91影视

The goal is to demonstrate that if

${鈭�}_C鈥�\mathbf{F}\mathbf{脳}d\mathbf{r}=0$

That does not imply that it is conservative.

Consider the vector field,

$\mathbf{F}(x,y,z)=x\mathbf{i}鈭�y\mathbf{j}+2^z\ln 鈦�(z+1)\mathbf{k}.$

Show that ${鈭�}_C\mathbf{F}\mathbf{脳}d\mathbf{r}=0$for the above vector field with C is the unit circle.

demonstrating that the given curve C is parameterized

$\mathbf{r}\left(t\right)=鉄�\cos 鈦�t,\sin 鈦�t,0鉄�\text{for}0鈮�t鈮�2蟺$

It will be derived in the manner specified.,

${\mathbf{r}}^{\mathrm{鈥�}}\left(t\right)=\frac{d}{dt}\mathbf{r}\left(t\right)$

$=\frac{d}{dt}鉄�\cos 鈦�t,\sin 鈦�t,0鉄�$

$=鉄�鈭�\sin 鈦�t,\cos 鈦�t,0鉄�$

For $\mathbf{r}\left(t\right)=鉄�\cos t,\sin t,0鉄�$, rewrite the vector field $\mathbf{F}(x,y,z)=x\mathbf{i}-y\mathbf{j}+2^z\ln (z+1)\mathbf{k}$as follows:

$\mathbf{F}\left(x\right(t),y(t),z(t\left)\right)=\cos 鈦�t\mathbf{i}鈭�\sin 鈦�t\mathbf{j}+2^0\ln 鈦�(0+1)\mathbf{k}$

$=\cos 鈦�t\mathbf{i}鈭�\sin 鈦�t\mathbf{j}+1脳\ln 鈦�1\mathbf{k}$

role="math" localid="1650798984047" $=\cos 鈦�t\mathbf{i}鈭�\sin 鈦�t\mathbf{j}+1脳0\mathbf{k}$

$=鉄�\cos 鈦�t,鈭�\sin 鈦�t,0鉄�.$

$\mathbf{F}(x,y,z)=x\mathbf{i}-y\mathbf{j}+2^z\ln (z+1)\mathbf{k}$The required line integral will then be computed.,

localid="1650799107105" $={鈭�}_f鈥�\mathbf{F}\left(x\right(t),y(t),z(t\left)\right)脳{\mathbf{r}}^{\mathrm{鈥�}}\left(t\right)dt$

$={鈭�}_0^{2蟺}鈥�鉄�\cos 鈦�t,鈭�\sin 鈦�t,0鉄�鈰�鉄�鈭�\sin 鈦�t,\cos 鈦�t,0鉄�dt$

Substitute value

$={鈭�}_0^{2蟺}鈥�(鈭�\sin 鈦�t\cos 鈦�t鈭�\sin 鈦�t\cos 鈦�t+0)dt$

$={鈭�}_0^{2蟺}鈥�鈭�2\sin 鈦�t\cos 鈦�tdt$

$={鈭�}_0^{2蟺}鈥�鈭�\sin 鈦�2tdt$

$={\left[\frac{\cos 鈦�2t}{2}\right]}_0^{2蟺}$

$=\frac{\cos 鈦�4蟺}{2}鈭�\frac{\cos 鈦�0}{2}$

$=\frac{1}{2}鈭�\frac{1}{2}$

$=0$

As a result, it is demonstrated that, for the vector field $\mathbf{F}(x,y,z)=x\mathbf{i}-y\mathbf{j}+2^z\ln (z+1)\mathbf{k}$with C is the unit circle, localid="1650799328348" ${鈭�}_C\mathbf{F}\mathbf{脳}d\mathbf{r}=0$.

Take note of the fact that this vector field$\mathbf{F}(x,y,z)=x\mathbf{i}-y\mathbf{j}+2^z\ln (z+1)\mathbf{k}$Because the domain of this vector field is not simply connected, it is not conservative.

Next, determine the vector field's curl. $\mathbf{F}(x,y,z)=x\mathbf{i}-y\mathbf{j}+2^z\ln (z+1)\mathbf{k}$

Next, determine the vector field's curl.$\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$is defined as follows:

$\mathrm{curl}鈦�\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\mathrm{鈭�}}{\mathrm{鈭�}x}& \frac{\mathrm{鈭�}}{\mathrm{鈭�}y}& \frac{\mathrm{鈭�}}{\mathrm{鈭�}z}\\ F_1(x,y,z)& F_2(x,y,z)& F_3(x,y,z)\end{array}\right|$

$=\left[\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}y}鈭�\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}z}\right]\mathbf{i}鈭�\left[\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}x}鈭�\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}z}\right]\mathbf{j}+\left[\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}x}鈭�\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}y}\right]\mathbf{k}$

Next, determine the vector field's curl.

$\mathbf{F}(x,y,z)=x\mathbf{i}鈭�y\mathbf{j}+2^z\ln 鈦�(z+1)\mathbf{k}$is,

$=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\mathrm{鈭�}}{\mathrm{鈭�}x}& \frac{\mathrm{鈭�}}{\mathrm{鈭�}y}& \frac{\mathrm{鈭�}}{\mathrm{鈭�}z}\\ x& 鈭�y& 2^z\ln 鈦�(z+1)\end{array}\right|$

$=\left[\frac{\mathrm{鈭�}}{\mathrm{鈭�}y}\left(2^z\ln 鈦�(z+1)\right)鈭�\frac{\mathrm{鈭�}}{\mathrm{鈭�}z}(鈭�y)\right]\mathbf{i}鈭�\left[\frac{\mathrm{鈭�}}{\mathrm{鈭�}x}\left(2^z\ln 鈦�(z+1)\right)鈭�\frac{\mathrm{鈭�}}{\mathrm{鈭�}z}\left(x\right)\right]\mathbf{j}+\left[\frac{\mathrm{鈭�}}{\mathrm{鈭�}x}(鈭�y)鈭�\frac{\mathrm{鈭�}}{\mathrm{鈭�}y}\left(x\right)\right]\mathbf{k}$

$=[0鈭�0]\mathbf{i}鈭�[0鈭�0]\mathbf{j}+[0鈭�0]\mathbf{k}$

$=0i+0j+0k$

$=0$

As a result, for non-conservative vector fields $\mathbf{F}(x,y,z)=x\mathbf{i}-y\mathbf{j}+2^z\ln (z+1)\mathbf{k}$,

localid="1650800657470" ${鈭�}_C鈥�\mathbf{F}\mathbf{脳}d\mathbf{r}=0$and

curl F=0

Thus, it is proved that, if localid="1650800648485" ${鈭�}_C\mathbf{F}\mathbf{脳}d\mathbf{r}=0$, it does not follow that$\mathrm{F}$is conservative.