Use Stokes' Theorem to evaluate |c F * dr where C is oriented counterclockwise as viewed from above.F(x, y, z) = xyi + 5zj + 7yk, where C is the curve of intersection of the plane x + z = 3 and the cylinder x2 + y2 = 144.

Respuesta :

Compute the curl of [tex]\vec F[/tex]:

[tex]\vec F(x,y,z)=xy\,\vec\imath+5z\,\vec\jmath+7y\,\vec k\implies\nabla\times\vec F(x,y,z)=2\,\vec\imath-x\,\vec k[/tex]

By Stoke's theorem, the (line) integral of [tex]\vec F[/tex] along [tex]C[/tex] is equivalent to the (surface) integral (or flux) of [tex]\nabla\times\vec F[/tex] across [tex]S[/tex], the oriented surface with boundary [tex]C[/tex].

Take [tex]S[/tex] to be the part of the plane [tex]x+z=3[/tex] within the cylinder [tex]x^2+y^2=144[/tex]. Parameterize [tex]S[/tex] by

[tex]\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(3-u\cos v)\,\vec k[/tex]

with [tex]0\le u\le12[/tex] and [tex]0\le v\le2\pi[/tex]. Take the normal vector to [tex]S[/tex] to be

[tex]\vec s_u\times\vec s_v=u\,\vec\imath+u\,\vec k[/tex]

Then the integral is

[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv[/tex]

[tex]=\displaystyle\int_0^{2\pi}\int_0^{12}(2\,\vec\imath-u\cos v\,\vec k)\cdot(u\,\vec\imath+u\,\vec k)\,\mathrm du\,\mathrm dv[/tex]

[tex]=\displaystyle\int_0^{2\pi}\int_0^{12}(2u-u^2\cos v)\,\mathrm du\,\mathrm dv=\boxed{288\pi}[/tex]

After evaluating the given integral the value obtained is 288[tex]\pi[/tex] and this can be determined by using the Stokes theorem.

Given :

F(x, y, z) = xyi + 5zj + 7yk, where C is the curve of intersection of the plane x + z = 3 and the cylinder x2 + y2 = 144.

Parameterize S as given below:

[tex]\rm \bar{s}(u,v)=u\;cosv\;i+u\;sinv\;j+(3-ucosv)\;k[/tex]

where the value of [tex]\rm 0\leq u\leq 12[/tex] and [tex]\rm 0\leq v\leq 2\pi[/tex].

Now, the integral can be evaluated as given below:

[tex]\rm \int_C\bar{F}.d\bar{r}=\int \int_S(\bigtriangledown \times \bar{F}).(\bar{s}_u\times (\bar{s}_v)\;du\;dv[/tex]

[tex]\rm \int_C\bar{F}.d\bar{r}=\int^{2\pi}_0 \int^{12}_0(2i-u\;cosv\;k).(u\;i+u\;k)\;du\;dv[/tex]

[tex]\rm \int_C\bar{F}.d\bar{r}=\int^{2\pi}_0 \int^{12}_0(2u-u^2cosv)\;du\;dv[/tex]

[tex]\rm \int_C\bar{F}.d\bar{r}=288\pi[/tex]

After evaluating the given integral the value obtained is 288[tex]\pi[/tex] and this can be determined by using the Stokes theorem.

For more information, refer to the link given below:

https://brainly.com/question/14093731

ACCESS MORE