91影视

The given information is,

$$\begin{gathered} {\text{F}}_1=-2yi+\left(z-2x\right)j+\left(y+z\right)k \\ {F}_2=yi+2xj \end{gathered}$$

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="1659348691097" $$\begin{gathered} 鈭�脳F_1=\left|\begin{array}{ccc}i& j& k\\ \frac{鈭�}{鈭�x}& \frac{鈭�}{鈭�y}& \frac{鈭�}{鈭�z}\\ -2y& z-2x& y+z\end{array}\right| \\ =\left(1-1\right)i-\left(0-0\right)j+\left(\left(-2\right)-\left(-2\right)\right)k \\ =0 \end{gathered}$$

So, $F_1$ is conservative.

$$\begin{gathered} 鈭�脳F_1=\left|\begin{array}{ccc}i& j& k\\ \frac{鈭�}{鈭�x}& \frac{鈭�}{鈭�y}& \frac{鈭�}{鈭�z}\\ y& 2x& 0\end{array}\right| \\ =\left(0-0\right)i-\left(0-0\right)j+\left(\left(-2\right)-\left(-1\right)\right)k \\ =k \end{gathered}$$

So, $F_2$ is not conservative

b) The work done is given by $鈭�F_2.dr=鈭�ydx+2xdy$.

Let parameterization be,

$$\begin{gathered} x=\cos 胃 \\ dx=-\sin 胃d胃 \\ y=2\sin 胃 \\ dy=2\cos 胃d胃 \end{gathered}$$

Thus,

$$\begin{gathered} 鈭�ydx+2xdy={鈭�}_0^{2\mathrm{蟺}}\left(-2{\sin }^2胃+4{\cos }^2胃\right)d胃 \\ ={鈭�}_0^{2\mathrm{蟺}}\left(1+3\cos 2胃\right)d胃 \end{gathered}$$

$$\begin{gathered} {\overline{)=2蟺+\left(\frac{3}{2}\sin 2胃\right)}}_0^{2\mathrm{蟺}} \\ =2\mathrm{蟺} \end{gathered}$$

The work done by $F_2$ over the path is $2\mathrm{蟺}$.$$\begin{gathered} 鈭�ydx+22xdy={鈭�}_0^{2\mathrm{蟺}}\left(-2{\sin }^2胃+4{\cos }^2胃\right)d胃 \\ ={鈭�}_0^{2\mathrm{蟺}}\left(1+3\cos 2胃\right)d胃 \\ {\overline{)=2\mathrm{蟺}+\left(\frac{3}{2}\sin 2\mathrm{胃}\right)}}_0^{2\mathrm{蟺}} \\ =2\mathrm{蟺} \\ \mathrm{The}\mathrm{work}\mathrm{done}\mathrm{by}{\mathrm{F}}_2\mathrm{over}\mathrm{the}\mathrm{path}\mathrm{is}2\mathrm{蟺}. \end{gathered}$$ .

c) It is known that $F_1=鈭�V$.

Now, integrate the below equation with respect to y.

$$\begin{gathered} V=鈭�\left(2x-z\right)dy \\ =\left(2x-z\right)y+g\left(x,z\right) \end{gathered}$$

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

Substitute W in equation (1).

localid="1659350094481" $$\begin{gathered} \frac{鈭�V}{鈭�x}=2y+\frac{鈭�g}{鈭�x} \\ =2y鈫�\frac{鈭�g}{鈭�x} \\ =0鈫�g\left(x,z\right) \\ =g\left(z\right)........\left(3\right) \end{gathered}$$

Where g is a function of z only.

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

$$\begin{gathered} \frac{鈭�V}{鈭�x}=2y+\frac{鈭�g}{鈭�x} \\ =-y-z鈫�\frac{鈭�g}{鈭�x} \\ =-z鈫�g\left(z\right) \\ =-\frac{1}{2}{z}^2+c \end{gathered}$$

Where c is an arbitrary constant.

Therefore,

.$V=\left(2x-z\right)y-\frac{1}{2}{z}^2$

d) The work done by $F_1$ is given by $鈭�F_1.dr=鈭�-2ydx+\left(z-2x\right)dy+\left(y+z\right)dz$.

Since, $F_1$ is conservative, so the work is the path independent, and therefore divide the path over two segments:

i) From $\left(0,1,0\right)to\left(0,2,0\right),$

$$\begin{gathered} dx=dz \\ =0 \end{gathered}$$

And

$$\begin{gathered} x=z \\ =0 \\ 鈭�-2ydx+\left(z-2x\right)dy+\left(y+z\right)dz=0 \end{gathered}$$

i) From$\left(0,2,0\right)to\left(0,2,5\right),$

$$\begin{gathered} dx=dy \\ =0 \end{gathered}$$

$$\begin{gathered} And \\ x=0 \\ y=2 \end{gathered}$$

$$\begin{gathered} 鈭�-2ydx+\left(z-2x\right)dy+\left(y+z\right)dz={鈭�}_0^5\left(2+z\right)dz \\ {\overline{)=\left(2z+\frac{1}{2}{z}^2\right)}}_0^5 \\ =10+\frac{25}{2} \\ =\frac{45}{2} \end{gathered}$$

The work done by $F_1$ over the path is $\frac{45}{2}$.

$$\begin{gathered} e)UseGreen鈥�sTheorem. \\ 鈭�{鈭�}_A\left(\frac{鈭�Q}{鈭�x}-\frac{鈭�P}{鈭�y}\right)dxdy=Pdx+Qdy \end{gathered}$$

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

It can be seen that

$$\begin{gathered} P=y鈫�\frac{鈭�P}{鈭�y} \\ =1 \\ Q=2x鈫�\frac{鈭�Q}{鈭�x} \\ =2 \end{gathered}$$

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

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

The area is enclosed by an ellipse parameterization by

$$\begin{gathered} x=a\cos 胃 \\ y=b\sin 胃 \end{gathered}$$

Is given by $ab蟺$.

In Part (b),

$$\begin{gathered} a=1 \\ b=2 \end{gathered}$$

So, the work done is just $2蟺$.

Hence, the solution to this problem is

a) $F_2$ is not conservative.

b) $W=2蟺$

c) $V=\left(2x-z\right)y-\frac{1}{2}{z}^2$

d) $W=\frac{45}{2}$

e) $W=2\mathrm{蟺}$