91Ó°ÊÓ

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

$\mathbf{F}(x,y,z)=\frac{2}{1−z}\mathbf{i}+\mathbf{j}−\frac{1}{z^2+1}\mathbf{k}$, with C the curve parameterized by role="math" localid="1651081777770" $x=\ln (1+t),y=\ln (1−t^2),z=t,$ for $3≤t≤7$.

Parameters:$x=\ln (1+t),y=\ln (1−t^2),z=t,dx=\frac{1}{1+t},dy=\frac{-2t}{1-t^2},dz=1$

Now the integral is${∫}_{C}f(x,y,z)\mathrm{d}s={∫}_a^{b}f\left(r\right(t\left)\right)\left[x\text{'}\left(t\right)+y\text{'}\left(t\right)+z\text{'}\left(t\right)\right]dt$

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

localid="1651125631754" $f\left(r\right(t\left)\right)=\frac{2}{1−t}\mathbf{i}+\mathbf{j}−\frac{1}{t^2+1}\mathbf{k}$

$$\begin{gathered} {∫}_{C}f(x,y,z)\mathrm{d}s={∫}_3^7\left(\frac{2}{1−t}\mathbf{i}+\mathbf{j}−\frac{1}{t^2+1}\mathbf{k}\right)\left(\frac{1}{1+t},\frac{-2t}{1-t^2},1\right)dt \\ {∫}_{C}f(x,y,z)\mathrm{d}s={∫}_3^7\left(\frac{2}{1−t}·\frac{1}{1+t}-\frac{2\mathrm{t}}{1-{\mathrm{t}}^2}−\frac{1}{t^2+1}·1\right)dt \\ {∫}_{C}f(x,y,z)\mathrm{d}s={∫}_3^7\left(\frac{2}{{1−t}^2}-\frac{2\mathrm{t}}{1-{\mathrm{t}}^2}−\frac{1}{t^2+1}\right)dt \\ {∫}_{C}f(x,y,z)\mathrm{d}s={∫}_3^7\frac{2}{{1−t}^2}dt-{∫}_3^7\frac{2\mathrm{t}}{1-{\mathrm{t}}^2}dt−{∫}_3^7\frac{1}{t^2+1}dt \\ {∫}_{C}f(x,y,z)\mathrm{d}s={∫}_3^7\frac{2}{{1−t}^2}(1-t)dt−{∫}_3^7\frac{1}{t^2+1}dt \\ {∫}_{C}f(x,y,z)\mathrm{d}s={∫}_3^7\frac{2}{1+t}dt−{∫}_3^7\frac{1}{t^2+1}dt \\ {∫}_{C}f(x,y,z)\mathrm{d}s=2I_1-I_2 \end{gathered}$$

$$\begin{gathered} I_1={∫}_3^7\frac{1}{1+t}dt \\ Let1+t=u \\ dt=du \\ {I}_1={∫}_3^7\frac{1}{u}du \\ {I}_1={\left[\ln \left(u\right)\right]}_3^7 \\ {I}_1={\left[\ln \left(1+t\right)\right]}_3^7 \\ {I}_1=\left[\ln \left(1+7\right)-\ln \left(1+3\right)\right] \\ {I}_1=\left[\ln 8-\ln 4\right] \end{gathered}$$

$$\begin{gathered} I_2={∫}_3^7\frac{1}{t^2+1}dt \\ {I}_2={\left[{\tan }^{-1}\left(t\right)\right]}_3^7 \\ {I}_2=\left[{\tan }^{-1}\left(7\right)-{\tan }^{-1}\left(3\right)\right] \end{gathered}$$

$$\begin{gathered} {∫}_{C}f(x,y,z)\mathrm{d}s=2\left[\ln \left(8\right)-\ln \left(4\right)\right]-\left[{\tan }^{-1}\left(7\right)-{\tan }^{-1}\left(3\right)\right] \\ {∫}_{C}f(x,y,z)\mathrm{d}s=2\left[2.079-1.387\right]-\left[1.43-1.25\right] \\ {∫}_{C}f(x,y,z)\mathrm{d}s=2×0.692-0.18 \\ {∫}_{C}f(x,y,z)\mathrm{d}s=1.384-0.18 \\ {∫}_{C}f(x,y,z)\mathrm{d}s=1.203 \end{gathered}$$