Calculate the circulation, F · dr, C in two ways, directly and using Stokes' Theorem. F = y i + z j + xk and C is the boundary of S, the paraboloid z = 4 − (x2 + y2), z ≥ 0 oriented upward. (Hint: Use polar coordinates.) F · dr C =

[tex]C[/tex], the boundary of [tex]S[/tex], is a circle in the [tex]x,y[/tex] plane centered at the origin and with radius 2, hence we can parameterize it by[tex]\vec r(t)=2\cos t\,\vec\imath+2\sin t\,\vec\jmath[/tex]with [tex]0\le t\le2\pi[/tex]. Then the line integral is[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}(2\sin t\,\vec\imath+2\cos t\,\vec k)\cdot(-2\sin t\,\vec\imath+2\cos t\,\vec\jmath)\,\mathrm dt[/tex][tex]=\displaystyle\int_0^{2\pi}-4\sin^2t\,\mathrm dt[/tex][tex]=\displaystyle-2\int_0^{2\pi}(1-\cos2t)\,\mathrm dt=\boxed{-4\pi}[/tex]By Stokes' theorem, the line integral of [tex]\vec F[/tex] along [tex]C[/tex] is equal to the surface integral of the curl of [tex]\vec F[/tex] across [tex]S[/tex]:[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S[/tex]Parameterize [tex]S[/tex] by[tex]\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(4-u^2)\,\vec k[/tex]with [tex]0\le u\le2[/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=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k[/tex]The curl is[tex]\nabla\times\vec F=-\vec\imath-\vec\jmath-\vec k[/tex]Then the surface integral is[tex]\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^2(-\vec\imath-\vec\jmath-\vec k)\cdot(2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k)\,\mathrm du\,\mathrm dv[/tex][tex]=\displaystyle-\int_0^{2\pi}\int_0^2(2u^2\cos v+2u^2\sin v+u)\,\mathrm du\,\mathrm dv=\boxed{-4\pi}[/tex]