In Exercises 21–28, evaluate the multivariate line integral of the given function over the specified curve.

$g(x,y,z)=xyz$, with C the curve parameterized by $\mathbf{r}\left(t\right)=\left(\frac{\sqrt{2}}{3}{t}^3,t^2,t\right)\mathrm{for}1\le t\le 4.$

In the given exercises we have to evaluate the multivariate line integral of the given function over the specified curve.

$g(x,y,z)=xyz$, with C the curve parameterized by $\mathbf{r}\left(t\right)=\left(\frac{\sqrt{2}}{3}{t}^3,t^2,t\right)\mathrm{for}1\le t\le 4.$

$$\begin{gathered} \mathbf{r}\left(t\right)=\left(\frac{\sqrt{2}}{3}{t}^3,t^2,t\right),So \\ \mathbf{r}\mathbf{\text{'}}\left(t\right)=\left(\frac{\sqrt{2}}{3}3t^2,2t,1\right) \\ \mathbf{r}\mathbf{\text{'}}\left(t\right)=\left(\sqrt{2}{t}^2,2t,1\right) \end{gathered}$$

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

$$\begin{gathered} g\left(r\right(t\left)\right)=xyz \\ g\left(r\right(t\left)\right)=\frac{\sqrt{2}}{3}{t}^3·t^2·t \\ g\left(r\right(t\left)\right)=\frac{\sqrt{2}}{3}{t}^{3+2+1} \\ g\left(r\right(t\left)\right)=\frac{\sqrt{2}}{3}{t}^6 \end{gathered}$$

$$\begin{gathered} {\int }_{C}g(x,y,z)\mathrm{ds}=\frac{\sqrt{2}}{3}{\int }_1^4{t}^6\sqrt{{\left(\sqrt{2}t\right)}^2+{\left(2t\right)}^2+{\left(1\right)}^2}dt \\ {\int }_{C}g(x,y,z)\mathrm{ds}=\frac{\sqrt{2}}{3}{\int }_1^4{t}^6\sqrt{{\left(\sqrt{2}{t}^2\right)}^2+{4t}^2+{\left(1\right)}^2}dt \\ {\int }_{C}g(x,y,z)\mathrm{ds}=\frac{\sqrt{2}}{3}{\int }_1^4{t}^6\sqrt{{{\left(\sqrt{2}{t}^2\right)}^2+2·1·(\sqrt{2}t)}^2+{\left(1\right)}^2}dt \\ {\int }_{C}g(x,y,z)\mathrm{ds}=\frac{\sqrt{2}}{3}{\int }_1^4{t}^6\sqrt{{(\sqrt{2}{t}^2+1)}^2}dt \\ {\int }_{C}g(x,y,z)\mathrm{ds}=\frac{\sqrt{2}}{3}{\int }_1^4{t}^6(\sqrt{2}{t}^2+1)dt \\ {\int }_{C}g(x,y,z)\mathrm{ds}=\frac{\sqrt{2}}{3}\left[{\int }_1^4\sqrt{2}{t}^2·t^{6}dt+{\int }_1^4{t}^{6}dt\right] \\ {\int }_{C}g(x,y,z)\mathrm{ds}=\frac{\sqrt{2}}{3}\left[\sqrt{2}{\int }_1^4{t}^{8}dt+{\int }_1^4{t}^{6}dt\right] \\ {\int }_{C}g(x,y,z)\mathrm{ds}=\frac{\sqrt{2}}{3}{\left[\sqrt{2}\left[\frac{t^{8+1}}{8+1}\right]+\left[\frac{t^{6+1}}{6+1}\right]\right]}_1^4 \\ {\int }_{C}g(x,y,z)\mathrm{ds}=\frac{\sqrt{2}}{3}{\left[\sqrt{2}\left[\frac{t^9}{9}\right]+\left[\frac{t^7}{7}\right]\right]}_1^4 \end{gathered}$$

$$\begin{gathered} {\int }_{C}g(x,y,z)\mathrm{ds}=\frac{\sqrt{2}}{3}\left[\left(\sqrt{2}\left[\frac{{\left(4\right)}^9}{9}\right]+\left[\frac{{\left(4\right)}^7}{7}\right]\right)-\left(\sqrt{2}\left[\frac{{\left(1\right)}^9}{9}\right]+\left[\frac{{\left(1\right)}^7}{7}\right]\right)\right] \\ {\int }_{C}g(x,y,z)\mathrm{ds}\approx \frac{1.41}{3}\left[\left(1.41\times 29127.11+2340.57\right)-\left(1.41\times 0.111+0.143\right)\right] \\ {\int }_{C}g(x,y,z)\mathrm{ds}\approx 0.47\left[\left(41069.23+2340.57\right)-\left(0.157+0.143\right)\right] \\ {\int }_{C}g(x,y,z)\mathrm{ds}\approx 0.47\left[43409.8-0.3\right] \\ {\int }_{C}g(x,y,z)\mathrm{ds}\approx 0.47\times 43409.5 \\ {\int }_{C}g(x,y,z)\mathrm{ds}\approx 20402.465 \end{gathered}$$