91影视

The given information is,

$$\begin{gathered} F_1=2xzi+yj+x^{2}k \\ {F}_2=yi-xj \end{gathered}$$

The Green's theorem connects a line integral around a simple closed curve C to a double integral over the plane region D circumscribed by C in vector calculus. Stokes' theorem has a two-dimensional special case.

a)If $鈭�脳F=0$ then F is said to be conservative.

role="math" localid="1659331799869" $$\begin{gathered} 鈭�脳F_1=\left|\begin{array}{ccc}i& j& k\\ \frac{鈭�}{鈭�x}& \frac{鈭�}{鈭�y}& \frac{鈭�}{鈭�z}\\ 2xz& y& x^2\end{array}\right| \\ =\left(0-0\right)i-\left(2x-2x\right)j+\left(0-0\right)k \\ =0 \end{gathered}$$

So, $F_1$ is conservative.

$$\begin{gathered} 鈭�脳F_2=\left|\begin{array}{ccc}i& j& k\\ \frac{鈭�}{鈭�x}& \frac{鈭�}{鈭�y}& \frac{鈭�}{鈭�z}\\ y& -x& 0\end{array}\right| \\ =\left(0-0\right)i-\left(0-0\right)j+\left(\left(-1\right)-1\right)k \\ =-2k \end{gathered}$$

So, $F_2$ is not conservative

b) It is known that$F_1=鈭�W.$.

$$\begin{gathered} \frac{鈭�W}{鈭�x}=(F_1{)}_x-2xz.......(1) \\ \frac{鈭�W}{鈭�y}=(F_1{)}_y-y \\ \frac{鈭�W}{鈭�z}=(F_1{)}_z-x^2......(2) \end{gathered}$$

Now, integrate the below equation with respect to y.

$$\begin{gathered} W=鈭�ydy \\ =\frac{1}{2}{y}^2+g\left(x,z\right) \end{gathered}$$

Where g is an arbitrary function in x and z only.

Substitute W in equation (1).

role="math" localid="1659332223371" $$\begin{gathered} \frac{鈭�W}{鈭�x}=\frac{鈭�g}{鈭�x} \\ =2xz鈫�g\left(x,z\right) \\ =鈭�2xzdx \\ =x^{2}z+h\left(z\right).....\left(3\right) \end{gathered}$$

Where h is a function of z only.

Substitute value of equation (3) in equation (2).

$$\begin{gathered} \frac{鈭�W}{鈭�Z}=x^2+\frac{鈭�h}{鈭�z} \\ =x^2鈫�\frac{鈭�h}{鈭�z} \\ =0鈫�h\left(z\right) \\ =0 \end{gathered}$$

Where c is an arbitrary constant.

Therefore,

$W=\frac{1}{2}{y}^2+x^{2}z$

(c)$鈭�F_2鈰�dx={鈭�}_{\left(0,1\right)}^{\left(1,0\right)}ydx-xdy$

Along the straight line (0,1) and (1,0), the relation between x and y is given by the equation $y=1-x鈫�dy=-dx$.

$$\begin{gathered} {鈭�}_{\left(0,1\right)}^{\left(1,0\right)}ydx-xdy={鈭�}_0^1\left(1-x+x\right)dx \\ ={\overline{)\left(c\right)}}_0^1 \\ =1 \end{gathered}$$

(d)

Use Green鈥檚 Theorem.

$鈭�{鈭�}_A(鈭�Q/鈭�x-鈭�P/鈭�y)dxdy={鈭�}_{鈭�}APdx+Qdy$

Where $鈭�A$is the boundary of the area A .

It can be seen that

$$\begin{gathered} P=y鈫�\frac{鈭�P}{鈭�y} \\ =1 \\ Q=-x鈫�\frac{鈭�Q}{鈭�X} \\ =-1 \end{gathered}$$

Thus, the integral over some contour C the encloses area A is

$$\begin{gathered} {鈭�}_c\left(y\right)dx+\left(x\right)dy={鈭�}_A\left(-1-1\right)dxdy \\ =-2{鈭�}_{A}dxdy \\ =-2A \end{gathered}$$

So, for the triangle with vertices $(0,0),(0,1),(1,0),$ its area is given by

role="math" localid="1659333082552" $$\begin{gathered} A=\frac{1}{2}\left(1\right)\left(1\right) \\ =\frac{1}{2} \\ {鈭�}_{triangle}{F}_2鈰�dr=-2A \\ =-1 \end{gathered}$$

But,

$$\begin{gathered} {鈭�}_{{}_{t}riangle}{F}_2鈰�dr={鈭�}_{(1,0)}^{(0,0)}{F}_2鈰�dr+{鈭�}_{(1,0)}^{(0,1)}{F}_2鈰�dr+{鈭�}_{(1,0)}^{(0,0)}{F}_2鈰�dr \\ ={鈭�}_0^1\left(0\right)dz+{鈭�}_{(0,1)}^{(1,0)}ydx-xdy+{鈭�}_1^0-\left(0\right)dy \\ ={鈭�}_{(0,1)}^{(1,0)}ydx-xdy \\ {鈭�}_{(0,1)}^{(1,0)}ydx-xdy=-1鈫�{鈭�}_{(1,0)}^{(0,1)}ydx-xdy \\ =1 \end{gathered}$$

Hence, the solutions of the problems are,

$$\begin{gathered} a)F_{2}isnotconservative. \\ b)W=\frac{1}{2}{y}^2+x^{2}z \\ c)I=1 \\ d)I=1 \end{gathered}$$