91Ó°ÊÓ

Vector field is $\mathbf{F}(x,y)=x{e}^{x^2+y^2}\mathbf{i}+y{e}^{x^2+y^2}\mathbf{j}$

Flux across smooth curve is defined as${∫}_C\mathbf{F}(x,y)·\mathbf{n}ds$

Applying line integral to evaluate integral

If parametrization of smooth curve is$\mathbf{r}\left(t\right)=⟨x\left(t\right),y\left(t\right),z\left(t\right)âŸ\copyright \text{for}a≤t≤b$

Unit Circle is parametrized as

$\mathbf{r}\left(t\right)=⟨\cos t,\sin tâŸ\copyright \text{for}0≤t≤2\mathrm{Ï€}$

Its derivative is ${\mathbf{r}}^{\text{'}}\left(t\right)=\frac{d}{dt}\mathbf{r}\left(t\right)$

$=⟨-\sin t,\cos tâŸ\copyright$

Also

$||{\mathbf{r}}^{\text{'}}\left(t\right)||=‖⟨-\sin t,\cos tâŸ\copyright ‖$

$=\sqrt{(-\sin t{)}^2+(\cos t{)}^2}$

$=\sqrt{{\sin }^{2}t+{\cos }^{2}t}$

$=1$

The unit vector on unit circle is $\mathbf{n}=x\mathbf{i}+y\mathbf{j}$

Hence, $\mathbf{F}(x,y)·\mathbf{n}=\left(x{e}^{x^2+y^2}\mathbf{i}+y{e}^{x^2+y^2}\mathbf{j}\right)·(x\mathbf{i}+y\mathbf{j})$

$=x^2{e}^{x^2+y^2}+y^2{e}^{x^2+y^2}$

$=\left(x^2+y^2\right)e^{x^2+y^2}$

For $\mathbf{r}\left(t\right)=⟨\cos t,\sin tâŸ\copyright$

$\mathbf{F}(x,y)·\mathbf{n}=\left(x^2+y^2\right)e^{x^2+y^2}$

$\mathbf{F}\left(x\right(t),y(t\left)\right)·\mathbf{n}=\left({\cos }^{2}t+{\sin }^{2}t\right)e^{{\cos }^{2}t+{\sin }^{2}t}$

role="math" localid="1653950650385" $=e$

Put $\mathbf{F}\left(x\right(t),y(t\left)\right)·\mathbf{n}=e\text{and}||{\mathbf{r}}^{\text{'}}\left(t\right)||=1\text{for}0≤t≤2\mathrm{π}$in above equation

${∫}_C\mathbf{F}(x,y)·\mathbf{n}ds={∫}_a^b\left(\mathbf{F}\right(x,y)·\mathbf{n})||{\mathbf{r}}^{\text{'}}\left(t\right)||dt$

$={∫}^{2\mathrm{π}}e·1dt$

$=e{∫}_0^{2\mathrm{π}}dt$

$=e[t{]}_0^{2\mathrm{Ï€}}$

$=e[2\mathrm{Ï€}-0]$

$=2\mathrm{Ï€}e$

Hence, required flux is$2\mathrm{Ï€}e$