$\iint \mathrm{P}\times \mathrm{nd\sigma }$over the upper half of the sphere $\mathrm{r}=1$, if $\mathrm{P}=\mathrm{curl}(\mathrm{jx}-\mathrm{kz})$.

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

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

"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’ theorem ${\iint }_{\mathrm{\sigma }} (\mathrm{\nabla }\times \mathrm{A})\times \mathrm{nd\sigma }={\text{∮}}_{\mathrm{\partial }\mathrm{\sigma }}\mathrm{A}\times \mathrm{dr}$, where$\mathrm{\sigma }$is the upper half of the unit sphere, so$\mathrm{\partial }\mathrm{\sigma }$is the unit circle in the $\mathrm{xy}$-plane. Since this is an approximate boundary of the surface,

${\iint }_{\mathrm{\sigma }} \mathrm{P}\times \mathrm{nd\sigma }={\text{∮}}_{\mathrm{\partial }\mathrm{\sigma }}(\mathrm{xj}-\mathrm{zk})\times \mathrm{dr}$

$=\int \mathrm{xdy}-\mathrm{zdz}$

Over the unit circle in the -plane,

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

localid="1657352075788" $=0$

Use the parameterization as shown below.

$\mathrm{x}=\cos \mathrm{\theta }$

$\mathrm{dx}=-\sin \mathrm{\theta d\theta }$

$\mathrm{y}=2\sin \mathrm{\theta }$$\mathrm{y}=2\sin \mathrm{\theta }$

$\mathrm{dy}=2\cos \mathrm{\theta d\theta }$

$\int \mathrm{xdy}-\mathrm{zdz}={\int }_0^{2\mathrm{\pi }} {\cos }^2\mathrm{\theta d\theta }$

$={\int }_0^{2\mathrm{\pi }} \frac{1}{2}(1-\cos 2\mathrm{\theta })\mathrm{d\theta }$

$={\left.\frac{1}{2}\left(\mathrm{\theta }-\frac{1}{2}\sin 2\mathrm{\theta }\right)\right|}_0^{2\mathrm{\pi }}$

$=\mathrm{\pi }$

Hence, the solution is ${\iint }_{\mathrm{\sigma }} \mathrm{P}\times \mathrm{nd\sigma }=\mathrm{\pi }$.