Evaluate the line integrals in Exercises 45–50.

${\int }_{\mathrm{C}}\mathbf{F}(x,y)·d\mathbf{r},\mathrm{where}\mathbf{F}(x,y)=-y\mathbf{i}+x\mathbf{j}$ and C is the spiral $x=t\cos t,y=t\sin t,\mathrm{for}\mathrm{\pi }\le t\le 2\mathrm{\pi }$.

We have to evaluate the line integrals in given exercise.

${\int }_{\mathrm{C}}\mathbf{F}(x,y)·d\mathbf{r},\mathrm{where}\mathbf{F}(x,y)=-y\mathbf{i}+x\mathbf{j}$ and C is the spiral role="math" $x=t\cos t,y=t\sin t,\mathrm{for}\mathrm{\pi }\le t\le 2\mathrm{\pi }$.

Here we have $\mathbf{F}\left(x\right(t),y(t\left)\right)=(-t\sin t,t\cos t)$and

$$\begin{gathered} \frac{\mathbf{d}\mathbf{r}}{\mathbf{d}\mathbf{t}}=\frac{d}{d\mathrm{t}}(\mathrm{tcost},\mathrm{tsint}) \\ \frac{d\mathrm{r}}{d\mathrm{t}}=\left(t\frac{d}{d\mathrm{t}}\mathrm{cost}+\cos t\frac{d}{d\mathrm{t}}t,t\frac{d}{d\mathrm{t}}\mathrm{sint}+\sin t\frac{d}{d\mathrm{t}}t\right) \\ \frac{d\mathrm{r}}{d\mathrm{t}}=\left(-t\sin t+\cos t·1,t\cos t+\sin t·1\right) \\ \frac{d\mathrm{r}}{d\mathrm{t}}=\left(-t\sin t+\cos t,t\cos t+\sin t\right) \\ d\mathbf{r}=\left(-t\sin t+\cos t,t\cos t+\sin t\right)dt \end{gathered}$$

$$\begin{gathered} {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}={\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}(-\mathrm{tsint})(-\mathrm{tsint}+\mathrm{cost})+\left(\mathrm{tcost}\right)(\mathrm{tcost}+\mathrm{sint})\mathrm{dt} \\ {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}={\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}-\mathrm{tsint}(-\mathrm{tsint})+\mathrm{cost}(-\mathrm{tsint})+\mathrm{tcost}\left(\mathrm{tcost}\right)+\mathrm{sint}\left(\mathrm{tcost}\right)\mathrm{dt} \\ {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}={\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}\left({\mathrm{t}}^2{\sin }^2\mathrm{t}-\mathrm{tsintcost}+{\mathrm{t}}^2{\cos }^2\mathrm{t}+\mathrm{tsintcost}\right)\mathrm{dt} \\ {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}={\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}\left({\mathrm{t}}^2{\sin }^2\mathrm{t}+{\mathrm{t}}^2{\cos }^2\mathrm{t}\right)\mathrm{dt} \\ {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}={\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}{\mathrm{t}}^2\left({\sin }^2\mathrm{t}+{\cos }^2\mathrm{t}\right)\mathrm{dt} \\ {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}={\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}{\mathrm{t}}^2·1\mathrm{dt} \\ {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}={\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}{\mathrm{t}}^2\mathrm{dt} \end{gathered}$$

$$\begin{gathered} {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}={\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}{\mathrm{t}}^2\mathrm{dt} \\ {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}={\left[\frac{{\mathrm{t}}^{2+1}}{2+1}\right]}_{\mathrm{\pi }}^{2\mathrm{\pi }} \\ {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}={\left[\frac{{\mathrm{t}}^3}{3}\right]}_{\mathrm{\pi }}^{2\mathrm{\pi }} \\ {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}=\left[\frac{{\left(2\mathrm{\pi }\right)}^3}{3}-\frac{{\left(\mathrm{\pi }\right)}^3}{3}\right] \\ {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}=\left[\frac{8{\mathrm{\pi }}^3}{3}-\frac{{\mathrm{\pi }}^3}{3}\right] \\ {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}=\frac{8{\mathrm{\pi }}^3-{\mathrm{\pi }}^3}{3} \\ {\int }_{\mathrm{C}}\mathbf{F}(\mathrm{x},\mathrm{y})·\mathrm{d}\mathbf{r}=\frac{7{\mathrm{\pi }}^3}{3} \end{gathered}$$