91Ó°ÊÓ

The line integral of a vector $v$along a route $dI$is defined as$∫v.dI$. 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 .$(0,0,0)→(1,0,0)→(1,1,0)→(1,1,1)$

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

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

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

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

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

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

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

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

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

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:

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

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

$(0,0,0)→(0,0,1)→(0,1,1)→(1,1,1)$

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

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

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

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

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

localid="1657361157696" $$\begin{gathered} {∫}_{\left(2\right)}vdl={∫}_{\left(2\right)}(x^{2}i+2yxj+y^{2}k)\left(dxj\right) \\ ={∫}_0^1(0+2y(1)+0)\left(dyj\right) \\ =2{{\left(\frac{y^2}{2}\right)}^1}_0 \\ =1 \end{gathered}$$

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

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

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

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:

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

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

Another route (c) is defined as

Since all the paths are changing from $(0,0,0)→(1,1,1)$, so $dl=dxi+dyj+dzk$. In this path variations along all direction is same, that is $dx=dy=dz$. Thus $x=y=z$.

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

$$\begin{gathered} {∫}_{\left(2\right)}vdl={∫}_{\left(2\right)}(x^{2}i+2yxj+y^{2}k)(dxi+dyj+dzk) \\ ={∫}_0^1{x}^{2}dx+{∫}_0^{1}2y^{2}dx+{∫}_0^1{z}^{2}dx \\ =\frac{1}{3}+\frac{2}{3}+\frac{1}{3} \\ =\frac{4}{3} \end{gathered}$$

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.