Given

$$\begin{gathered} {\mathit{F}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathit{x}\mathit{z}\mathit{i}\mathbf{+}\mathit{y}\mathit{j}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathit{k} \\ {\mathit{F}}_{\mathbf{2}}\mathbf{=}\mathit{y}\mathit{i}\mathbf{-}\mathit{x}\mathit{j}\mathbf{:} \end{gathered}$$

(a) Which F , if either, is conservative?

(b) If one of the given ’s is conservative, find a function Wso that $\mathit{F}\mathbf{=}\mathbf{\nabla }\mathit{W}\mathbf{.}$

(c) If one of the F’s is non conservative, use it to evaluate $\mathbf{\int }\mathit{F}$ along the straight line from $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{}\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{.}\mathbf{}$

(d) Do part (c) by applying Green’s theorem to the triangle with vertices $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{.}$$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{.}$.

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 $\nabla \times F=0$ then F is said to be conservative.

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

So, $F_1$ is conservative.

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

So, $F_2$ is not conservative

b) It is known that$F_1=\nabla W.$.

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

Now, integrate the below equation with respect to y.

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

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

Substitute W in equation (1).

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

Where h is a function of z only.

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

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

Where c is an arbitrary constant.

Therefore,

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

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

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

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

(d)

Use Green’s Theorem.

$\int {\int }_A(\partial Q/\partial x-\partial P/\partial y)dxdy={\oint }_{\partial }APdx+Qdy$

Where $\partial A$is the boundary of the area A .

It can be seen that

$$\begin{gathered} P=y\to \frac{\partial P}{\partial y} \\ =1 \\ Q=-x\to \frac{\partial Q}{\partial X} \\ =-1 \end{gathered}$$

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

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

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

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

But,

$$\begin{gathered} {\oint }_{{}_{t}riangle}{F}_2\cdot dr={\int }_{(1,0)}^{(0,0)}{F}_2\cdot dr+{\int }_{(1,0)}^{(0,1)}{F}_2\cdot dr+{\int }_{(1,0)}^{(0,0)}{F}_2\cdot dr \\ ={\int }_0^1\left(0\right)dz+{\int }_{(0,1)}^{(1,0)}ydx-xdy+{\int }_1^0-\left(0\right)dy \\ ={\int }_{(0,1)}^{(1,0)}ydx-xdy \\ {\int }_{(0,1)}^{(1,0)}ydx-xdy=-1\to {\int }_{(1,0)}^{(0,1)}ydx-xdy \\ =1 \end{gathered}$$

Hence, the solutions of the problems are,

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