91Ó°ÊÓ

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 $b=(1,1,1)$.

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 ${∫}_l=∇f-dl$.

(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 $\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{→}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{→}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{→}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}$.

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 $dl=dxi$.

The gradient of the function T is computed as follows

$$\begin{gathered} ∇Tdl=\left(\frac{∂}{∂x}i+\frac{∂}{∂y}j+\frac{∂}{∂z}k\right)(x^2+4xy+2y{z}^2) \\ =(2x+4y)i+(4x+2z^2)j+\left(6y{z}^2\right)k \end{gathered}$$

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

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

Solve further as,

localid="1657516376826" $$\begin{gathered} {∫}_{\left(1\right)}∇Tdl={∫}_0^1\left(2x\right)dx \\ ={\left(x^2\right)}_0^1 \\ =1 \end{gathered}$$

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

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

localid="1657516739534" $$\begin{gathered} {∫}_{\left(2\right)}∇Tdl={∫}_{\left(2\right)}\left((2x+4y)i+(4x+2z^2)j+\left(6y{z}^2\right)k\right)(dyj) \\ ={∫}_0^1(4x+2z^2)dx \\ ={∫}_0^1\left(4\right(1)+2{\left(0\right)}^2)dx \\ ={∫}_0^{1}4dx \end{gathered}$$

Solve further as,

$$\begin{gathered} {∫}_{\left(2\right)}∇Tdl={∫}_0^{1}4dx \\ =4 \end{gathered}$$

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

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

localid="1657517404150" $$\begin{gathered} {∫}_{\left(3\right)}∇Tdl={∫}_{\left(3\right)}\left((2x+4y)i+(4x+2z^2)j+\left(6y{z}^2\right)k\right)(dzk) \\ ={∫}_0^1\left(6y{z}^2\right)dx \\ ={∫}_0^1\left(6\right(1\left)z^2\right)dx \\ ={∫}_0^{1}6z^{2}dx \end{gathered}$$

Solve further as,

localid="1657517652325" $$\begin{gathered} {∫}_{\left(3\right)}∇Tdl={∫}_0^{1}6z^{2}dx \\ =\frac{6z^2}{3} \\ ={\left(2z^3\right)}_0^1 \\ =2 \end{gathered}$$

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:

$$\begin{gathered} {∫}_{(0,0,0)→(1,1,1)}∇Tdl={∫}_{\left(1\right)}∇Tdl+{∫}_{\left(2\right)}∇Tdl+{∫}_{\left(3\right)}∇Tdl \\ =1+4+2 \\ =7 \end{gathered}$$

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

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

Thus, the integral of gradient of function $T$along the path $(0,0,0)→(1,0,0)→(1,1,0)→(1,1,1)$, in part (a) is obtained as ${∫}_a^b∇Tdl=7$

(b)

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 T , along the path $(0,0,0)→(1,0,0)$ is computed as:

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

The x and z coordinate are 0 and 1, respectively in the pathlocalid="1657518764897" $(0,0,1)→(0,1,1)$. Thus $x=0$ and z $z=1$. The path is changing only in y direction, so $dl=dyj$

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

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

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

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

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

Solve further as,

$$\begin{gathered} {∫}_{\left(3\right)}∇Tdl={∫}_0^1\left(2x+4\right)dx \\ ={\left(x^2+4x\right)}_0^1 \\ =(1+4) \\ =5 \end{gathered}$$

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:

$$\begin{gathered} {∫}_{(0,0,0)→(1,1,1)}∇Tdl={∫}_{\left(1\right)}∇Tdl+{∫}_{\left(2\right)}∇Tdl+{∫}_{\left(3\right)}∇Tdl \\ =0+2+5 \\ =7 \end{gathered}$$

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

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

Thus, the integral of gradients of function T along the path $(0,0,0)→(0,0,1)→(0,1,1)→(1,1,1)$in path (b) is obtained as ${∫}_a^b∇Tdl=7$

(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 $dxi+dyj+dzk.$

In this path variations $∫∇Tdl$can be calculated as.

localid="1657521255032" $$\begin{gathered} ∫∇Tdl={∫}_0^1\left((2x+4y)i+(4x+2z^2)j+\left(6y{z}^2\right)k\right)(dxi+dyj+dzk) \\ ={∫}_0^1(2x+4y)dx+(4x+2z^2)dy+\left(6y{z}^2\right)dz.........\left(1\right) \end{gathered}$$

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

localid="1657522281219" $$\begin{gathered} ∫∇Tdl={∫}_0^1(2x+4(x\left)\right)dx+(4x+2{\left(x^2\right)}^3)dx+\left(6\right(x\left)\right(x^2\left)\right(2xdx) \\ ={∫}_0^1(10x+14x^6)dx \\ ={(5x^2+2x^7)}_0^1 \\ =7 \end{gathered}$$

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

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

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