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Ugo [173]
4 years ago
10

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 o

f S, the paraboloid z = 4 − (x2 + y2), z ≥ 0 oriented upward. (Hint: Use polar coordinates.) F · dr C =
Mathematics
1 answer:
otez555 [7]4 years ago
3 0

C, the boundary of S, is a circle in the x,y plane centered at the origin and with radius 2, hence we can parameterize it by

\vec r(t)=2\cos t\,\vec\imath+2\sin t\,\vec\jmath

with 0\le t\le2\pi. Then the line integral is

\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

=\displaystyle\int_0^{2\pi}-4\sin^2t\,\mathrm dt

=\displaystyle-2\int_0^{2\pi}(1-\cos2t)\,\mathrm dt=\boxed{-4\pi}

By Stokes' theorem, the line integral of \vec F along C is equal to the surface integral of the curl of \vec F across S:

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

Parameterize S by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(4-u^2)\,\vec k

with 0\le u\le2 and 0\le v\le2\pi. Take the normal vector to S to be

\vec s_u\times\vec s_v=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k

The curl is

\nabla\times\vec F=-\vec\imath-\vec\jmath-\vec k

Then the surface integral is

\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

=\displaystyle-\int_0^{2\pi}\int_0^2(2u^2\cos v+2u^2\sin v+u)\,\mathrm du\,\mathrm dv=\boxed{-4\pi}

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