91Ó°ÊÓ

Use the Fundamental Theorem of Line Integrals, if applicable, to evaluate the integrals in the given exercises. Otherwise, show that the vector field is not conservative.

$\mathbf{F}(x,y)=(2^x\ln 2,2y)$, with C the straight line segment from $(3,7)\mathrm{to}(0,1)$.

$$\begin{gathered} \frac{d\mathbf{F}(x,y)}{dy}=\frac{d}{dy}\left(2^x\ln 2\right)\frac{d\mathbf{F}(x,y)}{dx}=\frac{d}{dx}\left(2y\right) \\ \frac{d\mathbf{F}(x,y)}{dy}=0\frac{d\mathbf{F}(x,y)}{dx}=0 \end{gathered}$$

Since, $\frac{d\mathbf{F}(x,y)}{dy}=\frac{d\mathbf{F}(x,y)}{dx}$, so the given function is conservative.

The points are$(3,7)\mathrm{to}(0,1)$

We write the line segment as a vector function:

$$\begin{gathered} \mathbf{r}=(3,7)+t(0-3,1-7) \\ \mathbf{r}=(3,7)+t(-3,-6) \\ 0≤t≤1,\mathrm{or}\mathrm{in}\mathrm{parametric}\mathrm{form} \\ x=3-3t,y=7-6t \\ x\text{'}=-3,\mathrm{y}\text{'}=-6 \end{gathered}$$

$$\begin{gathered} {∫}_{C}f(x,y,z)\mathrm{d}s={∫}_0^1(2^{3-3t}\ln 2,2(7-6t\left)\right)\left[-3,-6\right]dt \\ {∫}_{C}f(x,y,z)\mathrm{d}s={∫}_0^1-3·2^{3-3t}\ln 2+-6·2(7-6t))dt \\ {∫}_{C}f(x,y,z)\mathrm{d}s={∫}_0^1-3·2^{3-3t}\ln 2dt+{∫}_0^1-6·2(7-6t\left)\right)dt \\ {∫}_{C}f(x,y,z)\mathrm{d}s=-3\ln 2{∫}_0^1{2}^{3-3t}dt-12{∫}_0^1(7-6t))dt \\ {∫}_{C}f(x,y,z)\mathrm{d}s=-3\ln 2{∫}_0^1{2}^{3-3t}dt-12×7{∫}_0^{1}dt-6{∫}_0^{1}tdt \\ {∫}_{C}f(x,y,z)\mathrm{d}s=-3\ln 2{∫}_0^1{2}^{3-3t}dt-84{∫}_0^{1}dt-6{∫}_0^{1}tdt \\ {∫}_{C}f(x,y,z)\mathrm{d}s=-3\ln 2I_1-84I_2-6I_3 \end{gathered}$$

$$\begin{gathered} I_1={∫}_0^1{2}^{3-3t}dt \\ let3-3t=u \\ -3dt=du \\ dt=-\frac{1}{3}du \\ {I}_1=-\frac{1}{3}{∫}_0^1{2}^{u}du \\ {I}_1=-\frac{1}{3}{\left[\frac{2^u}{\ln \left(u\right)}\right]}_0^1 \\ {I}_1=-\frac{1}{3}{\left[\frac{2^{3-3t}}{\ln \left(3-3t\right)}\right]}_0^1 \\ {I}_1=-\frac{1}{3}\left[\frac{2^{3-3·1}}{\ln \left(3-3·1\right)}-\frac{2^{3-3·0}}{\ln \left(3-3·0\right)}\right] \\ {I}_1=-\frac{1}{3}\left[\frac{2^{3-3}}{\ln \left(3-3\right)}-\frac{2^{3-0}}{\ln \left(3-0\right)}\right] \\ {I}_1=-\frac{1}{3}\left[\frac{2^0}{\ln \left(0\right)}-\frac{2^3}{\ln \left(3\right)}\right] \\ {I}_1=-\frac{1}{3}\left[0-\frac{8}{\ln \left(3\right)}\right] \\ {I}_1=\frac{8}{3\ln \left(3\right)} \end{gathered}$$

$$\begin{gathered} I_2={∫}_0^{1}dt \\ {I}_2={\left(t\right)}_0^1 \\ {I}_2=(1-0) \\ {I}_2=0 \end{gathered}$$

$$\begin{gathered} I_3={∫}_0^{1}tdt \\ {I}_3={\left[\frac{t^{1+1}}{1+1}\right]}_0^1 \\ {I}_3={\left[\frac{t^2}{2}\right]}_0^1 \\ {I}_3=\left[\frac{1^2}{2}-\frac{0^2}{2}\right] \\ {I}_3=\frac{1}{2} \end{gathered}$$

$$\begin{gathered} {∫}_{C}f(x,y,z)\mathrm{d}s=-3\ln 2·\frac{8}{3\ln \left(3\right)}-84·0-6·\frac{1}{2} \\ {∫}_{C}f(x,y,z)\mathrm{d}s=-\frac{24\ln 2}{3\ln \left(3\right)}-0-3 \\ {∫}_{C}f(x,y,z)\mathrm{d}s=-\frac{24\ln 2}{3\ln \left(3\right)}-3 \end{gathered}$$