Is F = yi+xzj+zk conservative? Evaluate $\mathbf{\int }\mathit{F}\mathbf{.}\mathit{d}\mathit{r}$from along the paths

(a) broken line (0,0,0)to (1,1,1) to (1,1,0) to (1,1,1)

(b) Straight line connecting the points.

The given information is,

F = yi + xzj + zk

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.

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

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

So, F is not conservative.

The integral is divided into 3 integrals.

$$\begin{gathered} \int F.dr=\int ydx+xzdy+zdz \\ i)From(0,0,0)to(1,0,0) \\ dy=dz \\ =0 \end{gathered}$$

And

$$\begin{gathered} y=z \\ =0 \end{gathered}$$

$\int ydx+xzdy+zdz=0$

ii) From (1,0,0) to (1,1,0)

dx = dz

= 0,

And

x = 1

z = 0

$\int ydx+xzdy+zdz=0$

iii) ) From (1,0,0) to (1,1,0)

dx = dy

= 0,

And

x = 1

y = 1

$$\begin{gathered} \int ydx+xzdy+zdz={\int }_0^{1}zdz \\ ={\overline{)\frac{1}{2}{x}^2}}_0^1 \\ =\frac{1}{2} \end{gathered}$$

The sum is $\frac{1}{2}$.

On the straight line connecting (0,0,0) to (1,1,1) it is given that

dx = dy

= dz

x = y

= z

Thus,

$$\begin{gathered} \int ydx+xzdy+zdz={\int }_0^1\left(2x+x^2\right)dx \\ =(x^2+{\overline{)\frac{1}{3}{x}^3)}}_0^1 \\ =\frac{4}{3} \end{gathered}$$

Hence, the solution to this problem is

F is not conservative.

a) The sum is $\frac{1}{2}$

b) $\int ydx+xzdy+zdz=\frac{4}{3}$