Respuesta :
Stokes' theorem says the integral of the curl of [tex]\vec F[/tex] over [tex]S[/tex] is equal to the integral of [tex]\vec F[/tex] along the boundary of [tex]S[/tex], with counterclockwise orientation (when viewed from above). This boundary is the circle [tex]x^2+y^2=1[/tex] set in the plane [tex]z=1[/tex].
Parameterize this path by
[tex]\vec r(t)=\cos t\,\vec\imath+\sin t\,\vec\jmath+\vec k[/tex]
with [tex]0\le t\le2\pi[/tex]. Then
[tex]\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S=\int_{\partial S}\vec F\cdot\mathrm d\vec r[/tex]
[tex]=\displaystyle\int_0^{2\pi}(-3\sin t\,\vec\imath+3\cos t\,\vec\jmath+14\,\vec k)\cdot(-\sin t\,\vec\imath+\cos t\,\vec\jmath)\,\mathrm dt[/tex]
[tex]=\displaystyle3\int_0^{2\pi}\mathrm dt=\boxed{6\pi}[/tex]
The value of the given integral is 6[tex]\pi[/tex] and this can be determined by using the stokes theorem and also by using the given data.
Given :
[tex]\rm F(x,y,z) = -3yz\;i+3xz\;j+14(x^2+y^2)z \; k[/tex]
First, parameterize the path as given below:
[tex]\rm \bar{r}(t) = cos\;t \;i+sin\;t\;j+k[/tex]
where [tex]\rm 0\leq t\leq 2\pi[/tex].
Now, according to the Stokes theorem, the integral becomes:
[tex]\rm \int\int_S(\bigtriangledown \times \bar{F}).d\bar{S}=\int_{\delta S} \bar{F} .d\bar{r}[/tex]
[tex]\rm \int\int_S(\bigtriangledown \times \bar{F}).d\bar{S}=\int^{2\pi}_0 (-3sint \;i+3cost\;j+14k).(-sint\;i+cost\;j)\; dt[/tex]
[tex]\rm \int\int_S(\bigtriangledown \times \bar{F}).d\bar{S}=\int^{2\pi}_0dt=6\pi[/tex]
The value of the given integral is 6[tex]\pi[/tex] and this can be determined by using the stokes theorem and also by using the given data.
For more information, refer to the link given below:
https://brainly.com/question/24308099
