Q29P - Introduction to Electrodynamics

Question

Calculate the line integral of the function $\mathbf{v}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{i}\mathbf{+}\mathbf{2}\mathbf{yx}\mathbf{ }\mathbf{j}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{k}$ from the origin to the point (1,1,1) by three different routes:

(a)   $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{.}$

(b)  $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{.}$

(c) The direct straight line.

(d) What is the line integral around the closed loop that goes out along path (a) and back along path (b)?

Step-by-Step Solution

Answer

(a) The line integral of the vector $v$  through the route  as  $(0,0,0)\to (1,0,0)\to (1,1,0)\to (1,1,1)$ as $\frac{4}{3}$

(b) The line integral of the vector $v$  through the route  as  $(0,0,0)\to (0,0,1)\to (0,1,1)\to (1,1,1)$ as $\frac{4}{3}$

(c) The line integral of the vector $v$  through the route  as  $(0,0,0)\to (1,1,1)$ as$\frac{4}{3}$

(d) The line integral of any vector around closed path is always 0.

Step 1: Define the line integral

The line integral of a vector $v$ along a route  $dI$ is defined as $\int v.dI$ . The line integral of the function $v$  which is defined as  $v=x^{2}i+2yzj+y^2$ is to be computed from origin to point (1,1,1) . Thus the route of the line integral is defined as . $(0,0,0)\to (1,0,0)\to (1,1,0)\to (1,1,1)$

Step 2: Compute the line integral of the vector v from ( 0 , 0 , 0 ) → ( 1 . 0 . 0 )

The y and z coordinate is 0 in the path $(0,0,0)\to (1,0,0)$ . Thus $y=0$  and $z=0$ . The path is changing only in x direction , so  $dI=dxi$

 

The integral of vector $\overrightarrow{v}$ , along the path  $(0,0,0)\to (1,0,0)$ is computed as:

$$\begin{gathered} {\int }_{\left(1\right)}vdl={\int }_{\left(1\right)}(x^{2}i+2yxj+y^{2}k)\left(dxi\right) \\ ={\int }_0^1{x}^{2}dx \\ =\frac{1}{3} \end{gathered}$$

Step 3: Compute the line integral of the vector v from ( 1 , 0 , 0 ) → ( 1 , 1 , 0 )

The x and z coordinate is 0 in the path $(1,0,0)\to (1,1,0)$ . Thus $x=0$and $z=0$ . The path is changing only in y direction , so $dl=dzk$

 

The integral of vector $\overrightarrow{v}$ , along the path  $(1,0,0)\to (1,1,0)$ is computed as:

$$\begin{gathered} {\int }_{\left(2\right)}vdl={\int }_{\left(2\right)}(x^{2}i+2yxj+y^{2}k)\left(dyj\right) \\ ={\int }_0^1\left({\left(1\right)}^{2}i+0+0\right)\left(dyj\right) \\ =0 \end{gathered}$$

Step 4: Compute the line integral of the vector v from ( 1 , 1 , 0 ) → ( 1 , 1 , 1 )

The x and z coordinate is 1 in the path $(1,1,0)\to (1,1,1)$ . Thus $x=1$and $z=1$. The path is changing only in z direction , so $dl=dzk$

 

The integral of vector $\overrightarrow{v}$ , along the path $(1,1,0)\to (1,1,1)$  is computed as:

 $$\begin{gathered} {\int }_{\left(3\right)}vdl={\int }_{\left(3\right)}(x^{2}i+2yxj+y^{2}k)\left(dzk\right) \\ ={\int }_0^1\left(0+0+1\right)\left(dz\right) \\ =1 \end{gathered}$$

Thus the net value of line integral from the origin to point $(1,1,1)$ is the sum of the line integral through path (1), (2), and (3), as follows:

$$\begin{gathered} {\int }_{(0,0,0)\to (1,1,1)}vdl={\int }_{\left(1\right)}vdl+{\int }_{\left(2\right)}vdl+{\int }_{\left(3\right)}vdl \\ =\frac{1}{3}+0+1 \\ =\frac{4}{3} \end{gathered}$$

Therefore for route (a), the line integral is obtained as $\frac{4}{3}$ . Another route (b) is defined as  

$(0,0,0)\to (0,0,1)\to (0,1,1)\to (1,1,1)$

Step 5: Compute the line integral of the vector v from ( 0 , 0 , 0 ) → ( 0 , 0 , 1 )

The y and x coordinate is 0 in the path $(0,0,0)\to (0,0,1)$ . Thus  $y=0$ and $x=0$ . The path is changing only in z direction , so $dl=dzk.$

 

The integral of vector $\overrightarrow{v}$ , along the path $(0,0,0)\to (0,0,1)$  is computed as:

$$\begin{gathered} {\int }_{\left(1\right)}vdl={\int }_{\left(1\right)}(x^{2}i+2yxj+y^{2}k)\left(dzk\right) \\ ={\int }_0^{1}0+0+y^{2}dz \\ =0 \end{gathered}$$

Step 6: Compute the line integral of the vector v from ( 0 , 0 , 1 ) → ( 0 , 1 , 1 )

The x and z coordinate are 0 and 1, respectively in the path $(0,0,1)\to (0,1,1)$ . Thus $x=0$  and $z=1$ . The path is changing only in y direction , so  $dl=dy j$

 

The integral of vector $\overrightarrow{v}$ , along the path $(0,0,1)\to (0,1,1)$  is computed as:

$$\begin{gathered} {\int }_{\left(2\right)}vdl={\int }_{\left(2\right)}(x^{2}i+2yxj+y^{2}k)\left(dxj\right) \\ ={\int }_0^1(0+2y(1)+0)\left(dyj\right) \\ =2{{\left(\frac{y^2}{2}\right)}^1}_0 \\ =1 \end{gathered}$$ 

Step 7: Compute the line integral of the vector v from ( 0 , 1 , 1 ) → ( 1 , 1 , 1 )

The y and z coordinate is 1 in the path $(0,1,1)\to (1,1,1)$ . Thus $y=1$ and $z=1$. The path is changing only in x direction , so$dl=dxi$  

 

The integral of vector $\overrightarrow{v}$, along the path $(0,1,1)\to (1,1,1)$  is computed as:

$$\begin{gathered} {\int }_{\left(3\right)}vdl={\int }_{\left(3\right)}(x^{2}i+2yxj+y^{2}k)\left(dxi\right) \\ ={\int }_0^1\left(x^{2}i+0+0\right)\left(dx\right) \\ ={{\left(\frac{x^3}{3}\right)}^1}_0 \\ =\frac{1}{3} \end{gathered}$$

Thus the net value of line integral from the origin to point $(1,1,1)$ is the sum of the line integral through path (1), (2), and (3), as follows:

$$\begin{gathered} {\int }_{(0,0,0)\to (1,1,1)}vdl={\int }_{\left(1\right)}vdl+{\int }_{\left(2\right)}vdl+{\int }_{\left(3\right)}vdl \\ =0+1+\frac{1}{3} \\ =\frac{4}{3} \end{gathered}$$

Therefore for route (b), the line integral is obtained as $\frac{4}{3}$

Another route (c) is defined as  

Step 8: Compute the line integral of the vector v from ( 0 , 0 , 0 ) → ( 1 , 1 , 1 )

Since all the paths are changing from $(0,0,0)\to (1,1,1)$ , so $dl=dx i +dy j+dz k$ . In this path variations along all direction is same, that is $dx=dy=dz$ . Thus $x=y=z$ .

 

The integral of vector $\overrightarrow{v}$ , along the path $(0,0,0)\to (1,1,1)$  is computed as:

$$\begin{gathered} {\int }_{\left(2\right)}vdl={\int }_{\left(2\right)}(x^{2}i+2yxj+y^{2}k)(dx i+dy j+dz k) \\ ={\int }_0^1{x}^{2}dx+{\int }_0^{1}2y^{2}dx+{\int }_0^1{z}^{2}dx \\ =\frac{1}{3}+\frac{2}{3}+\frac{1}{3} \\ =\frac{4}{3} \end{gathered}$$

Thus line integral of the vector v  which computed from origin to point $(1,1,1)$ , is obtained to be same as $\frac{4}{3}$  through all the routes.