LaTeX Wiki
Advertisement

\(\textbf{Part 2}\)

\( \int_{c}\underline{F}.d\underline{r} = \int_{c_1}\underline{F}.d\underline{r} + \int_{c_2}\underline{F}.d\underline{r} + \int_{c_3}\underline{F}.d\underline{r} + \int_{c_4}\underline{F}.d\underline{r}\)

c1:

\(  \underline{r}(t)  =  \langle t ,  1 \rangle   ,  t  \in [1,-1]\)

\(  \underline{r}'(t) =  \langle 1 , 0 \rangle\)

\( \int_{c_1}\underline{F}.d\underline{r} = \int_{1}^{-1}  \langle \frac{-1}{t^2+1} , \frac{t}{t^2+1} \rangle.   \langle 1 , 0 \rangle dt \)

\(= \int_{1}^{-1}  \frac{-1}{t^2+1} dt =  -\tan^{-1}{t}\Biggr|_{1}^{-1}\)

\(= \frac{\pi}{2}\)

c2:

\(  \underline{r}(t)  =  \langle -1 ,  t \rangle   ,  t  \in [1,-1]\)

\(  \underline{r}'(t) =  \langle 0 , 1 \rangle\)

\( \int_{c_2}\underline{F}.d\underline{r} = \int_{1}^{-1}  \langle \frac{-t}{t^2+1} , \frac{-1}{t^2+1} \rangle.   \langle 0 , 1 \rangle dt \)

\(= \int_{1}^{-1}  \frac{-1}{t^2+1} dt =  -\tan^{-1}{t}\Biggr|_{1}^{-1}\)

\(= \frac{\pi}{2}\)

c3:

\(  \underline{r}(t)  =  \langle t ,  -1 \rangle   ,  t  \in [-1,1]\)

\(  \underline{r}'(t) =  \langle 1 , 0 \rangle\)

\( \int_{c_3}\underline{F}.d\underline{r} = \int_{-1}^{1}  \langle \frac{1}{t^2+1} , \frac{t}{t^2+1} \rangle.   \langle 1 , 0 \rangle dt \)

\(= \int_{-1}^{1}  \frac{1}{t^2+1} dt =  \tan^{-1}{t}\Biggr|_{-1}^{1}\)

\(= \frac{\pi}{2}\)

c4:

\(  \underline{r}(t)  =  \langle 1 ,  t \rangle   ,  t  \in [-1,1]\)

\(  \underline{r}'(t) =  \langle 0 , 1 \rangle\)

\( \int_{c_4}\underline{F}.d\underline{r} = \int_{-1}^{1}  \langle \frac{-t}{t^2+1} , \frac{1}{t^2+1} \rangle.   \langle 0 , 1 \rangle dt \)

\(= \int_{-1}^{1}  \frac{1}{t^2+1} dt =  \tan^{-1}{t}\Biggr|_{-1}^{1}\)

\(= \frac{\pi}{2}\)

\( \int_{c}\underline{F}.d\underline{r} = 4 \times \frac{\pi}{2}   =  2\pi \)

Advertisement