Evaluate the line integrals in Exercises 45–50.

${\int }_{C}f(x,y)ds,\mathrm{where}f(x,y)={(x^2+y^2)}^{1/2}$ and C is the curve parameterized by $x=t\sin t,y=t\cos t,\mathrm{for}0\le t\le 4\mathrm{\pi }$.

We have to evaluate the line integrals in given exercise.

${\int }_{C}f(x,y)ds,\mathrm{where}f(x,y)={(x^2+y^2)}^{1/2}$ and C is the curve parameterized by $x=t\sin t,y=t\cos t,\mathrm{for}0\le t\le 4\mathrm{\pi }$.

Here we have

$$\begin{gathered} f\left(x\right(t),y(t\left)\right)={({\left(t\cos t\right)}^2+{\left(t\sin t\right)}^2)}^{1/2} \\ f\left(x\right(t),y(t\left)\right)={(t^2{\cos }^{2}t+t^2{\sin }^{2}t)}^{1/2} \\ f\left(x\right(t),y(t\left)\right)={\left[t^2({\cos }^{2}t+{\sin }^{2}t)\right]}^{1/2} \\ f\left(x\right(t),y(t\left)\right)={\left[t^2·1\right]}^{1/2} \\ f\left(x\right(t),y(t\left)\right)={\left[t^2\right]}^{1/2} \\ f\left(x\right(t),y(t\left)\right)=t \end{gathered}$$

and

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

$$\begin{gathered} {\int }_{C}f(x,y)ds={\int }_0^{4\mathrm{\pi }}t\sqrt{t^2{\sin }^{2}t+{\cos }^{2}t-2t\sin t\cos t+t^2{\cos }^{2}t+{\sin }^{2}t-2t\sin t\cos t}\mathrm{d}t \\ {\int }_{C}f(x,y)ds={\int }_0^{4\mathrm{\pi }}t\sqrt{t^2{\sin }^{2}t+{\cos }^{2}t+t^2{\cos }^{2}t+{\sin }^{2}t}\mathrm{d}t \\ {\int }_{C}f(x,y)ds={\int }_0^{4\mathrm{\pi }}t\sqrt{t^2({\sin }^{2}t+{\cos }^{2}t)+({\cos }^{2}t+{\sin }^{2}t)}\mathrm{d}t \\ {\int }_{C}f(x,y)ds={\int }_0^{4\mathrm{\pi }}t\sqrt{t^2+1}\mathrm{d}t \end{gathered}$$

$$\begin{gathered} {\int }_{C}f(x,y)ds={\int }_0^{4\mathrm{\pi }}t\sqrt{t^2+1}\mathrm{d}t \\ \mathrm{let}{t}^2+1=u \\ 2tdt=du \\ tdt=\frac{1}{2}du \\ {\int }_{C}f(x,y)ds=\frac{1}{2}{\int }_0^{4\mathrm{\pi }}\sqrt{u}\mathrm{du} \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{ds}=\frac{1}{2}{\int }_0^{4\mathrm{\pi }}{\mathrm{u}}^{1/2}\mathrm{du} \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{ds}=\frac{1}{2}{\left[\frac{{\mathrm{u}}^{1/2+1}}{1/2+1}\right]}_0^{4\mathrm{\pi }} \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{ds}=\frac{1}{2}{\left[\frac{{\mathrm{u}}^{3/2}}{3/2}\right]}_0^{4\mathrm{\pi }} \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{ds}=\frac{1}{2}·\frac{2}{3}{\left[{({\mathrm{t}}^2+1)}^{3/2}\right]}_0^{4\mathrm{\pi }} \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{ds}=\frac{1}{3}\left[{\left\{{\left(4\mathrm{\pi }\right)}^2+1\right\}}^{3/2}-{\left\{{\left(0\right)}^2+1\right\}}^{3/2}\right] \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{ds}=\frac{1}{3}\left[{({16\mathrm{\pi }}^2+1)}^{3/2}-{\left\{1\right\}}^{3/2}\right] \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{ds}=\frac{1}{3}\left[{({16\mathrm{\pi }}^2+1)}^{3/2}-1\right] \end{gathered}$$