Q32P - Introduction to Electrodynamics

Question

Check the fundamental theorem for gradients, using $\mathbf{T}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{xy}\mathbf{+}\mathbf{2}{\mathbf{yz}}^{\mathbf{3}}$ the points $\mathbf{a}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{b}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}$ and the three paths in Fig. 1.28.

 $$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{.} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{.} \end{gathered}$$ 

  (c) The parabolic path $\mathbf{z}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{,}\mathbf{y}\mathbf{=}\mathbf{x}$

Step-by-Step Solution

Answer

 

(a) The integral of gradient of function T along the path $\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}$ , in path (a) is obtained as ${\int }_a^b\nabla Tdl=7.$ 

(b) The integral of gradient of function T along the path$\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}$ , in path (a) is obtained as ${\int }_a^b\nabla Tdl=7.$

(c) The integral of gradient of function  along the parabolic path in path (c), is obtained as  ${\int }_a^b\nabla Tdl=7.$

Step 1: Describe the given information

The given function is $T=x^2+4xy+2y{z}^2$. The stokes theorem is toe verified along the source point $a=(0.0.0)$  and the destination point $b=(1,1,1)$.

Step 2: Describe the integral of gradient of the function

The integral of gradient of a function  $f(x,y,z)$ over an open surface area is equal to the line integral of the gradient of the function, as ${\int }_l=\nabla f-dl$.

Step 3: Compute the integral of gradient for route in part (a)

(a)

The line integral of the function T which is defined as $T=x^2+4xy+2y{z}^2$ is to be computed from origin to point$(1,1,1)$. Thus the route of the line integral is defined as $\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}$.

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 $dl=dx i$ .

 

The gradient of the function T  is computed as follows

 $$\begin{gathered} \nabla T dl=\left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right)(x^2+4xy+2y{z}^2) \\ =(2x+4y)i+(4x+2z^2)j+\left(6y{z}^2\right)k \end{gathered}$$

 

The integral of function  T along the path $(0,0,0)\to (1,0,0)$  is computed as: 

$$\begin{gathered} {\int }_{\left(1\right)}\nabla T dl={\int }_{\left(1\right)}\left((2x+4y)i+(4x+2z^2)j+\left(6y{z}^2\right)k\right)(dx i) \\ ={\int }_0^1(2x+4y)dx \\ ={\int }_0^1(2x+4(0\left)\right)dx \\ ={\int }_0^1\left(2x\right)dx \end{gathered}$$

 

Solve further as,

 $$\begin{gathered} {\int }_{\left(1\right)}\nabla T dl={\int }_0^1\left(2x\right)dx \\ ={\left(x^2\right)}_0^1 \\ =1 \end{gathered}$$

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

 

The integral of vector T , along the path $(1,0,0)\to (1,1,0)$  is computed as: 

 $$\begin{gathered} {\int }_{\left(2\right)}\nabla Tdl={\int }_{\left(2\right)}\left((2x+4y)i+(4x+2z^2)j+\left(6y{z}^2\right)k\right)(dy j) \\ ={\int }_0^1(4x+2z^2)dx \\ ={\int }_0^1\left(4\right(1)+2{\left(0\right)}^2)dx \\ ={\int }_0^{1}4dx \end{gathered}$$

Solve further as,

 $$\begin{gathered} {\int }_{\left(2\right)}\nabla T dl={\int }_0^{1}4dx \\ =4 \end{gathered}$$

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

 

The integral of vector $T$ , along the path  $(1,1,0)\to (1,1,1)$ is computed as:

  $$\begin{gathered} {\int }_{\left(3\right)}\nabla Tdl={\int }_{\left(3\right)}\left((2x+4y)i+(4x+2z^2)j+\left(6y{z}^2\right)k\right)(dz k) \\ ={\int }_0^1\left(6y{z}^2\right)dx \\ ={\int }_0^1\left(6\right(1\left)z^2\right)dx \\ ={\int }_0^{1}6z^{2}dx \end{gathered}$$

Solve further as,

$$\begin{gathered} {\int }_{\left(3\right)}\nabla T dl={\int }_0^{1}6z^{2}dx \\ =\frac{6z^2}{3} \\ ={\left(2z^3\right)}_0^1 \\ =2 \end{gathered}$$ 

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

$$\begin{gathered} {\int }_{(0,0,0)\to (1,1,1)}\nabla T dl={\int }_{\left(1\right)}\nabla T dl+{\int }_{\left(2\right)}\nabla T dl+{\int }_{\left(3\right)}\nabla T dl \\ =1+4+2 \\ =7 \end{gathered}$$

Compute the value of integral of gradient using difference of source and destination point as,

$$\begin{gathered} {\int }_a^b\nabla T dl=T\left(b\right)-T\left(a\right) \\ ={\left|x^2+4xy+2y{z}^3\right|}_{(1,1,1)}-{\left|x^2+4xy+2y{z}^3\right|}_{(0,0,0)} \\ =7-0 \\ =7 \end{gathered}$$

Thus, the integral of gradient of function $T$ along the path $(0,0,0)\to (1,0,0)\to (1,1,0)\to (1,1,1)$, in part (a) is obtained as  ${\int }_a^b\nabla T dl=7$

Step 4: Compute the integral of gradient for route in part (b)

(b)

Another route (b) is defined as $(0,0,0)\to (0,0,1)\to (0,1,1)\to (1,1,1)$. 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 $dl=dz k$.

 

The integral of vector T , along the path $(0,0,0)\to (1,0,0)$  is computed as: 

 

 $$\begin{gathered} {\int }_{\left(1\right)}\nabla Tdl={\int }_{\left(1\right)}\left((2x+4y)i+(4x+2z^2)j+\left(6y{z}^2\right)k\right)(dz k) \\ ={\int }_0^1\left(6y{z}^2\right)dx \\ ={\int }_0^1\left(6\right(0\left)z^2\right)dx \\ =0 \end{gathered}$$

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 $z=1$ . The path is changing only in y direction, so  $dl=dy j$

 

The integral of vector T along the path $(0,0,1)\to (0,1,1)$  is computed as: 

 $$\begin{gathered} {\int }_{\left(2\right)}\nabla Tdl={\int }_{\left(2\right)}\left((2x+4y)i+(4x+2z^2)j+\left(6y{z}^2\right)k\right)(dy j) \\ ={\int }_0^1(4x+2z^2)dx \\ ={\int }_0^1\left(4\right(1)+2{\left(1\right)}^2)dx \\ =2 \end{gathered}$$

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  $dl=dx i$

 

The integral of vector T , along the path $(0,1,1)\to (1,1,1)$ is computed as:

 $$\begin{gathered} {\int }_{\left(3\right)}\nabla T dl={\int }_{\left(3\right)}\left((2x+4y)i+(4x+2z^2)j+\left(6y{z}^2\right)k\right)(dx i) \\ ={\int }_0^1(2x+4y)dx \\ ={\int }_0^1(2x+4(1\left)\right)dx \\ ={\int }_0^1(2x+4)dx \end{gathered}$$

Solve further as,

 $$\begin{gathered} {\int }_{\left(3\right)}\nabla T dl={\int }_0^1\left(2x+4\right)dx \\ ={\left(x^2+4x\right)}_0^1 \\ =(1+4) \\ =5 \end{gathered}$$

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

 $$\begin{gathered} {\int }_{(0,0,0)\to (1,1,1)}\nabla T dl={\int }_{\left(1\right)}\nabla T dl+{\int }_{\left(2\right)}\nabla T dl+{\int }_{\left(3\right)}\nabla T dl \\ =0+2+5 \\ =7 \end{gathered}$$

Compute the value of integral of gradient using difference of source and destination point as,

 $$\begin{gathered} {\int }_a^b\nabla T dl=T\left(b\right)-T\left(a\right) \\ ={\left|x^2+4xy+2y{z}^3\right|}_{(1,1,1)}-{\left|x^2+4xy+2y{z}^3\right|}_{(0,0,0)} \\ =7-0 \\ =7 \end{gathered}$$

Thus, the integral of gradients of function T along the path $(0,0,0)\to (0,0,1)\to (0,1,1)\to (1,1,1)$ in path (b) is obtained as  ${\int }_a^b\nabla T dl=7$

Step 5: Compute the integral of gradient for parabolic route in part (c)

(c)

The coordinates of parabolic path are given as 

           $$\begin{gathered} z=x^2 \\ y=x \end{gathered}$$

Differentiate both the equations with respect to x as:

         $$\begin{gathered} dz=2xdx \\ dy=dx \end{gathered}$$

The gradient of the function T is obtained as $(2x+4y)i+(4x+2z^2)j+\left(6y{z}^2\right)k$ since all the paths are changing in the parabolic path, so $dx i+dy j+dz k.$

 

In this path variations $\int \nabla Tdl$ can be calculated as.

$$\begin{gathered} \int \nabla T dl={\int }_0^1\left((2x+4y)i+(4x+2z^2)j+\left(6y{z}^2\right)k\right)(dx i+dy j+dz k) \\ ={\int }_0^1(2x+4y)dx+(4x+2z^2)dy+\left(6y{z}^2\right)dz .........\left(1\right) \end{gathered}$$ 

Substitute  $x^2$ for z , x for y ,  2 xdx for dz ,and dx for dy  into equation (1)

 $$\begin{gathered} \int \nabla T dl={\int }_0^1(2x+4(x\left)\right)dx+(4x+2{\left(x^2\right)}^3)dx+\left(6\right(x\left)\right(x^2\left)\right(2xdx) \\ ={\int }_0^1(10x+14x^6)dx \\ ={(5x^2+2x^7)}_0^1 \\ =7 \end{gathered}$$

Compute the value of integral of gradient using difference of source and destination point as,

$$\begin{gathered} {\int }_a^b\nabla T dl=T\left(b\right)-T\left(a\right) \\ ={\left|x^2+4xy+2y{z}^3\right|}_{(1.1,1)}-{\left|x^2+4xy+2y{z}^3\right|}_{(0,0,0)} \\ =7-0 \\ =7 \end{gathered}$$

Thus, the integral of gradient of function $T$ along the parabolic path in part (c), is obtained as  ${\int }_a^b\nabla T dl=7.$