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) role="math" localid="1657357520925" $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{.}$

(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 outalong path (a) and backalong path (b)?

Short 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-by-step solution

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

 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:

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

 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:

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

 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

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

 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:

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

 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 $dl=dyj$

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

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

 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

 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 $dl=dxi+dyj+dzk$. In this path variations along all direction is same, that is $dx=dy=dz$. Thus $x=y=z$.

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

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

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.