int }_C\mathbf{F}\left(x\right(t),y(t),z(t\left)\right)·{\mathbf{r}}^{\text{'}}\left(t\right)dt$

$={\int }_0^{\pi /4}\left[\left(\sin t{\cos }^{2}t-{\sin }^{3}t\right)(-\sin t)+\left({\cos }^{3}t-{\sin }^{2}t\cos t\right)\cos t\right.$

$+\left(\cos t\sin t\right)(-4\sin t\cos t)]dt$

Apply dot product

$={\int }_0^{\pi /4}\left[-{\sin }^{2}t{\cos }^{2}t+{\sin }^{4}t+{\cos }^{4}t-{\sin }^{2}t{\cos }^{2}t-4{\sin }^{2}t{\cos }^{2}t\right]dt$

$={\int }_0^{\pi /4}\left[{\sin }^{4}t+{\cos }^{4}t+2{\sin }^{2}t{\cos }^{2}t-8{\sin }^{2}t{\cos }^{2}t\right]dt$

$={\int }_0^{\pi /4}\left[{\left({\sin }^{2}t+{\cos }^{2}t\right)}^2-8{\sin }^{2}t{\cos }^{2}t\right]dt$

$={\int }_0^{\pi /4}\left[1-2(2\sin t\cos t{)}^2\right]dt$

$={\int }_0^{\pi /4}\left[1-2{\sin }^{2}2t\right]dt$

$={\int }_0^{\pi /4}\cos 4tdt$

$=\frac{\sin \pi }{4}-\frac{\sin 0}{4}$

$=0$

Hence, Work done

${\int }_C\mathbf{F}(x,y,z)·d\mathbf{r}=0.$