91影视

The given information is$\mathrm{r}=1$

$\mathrm{P}=\mathrm{curl}(\mathrm{jx}鈭�\mathrm{kz})$

"The surface integral of the curl of a function over a surface limited by a closed surface is equivalent to the line integral of the particular vector function around that surface," according to Stokes' theorem.

Use Stokes鈥� theorem ${鈭�}_{\mathrm{蟽}}鈥�(\mathrm{鈭�}脳\mathrm{A})脳\mathrm{苍诲蟽}={\text{鈭�}}_{\mathrm{鈭�}\mathrm{蟽}}\mathrm{A}脳\mathrm{dr}$, where$\mathrm{蟽}$is the upper half of the unit sphere, so$\mathrm{鈭�}\mathrm{蟽}$is the unit circle in the $\mathrm{xy}$-plane. Since this is an approximate boundary of the surface,

${鈭�}_{\mathrm{蟽}}鈥�\mathrm{P}脳\mathrm{苍诲蟽}={\text{鈭�}}_{\mathrm{鈭�}\mathrm{蟽}}(\mathrm{xj}鈭�\mathrm{zk})脳\mathrm{dr}$

$=鈭�\mathrm{xdy}鈭�\mathrm{zdz}$

Over the unit circle in the -plane,

$\mathrm{z}=\mathrm{dz}$

localid="1657352075788" $=0$

Use the parameterization as shown below.

$\mathrm{x}=\cos 鈦�\mathrm{胃}$

$\mathrm{dx}=鈭�\sin 鈦�\mathrm{胃诲胃}$

$\mathrm{y}=2\sin 鈦�\mathrm{胃}$$\mathrm{y}=2\sin 鈦�\mathrm{胃}$

$\mathrm{dy}=2\cos 鈦�\mathrm{胃诲胃}$

$鈭�\mathrm{xdy}鈭�\mathrm{zdz}={鈭�}_0^{2\mathrm{蟺}}鈥�{\cos }^2鈦�\mathrm{胃诲胃}$

$={鈭�}_0^{2\mathrm{蟺}}鈥�\frac{1}{2}(1鈭�\cos 鈦�2\mathrm{胃})\mathrm{诲胃}$

$={\left.\frac{1}{2}\left(\mathrm{胃}鈭�\frac{1}{2}\sin 鈦�2\mathrm{胃}\right)\right|}_0^{2\mathrm{蟺}}$

$=\mathrm{蟺}$

Hence, the solution is ${鈭�}_{\mathrm{蟽}}鈥�\mathrm{P}脳\mathrm{苍诲蟽}=\mathrm{蟺}$.