Chapter MA, Problem 2.8MA

(i)

To determine

To evaluate: The line integral $V_{AB}$ along the line that joins the points A and B. 

Answer to Problem 2.8MA

The line integral $V_{AB}$ along the line that joins the points A and B is –1039.

Explanation of Solution

The vector is given as $E=\left(4xy+z\right)a_x+2x^2{a}_y+x{a}_z$ and the points are  $A\left(3,7,1\right)$ to $B\left(8,9,2\right)$ that forms the line segment.

Obtain the parameterization of the line as follows.

$\begin{array}{c}\overrightarrow{l}\left(t\right)=\left(1-t\right)\left(3a_x+7a_y+a_z\right)+t\left(8a_x+9a_y+2a_z\right)\\ =\left[3\left(1-t\right)+8t\right]a_x+\left[7\left(1-t\right)+9t\right]a_y+\left[1-t+2t\right]a_z\\ =\left(3-3t+8t\right)a_x+\left(7-7t+9t\right)a_y+\left(1-t+2t\right)a_z\\ =\left(3+5t\right)a_x+\left(7+2t\right)a_y+\left(1+t\right)a_z\end{array}$

The vector along the line segment becomes

$\begin{array}{c}E=\left(4\left(3+5t\right)\left(7+2t\right)+\left(1+t\right)\right)a_x+2{\left(3+5t\right)}^2{a}_y+\left(3+5t\right)a_z\\ =\left(4\left(21+6t+35t+10t^2\right)+\left(1+t\right)\right)a_x+2\left(9+30t+25t^2\right)a_y+\left(3+5t\right)a_z\\ =\left(84+24t+140t+40t^2+1+t\right)a_x+18+60t+50t^2{a}_y+\left(3+5t\right)a_z\\ =\left(85+165t+40t^2\right)a_x+\left(18+60t+50t^2\right)a_y+\left(3+5t\right)a_z\end{array}$

Obtain the derivative of the parameterization as $dl={\overrightarrow{l}}^{\prime }\left(t\right)=\left[5a_x+2a_y+a_z\right]dt$.

Perform the line integration as follows.

$\begin{array}{c}{\displaystyle {\int }_C^{}E\cdot dl}={\displaystyle {\int }_0^1\left[\left(85+165t+40t^2\right)a_x+\left(18+60t+50t^2\right)a_y+\left(3+5t\right)a_z\right]\cdot \left[5a_x+2a_y+a_z\right]dt}\\ ={\displaystyle {\int }_0^1\left[5\left(85+165t+40t^2\right)+2\left(18+60t+50t^2\right)+\left(3+5t\right)\right]dt}\\ ={\displaystyle {\int }_0^{1}5\left(85+165t+40t^2\right)dt}+{\displaystyle {\int }_0^{1}2\left(18+60t+50t^2\right)dt}+{\displaystyle {\int }_0^1\left(3+5t\right)dt}\\ =5{\left[85t+165\frac{t^2}{2}+40\frac{t^3}{3}\right]}_0^1+2{\left[18t+60\frac{t^2}{2}+50\frac{t^3}{3}\right]}_0^1+{\left[3t+5\frac{t^2}{2}\right]}_0^1\end{array}$

On further simplification,

$\begin{array}{c}{\displaystyle {\int }_C^{}E\cdot dl}=5\left[85+165\frac{1}{2}+40\frac{1}{3}\right]+2\left[18+60\frac{1}{2}+50\frac{1}{3}\right]+\left[3+5\frac{1}{2}\right]\\ =425+\frac{825}{2}+\frac{200}{3}+36+60+\frac{100}{3}+3+\frac{5}{2}\\ =524+415+100\\ =1039\end{array}$

Since $V_{AB}=-{\displaystyle {\int }_C^{}E\cdot dl}$, the value of $V_{AB}$ becomes $V_{AB}=-1039$

Therefore, the line integral $V_{AB}$ along the line that joins the points A and B is –1039.

(ii)

To determine

To evaluate: The line integrals sum along the line that joins the points A and B with stopovers C and D are $C\left(8,7,1\right)$ and $D\left(8,9,1\right)$. 

Answer to Problem 2.8MA

The line integrals sum along the line that joins the points A and B with stopovers C and D are $C\left(8,7,1\right)$ and $D\left(8,9,1\right)$ is –2063.

Explanation of Solution

The vector is given as $E=\left(4xy+z\right)a_x+2x^2{a}_y+x{a}_z$ and the points are  $A\left(3,7,1\right)$ to $B\left(8,9,2\right)$ that forms the line segment with stopovers C and D are $C\left(8,7,1\right)$ and $D\left(8,9,1\right)$.

Obtain the parameterization of the line A to C as follows.

$\begin{array}{c}\overrightarrow{l_1}\left(t\right)=\left(1-t\right)\left(3a_x+7a_y+a_z\right)+t\left(8a_x+7a_y+1a_z\right)\\ =\left[3\left(1-t\right)+8t\right]a_x+\left[7\left(1-t\right)+7t\right]a_y+\left[1-t+t\right]a_z\\ =\left(3-3t+8t\right)a_x+\left(7-7t+7t\right)a_y+\left(1\right)a_z\\ =\left(3+5t\right)a_x+7a_y+a_z\end{array}$

Obtain the parameterization of the line C to D as follows.

$\begin{array}{c}\overrightarrow{l_2}\left(t\right)=\left(1-t\right)\left(8a_x+7a_y+a_z\right)+t\left(8a_x+9a_y+a_z\right)\\ =\left[8\left(1-t\right)+8t\right]a_x+\left[7\left(1-t\right)+9t\right]a_y+\left[1-t+t\right]a_z\\ =\left(8-8t+8t\right)a_x+\left(7-7t+9t\right)a_y+\left(1\right)a_z\\ =8a_x+\left(7+2t\right)a_y+a_z\end{array}$

Obtain the parameterization of the line D to B as follows.

$\begin{array}{c}\overrightarrow{l_3}\left(t\right)=\left(1-t\right)\left(8a_x+9a_y+a_z\right)+t\left(8a_x+9a_y+2a_z\right)\\ =\left[8\left(1-t\right)+8t\right]a_x+\left[9\left(1-t\right)+9t\right]a_y+\left[1-t+2t\right]a_z\\ =\left(8-8t+8t\right)a_x+\left(9-9t+9t\right)a_y+\left(1+t\right)a_z\\ =8a_x+9a_y+\left(1+t\right)a_z\end{array}$

The vector along the line segment $\overrightarrow{l_1}$ becomes

$\begin{array}{c}{E}_1=\left(4\left(3+5t\right)\left(7\right)+1\right)a_x+2{\left(3+5t\right)}^2{a}_y+\left(3+5t\right)a_z\\ =\left(84+140t+1\right)a_x+2\left(9+30t+25t^2\right)a_y+\left(3+5t\right)a_z\\ =\left(140t+85\right)a_x+18+60t+50t^2{a}_y+\left(3+5t\right)a_z\end{array}$

The vector along the line segment $\overrightarrow{l_2}$ becomes

$\begin{array}{c}{E}_2=\left(4\left(8\right)\left(7+2t\right)+1\right)a_x+2{\left(8\right)}^2{a}_y+\left(8\right)a_z\\ =\left(224+64t+1\right)a_x+2\left(64\right)a_y+8a_z\\ =\left(64t+225\right)a_x+128a_y+8a_z\end{array}$

The vector along the line segment $\overrightarrow{l_3}$ becomes

$\begin{array}{c}{E}_3=\left(4\left(8\right)\left(9\right)+1+t\right)a_x+2{\left(8\right)}^2{a}_y+\left(8\right)a_z\\ =\left(288+t+1\right)a_x+2\left(64\right)a_y+8a_z\\ =\left(t+289\right)a_x+128a_y+8a_z\end{array}$

Obtain the derivative of the parameterizations as $d{l}_1={\overrightarrow{l_1}}^{\prime }\left(t\right)=5a_{x}dt$, $d{l}_2={\overrightarrow{l_2}}^{\prime }\left(t\right)=2a_{y}dt$ and $d{l}_3={\overrightarrow{l_3}}^{\prime }\left(t\right)=a_{z}dt$.

Perform the line integration as follows.

$\begin{array}{c}{\displaystyle {\int }_C^{}{E}_1\cdot d{l}_1}={\displaystyle {\int }_0^1\left[\left(140t+85\right)a_x+18+60t+50t^2{a}_y+\left(3+5t\right)a_z\right]\cdot \left[5a_x\right]dt}\\ ={\displaystyle {\int }_0^1\left[5\left(140t+85\right)\right]dt}\\ =5{\left[140\frac{t^2}{2}+85t\right]}_0^1\\ =5\left[70+85\right]\end{array}$

On further simplification,

$\begin{array}{c}{\displaystyle {\int }_C^{}{E}_1\cdot d{l}_1}=5\left[155\right]\\ =775\end{array}$

Perform the line integration as follows.

$\begin{array}{c}{\displaystyle {\int }_C^{}{E}_2\cdot d{l}_2}={\displaystyle {\int }_0^1\left[\left(64t+225\right)a_x+128a_y+8a_z\right]\cdot \left[2a_y\right]dt}\\ ={\displaystyle {\int }_0^1\left[2\left(128\right)\right]dt}\\ =5{\left[256t\right]}_0^1\\ =5\left[256\right]\end{array}$

On further simplification, the integral becomes ${\displaystyle {\int }_C^{}{E}_2\cdot d{l}_2}=1280$.

Perform the line integration as follows.

$\begin{array}{c}{\displaystyle {\int }_C^{}{E}_3\cdot d{l}_3}={\displaystyle {\int }_0^1\left[\left(t+289\right)a_x+128a_y+8a_z\right]\cdot \left[a_y\right]dt}\\ ={\displaystyle {\int }_0^1\left[8\right]dt}\\ ={\left[8t\right]}_0^1\\ =8\end{array}$

Substitute the values of line integrals in the given expression and obtain the result.

 $\begin{array}{c}-{\displaystyle {\int }_A^{B}E\cdot dl}-{\displaystyle {\int }_C^{D}E\cdot dl}-{\displaystyle {\int }_D^{B}E\cdot dl}=-{\displaystyle {\int }_C^{}{E}_1\cdot d{l}_1}-{\displaystyle {\int }_C^{}{E}_2\cdot d{l}_2}-{\displaystyle {\int }_C^{}{E}_3\cdot d{l}_3}\\ =-775-1280-8\\ =-2063\end{array}$

Therefore, the line integrals sum along the line that joins the points A and B with stopovers C and D are $C\left(8,7,1\right)$ and $D\left(8,9,1\right)$ is –2063.