Evaluate each of the vector field line integrals in Exercises 29–36 over the indicated curves.

$\mathbf{F}(x,y,z)=2z\mathbf{i}+\ln x\mathbf{j}+xz\mathbf{k}$, with C the curve parameterized by $(t,\ln t,t)\mathrm{for}1\le t\le e^2$.

We have to evaluate the vector field line integrals in given exercises over the indicated curves.

$\mathbf{F}(x,y,z)=2z\mathbf{i}+\ln x\mathbf{j}+xz\mathbf{k}$, with C the curve parameterized by $(t,\ln t,t)\mathrm{for}1\le t\le e^2$.

So the line integral is

${\int }_{C}f(x,y,z)\mathrm{d}s={\int }_a^{b}f\left(r\right(t)\sqrt{(x\text{'}{\left(t\right))}^2+(y\text{'}{\left(t\right))}^2+(z\text{'}{\left(t\right))}^2}dt$

$$\begin{gathered} \mathbf{r}\left(t\right)=(t,\ln t,t) \\ \mathbf{r}\mathbf{\text{'}}\left(t\right)=(1,\frac{1}{t},1) \end{gathered}$$

Now finding $f\left(r\right(t)$

localid="1650995848859" $$\begin{gathered} f\left(r\right(t)=2t\mathbf{i}+\ln (\ln t)\mathbf{j}+t·t\mathbf{k} \\ f(r\left(t\right)=2t\mathbf{i}+t\mathbf{j}+t^2\mathbf{k} \end{gathered}$$

$$\begin{gathered} {\int }_{C}f(x,y,z)\mathrm{d}s={\int }_1^{e^2}2\mathrm{t}\mathbf{i}+\mathrm{t}\mathbf{j}+{\mathrm{t}}^2\mathbf{k}\sqrt{{\left(1\right)}^2+{\left(\frac{1}{t}\right)}^2+{\left(1\right)}^2}dt \\ {\int }_{C}f(x,y,z)\mathrm{d}s={\int }_1^{e^2}2t\mathbf{i}+t\mathbf{j}+t^2\mathbf{k}\sqrt{1+\frac{1}{t^2}+1}dt \\ {\int }_{C}f(x,y,z)\mathrm{d}s={\int }_1^{e^2}2t\mathbf{i}+t\mathbf{j}+t^2\mathbf{k}\sqrt{2+\frac{1}{t^2}}dt \\ {\int }_{C}f(x,y,z)\mathrm{d}s={\int }_1^{e^2}(2t\mathbf{i}+t\mathbf{j}+t^2\mathbf{k}\mathbf{)}\sqrt{\frac{2t^2+1}{t^2}}dt \\ {\int }_{C}f(x,y,z)\mathrm{d}s={\int }_1^{e^2}(2t\mathbf{i}+t\mathbf{j}+t^2\mathbf{k}\mathbf{)}\frac{1}{t}\sqrt{2t^2+1}dt \\ {\int }_{C}f(x,y,z)\mathrm{d}s={\int }_1^{e^2}(2\mathbf{i}+\mathbf{j}+t\mathbf{k}\mathbf{)}\sqrt{2t^2+1}dt \\ {\int }_{C}f(x,y,z)\mathrm{d}s={\int }_1^{e^2}2\sqrt{2t^2+1}dt\mathbf{i}+{\int }_1^{e^2}\sqrt{2t^2+1}dt\mathbf{j}+{\int }_1^{e^2}t\sqrt{2t^2+1}dt\mathbf{k} \end{gathered}$$

$$\begin{gathered} {\int }_{C}f(x,y,z)\mathrm{d}s=2{\int }_1^{e^2}\sqrt{{\left(\sqrt{2}t\right)}^2+{\left(1\right)}^2}dt\mathbf{i}+{\int }_1^{e^2}\sqrt{{\left(\sqrt{2}t\right)}^2+{\left(1\right)}^2}dt\mathbf{j}+{\int }_1^{e^2}t\sqrt{2t^2+1}dt\mathbf{k} \\ {\int }_{C}f(x,y,z)\mathrm{d}s=2{\int }_1^{e^2}\sqrt{{\left(1\right)}^2+{\left(\sqrt{2}t\right)}^2}dt\mathbf{i}+{\int }_1^{e^2}\sqrt{{{\left(1\right)}^2+\left(\sqrt{2}t\right)}^2}dt\mathbf{j}+{\int }_1^{e^2}t\sqrt{2t^2+1}dt\mathbf{k} \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y},\mathrm{z})\mathrm{ds}=2I_{\mathit{1}}\mathbf{i}\mathit{+}{I}_{\mathit{2}}\mathbf{j}\mathit{+}{I}_{\mathit{3}}\mathbf{k} \\ \mathrm{Now}\mathrm{solving}\mathrm{the}{I}_{\mathit{1}} \\ {I}_{\mathit{1}}={\int }_1^{e^2}\sqrt{{{\left(1\right)}^2+\left(\sqrt{2}t\right)}^2}dt \\ {I}_{\mathit{1}}={\left[\frac{1}{2}\sqrt{2}t\sqrt{{{\left(1\right)}^2+\left(\sqrt{2}t\right)}^2}+\frac{1}{2}{\left(1\right)}^2{\log }_e\left|\sqrt{2}t+\sqrt{{{\left(1\right)}^2+\left(\sqrt{2}t\right)}^2}\right|\right]}_1^{e^2} \\ {I}_{\mathit{1}}={\left[\frac{1}{2}\sqrt{2}t\sqrt{{1+2t}^2}+\frac{1}{2}{\left(1\right)}^2{\log }_e\left|\sqrt{2}t+\sqrt{{1+2t}^2}\right|\right]}_1^{e^2} \end{gathered}$$

$$\begin{gathered} I_{\mathit{1}}=\left[\frac{1}{2}\sqrt{2}{e}^2\sqrt{{1+2\left(e^2\right)}^2}+\frac{1}{2}{\left(1\right)}^2{\log }_e\left|\sqrt{2}{e}^2+\sqrt{{1+2\left(e^2\right)}^2}\right|-\frac{1}{2}\sqrt{2}·1\sqrt{{1+2\left(1\right)}^2}+\frac{1}{2}{\left(1\right)}^2{\log }_e\left|\sqrt{2}·1+\sqrt{{1+2\left(1\right)}^2}\right|\right] \\ {I}_{\mathit{1}}=\left[\frac{1}{2}\sqrt{2}{e}^2\sqrt{1+2e^4}+\frac{1}{2}·1{\log }_e\left|\sqrt{2}{e}^2+\sqrt{1+2e^4}\right|-\frac{1}{2}\sqrt{2}\sqrt{1+2·1}+\frac{1}{2}·1{\log }_e\left|\sqrt{2}+\sqrt{1+2·1}\right|\right] \\ {I}_{\mathit{1}}=\left[\frac{1}{\sqrt{2}}{e}^2\sqrt{1+2e^4}+\frac{1}{2}{\log }_e\left|\sqrt{2}{e}^2+\sqrt{1+2e^4}\right|-\frac{1}{\sqrt{2}}\sqrt{3}+\frac{1}{2}{\log }_e\left|\sqrt{2}+\sqrt{3}\right|\right] \\ {I}_{\mathit{1}}=\left[\frac{1}{\sqrt{2}}{e}^2\sqrt{1+2e^4}+\frac{1}{2}{\log }_e\left|\sqrt{2}{e}^2+\sqrt{1+2e^4}\right|-\frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{2}{\log }_e\left|\sqrt{2}+\sqrt{3}\right|\right]=I_2 \end{gathered}$$

$$\begin{gathered} I_{\mathit{3}}={\int }_1^{e^2}t\sqrt{2t^2+1}dt \\ Let2t^2+1=u \\ 4tdt=du \\ tdt=\frac{1}{4}du \\ {I}_{\mathit{3}}=\frac{1}{4}{\int }_1^{e^2}\sqrt{u}dt \\ {I}_{\mathit{3}}=\frac{1}{4}{\int }_1^{e^2}{u}^{1/2}dt \\ {I}_{\mathit{3}}=\frac{1}{4}{\left[\frac{u^{1/2+1}}{1/2+1}\right]}_1^{e^2} \\ {I}_{\mathit{3}}=\frac{1}{4}{\left[\frac{u^{3/2}}{3/2}\right]}_1^{e^2} \\ {I}_{\mathit{3}}=\frac{1}{4}·\frac{2}{3}{\left[u^{3/2}\right]}_1^{e^2} \\ {I}_{\mathit{3}}=\frac{1}{6}\left[{\left(e^2\right)}^{3/2}-{\left(1\right)}^{3/2}\right] \\ {I}_{\mathit{3}}=\frac{1}{6}\left[e^3-1\right] \end{gathered}$$

$$\begin{gathered} {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y},\mathrm{z})\mathrm{ds}=2I_{\mathit{1}}\mathbf{i}\mathit{+}{I}_{\mathit{2}}\mathbf{j}\mathit{+}{I}_{\mathit{3}}\mathbf{k} \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y},\mathrm{z})\mathrm{ds}=2\left[\frac{1}{\sqrt{2}}{e}^2\sqrt{1+2e^4}+\frac{1}{2}{\log }_e\left|\sqrt{2}{e}^2+\sqrt{1+2e^4}\right|-\frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{2}{\log }_e\left|\sqrt{2}+\sqrt{3}\right|\right]\mathbf{i} \\ +\left[\frac{1}{\sqrt{2}}{e}^2\sqrt{1+2e^4}+\frac{1}{2}{\log }_e\left|\sqrt{2}{e}^2+\sqrt{1+2e^4}\right|-\frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{2}{\log }_e\left|\sqrt{2}+\sqrt{3}\right|\right]\mathbf{j}+\frac{1}{6}\left[e^3-1\right]\mathbf{k} \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y},\mathrm{z})\mathrm{ds}=\left[2·\frac{1}{\sqrt{2}}{e}^2\sqrt{1+2e^4}+2·\frac{1}{2}{\log }_e\left|\sqrt{2}{e}^2+\sqrt{1+2e^4}\right|-2·\frac{\sqrt{3}}{\sqrt{2}}+2·\frac{1}{2}{\log }_e\left|\sqrt{2}+\sqrt{3}\right|\right]\mathbf{i} \\ +\left[\frac{1}{\sqrt{2}}{e}^2\sqrt{1+2e^4}+\frac{1}{2}{\log }_e\left|\sqrt{2}{e}^2+\sqrt{1+2e^4}\right|-\frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{2}{\log }_e\left|\sqrt{2}+\sqrt{3}\right|\right]\mathbf{j}+\frac{1}{6}\left[e^3-1\right]\mathbf{k} \end{gathered}$$

$$\begin{gathered} {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y},\mathrm{z})\mathrm{ds}=\left[\sqrt{2}{e}^2\sqrt{1+2e^4}+{\log }_e\left|\sqrt{2}{e}^2+\sqrt{1+2e^4}\right|-\sqrt{6}+{\log }_e\left|\sqrt{2}+\sqrt{3}\right|\right]\mathbf{i} \\ +\left[\frac{1}{\sqrt{2}}{e}^2\sqrt{1+2e^4}+\frac{1}{2}{\log }_e\left|\sqrt{2}{e}^2+\sqrt{1+2e^4}\right|-\frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{2}{\log }_e\left|\sqrt{2}+\sqrt{3}\right|\right]\mathbf{j}+\frac{1}{6}\left[e^3-1\right]\mathbf{k} \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y},\mathrm{z})\mathrm{ds}\approx \left[1.41\times 7.39\sqrt{1+2\times 54.59}+{\log }_e\left|1.41\times 7.39+\sqrt{1+2\times 54.59}\right|-2.45+0.49\right]\mathbf{i} \\ +\left[0.71\times 7.39\sqrt{1+2\times 54.59}+0.5\times {\log }_e\left|1.41\times 7.39+\sqrt{1+2\times 54.59}\right|-1.21+0.5\times 0.49\right]\mathbf{j}+0.17\left[20.09-1\right]\mathbf{k} \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y},\mathrm{z})\mathrm{ds}\approx \left[1.41\times 7.39\times 110.18+2.08-2.45+0.49\right]\mathbf{i} \\ +\left[0.71\times 7.39\times 110.18+0.5\times 2.08-1.21+0.5\times 0.49\right]\mathbf{j}+0.17\left[20.09-1\right]\mathbf{k} \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y},\mathrm{z})\mathrm{ds}\approx (1148.06+2.08-2.45+0.49)\mathbf{i}+(578.10+1.04-1.21+0.245)\mathbf{j} \\ +0.17\times 19.09\mathbf{k} \\ {\int }_{\mathrm{C}}\mathrm{f}(\mathrm{x},\mathrm{y},\mathrm{z})\mathrm{ds}\approx 1148.18\mathbf{i}+578.175\mathbf{j}+3.2453\mathbf{k} \end{gathered}$$