Q56P - Introduction to Electrodynamics

Question

Compute the line integral of

 $v=6i+y^{2}j+\left(3y+z\right)k$

along the triangular path shown in Fig. 1.49. Check your answer using Stokes' theorem. [Answer: 8/3]

Step-by-Step Solution

Answer

The line integral of the given function is obtained as $\frac{8}{3}$. The left and right side of the stokes theorem are equal to $\frac{8}{3}$ .

Step 1: Describe the given information

The given function is  $v=6i+y^{2}j+\left(3y+z\right)k$ and the del operatoe is defined as $\nabla =\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k$  . The line integral of the function v which is defined as $v=x^{2}i+2yzj+y^{2}k$  is to be computed along the following path:

 

Step 2: Define Stokes theorem

The integral of curl of a vector function  v over an open surface area is equal to the line integral of the vector function, $\int \left(\nabla \times v\right)\cdot da=\underset{l}{\oint }v\cdot dl$

The line integral of a vector v along a route is defined as$\int v\cdot dl$ . 

Step 3: Compute the line integralof the vector v along path (i)

The x and z coordinate is 0 in the path (i). Thus $dx=dz=0$ and z = 0. The path is changing only in y direction, from 0 to 1 , so $dl=dyj$ .

 

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

 $\begin{array}{c}\underset{l}{\overset{}{\int }}vdl=\underset{0}{\overset{1}{\int }}\left(y{z}^2\right)\left(dy\right)\\ =\underset{0}{\overset{1}{\int }}\left(y{\left(0\right)}^2\right)\left(dy\right)\\ =0\end{array}$

Step 4: Compute the line integral of the vector v along path (ii)

The curve along path (ii) is a line, the equation of which can be written as $z-z_1=m\left(y-y_1\right)$  , where is the slope of the line, $z_1$  is the z coordinate, and  $y_1$ is the y coordinate of the line.

 

From the figure,$\left(0,2\right),\left(1,0\right)$ are the points lying on the line. So, th eslopeof the line can be obtained as

 $\begin{array}{c}m=\frac{2-0}{0-1}\\ =-2\end{array}$

 

 

Substitute  -2 for m , 0 for $z_1$ , 0 for $y_2$ , and 1 for $y_1$  into the equation .

 $z-z_1=m\left(y-y_1\right)$

 $\begin{array}{c}z-0=-2\left(y-1\right)\\ z=2\left(1-y\right)\end{array}$

 

The equation of the line is obtained as $z=2\left(1-y\right)$ .  Differentiate this equation with respect to y as,

 $\begin{array}{l}z=2\left(1-y\right)\\ dz=-2dy\end{array}$

 

 

Here,  z varies from 0 to 2 and y  varies from 1 to 0.

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

$\begin{array}{c}\underset{\left(2\right)}{\overset{}{\int }}vdl=\underset{}{\overset{}{\int }}\left(y{z}^2\right)dy+\left(3y+z\right)\left(dz\right)\\ =\underset{}{\overset{}{\int }}\left(4y{\left(1-y\right)}^2\right)dy+\left(3y+2\left(1-y\right)\right)\left(2dy\right)\\ ={\left[y^2+y^4-\frac{8y^3}{3}-4y\right]}_1^0\\ =\frac{14}{3}\end{array}$

 

Step 5: Compute the line integral of the vector along path (iii)

The x and y coordinate is 0 in the path (iii). The path is changing only in z direction,from 2 to 0, so $dl=dzk$ 

 

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

 $\begin{array}{c}\underset{\left(3\right)}{\overset{}{\int }}vdl=\underset{2}{\overset{0}{\int }}\left(3y+z\right)\left(dz\right)\\ =\underset{2}{\overset{0}{\int }}\left(3\left(0\right)+z\right)\left(dz\right)\\ ={\left[\frac{z^2}{2}\right]}_2^0\\ =-2\end{array}$

 

 

Thus the net value of line integral along the path given is the sum of the line integral through path (1), (2), and (3), as follows:

 $\begin{array}{c}\underset{l}{\overset{}{\int }}vdl=\underset{\left(1\right)}{\overset{}{\int }}vdl+\underset{\left(2\right)}{\overset{}{\int }}vdl+\underset{\left(3\right)}{\overset{}{\int }}vdl\\ =0+\frac{14}{3}-2\\ =\frac{8}{3}\end{array}$

 

 

Therefore for given route the line integral is obtained as $\frac{8}{3}$  .

 

Step 6: Compute the left side of strokes theorem

Compute the curl of vector v as

 $\begin{array}{c}\left(\nabla \times \mathbf{v}\right)=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 6& y{z}^2& 3y+z\end{array}\right|\\ =\left(\frac{\partial }{\partial y}\left(3y+z\right)-\frac{\partial }{\partial z}\left(y{z}^2\right)\right)i-\left(\frac{\partial }{\partial yx}\left(3y+z\right)-\frac{\partial }{\partial z}\left(6\right)\right)j+\left(\frac{\partial }{\partial x}\left(y{z}^2\right)-\frac{\partial }{\partial y}\left(6\right)\right)k\\ =\left(3-2yz\right)i-0-0\\ =\left(3-2yz\right)i\end{array}$

 

The area vector is given by $\overrightarrow{da}=dydzi$ as the open surface area lies in y-z plane. The left part of the strokes theorem is calculated as:

 $\begin{array}{c}\underset{S}{\overset{}{\int }}\left(\nabla \times \mathbf{v}\right)\cdot \overrightarrow{da}=\underset{S}{\int }\left(3-2yz\right)i\cdot \left(dydzi\right)\\ =\underset{0}{\overset{2\left(1-y\right)}{\int }}\underset{0}{\overset{y}{\int }}\left(3-2yz\right)dydz\\ =\underset{0}{\overset{1}{\int }}{\left[3z-y{z}^2\right]}_0^{2\left(1-y\right)}dy\\ =\underset{0}{\overset{1}{\int }}\left(3\left(2\left(1-y\right)\right)-y{\left(2\left(1-y\right)\right)}^2\right)dy\end{array}$

 

 

Solve further as,

$\begin{array}{c}\underset{S}{\overset{}{\int }}\left(\nabla \times \mathbf{v}\right)\cdot \overrightarrow{da}=\underset{0}{\overset{1}{\int }}\left(3\left(2\left(1-y\right)\right)-y{\left(2\left(1-y\right)\right)}^2\right)dy\\ =\underset{0}{\overset{1}{\int }}\left(6-10y-4y^3+8y^2\right)dy\\ ={\left[6y-5y^2-y^4+\frac{8y^3}{3}\right]}_0^1\\ =\frac{8}{3}\end{array}$ 

 

From equations (1) and (2), It can be concluded that the left and right side of the stokes theorem are equal to $\frac{8}{3}$ .