g Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = 7yi + xzj + (x + y)k, C is the curve of intersection of the plane z = y + 9 and the cylinder x2 + y2 = 1. Exp

Accepted Solution

By Stokes' theorem,[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S[/tex]where [tex]S[/tex] is any oriented surface with boundary [tex]C[/tex]. We have[tex]\vec F(x,y,z)=7y\,\vec\imath+xz\,\vec\jmath+(x+y)\,\vec k[/tex][tex]\implies\nabla\times\vec F(x,y,z)=(1-x)\,\vec\imath-\vec\jmath+(z-7)\,\vec k[/tex]Take [tex]S[/tex] to be the ellipse that lies in the plane [tex]z=y+9[/tex] with boundary on the cylinder [tex]x^2+y^2=1[/tex]. Parameterize [tex]S[/tex] by[tex]\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(u\sin v+9)\,\vec k[/tex]with [tex]0\le u\le1[/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\jmath+u\,\vec k[/tex]Then we have[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S[/tex][tex]=\displaystyle\int_0^{2\pi}\int_0^1\big((1-u\cos v)\,\vec\imath-\vec\jmath+(u\sin v+2)\,\vec k\big)\cdot\big(-u\,\vec\jmath+u\,\vec k\big)\,\mathrm du\,\mathrm dv[/tex][tex]=\displaystyle\int_0^{2\pi}\int_0^1(3u+u^2\sin v)\,\mathrm du\,\mathrm dv=\boxed{3\pi}[/tex]