Use Stokes’ Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 − y2 that lie

Question

Use Stokes’ Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 − y2 that lies above the xy-plane, oriented upward.

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Kaylee 2 weeks 2021-09-13T13:42:12+00:00 1 Answer 0

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    2021-09-13T13:43:59+00:00

    The vector field

    \vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k

    has curl

    \nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath

    Parameterize S by

    \vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k

    where

    \begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}

    with 0\le u\le3 and 0\le v\le2\pi.

    Take the normal vector to S to be

    \dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k

    Then by Stokes’ theorem we have

    \displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

    =\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv

    =\displaystyle\int_0^{2\pi}\int_0^3u^3(2u\cos^3v\sin(u^2-9)+\cos^3v\sin v+2u\sin^3v+\cos v\sin^3v)\,\mathrm du\,\mathrm dv

    which has a value of 0, since each component integral is 0:

    \displaystyle\int_0^{2\pi}\cos^3v\,\mathrm dv=0

    \displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0

    \displaystyle\int_0^{2\pi}\sin^3v\,\mathrm dv=0

    \displaystyle\int_0^{2\pi}\cos v\sin^3v\,\mathrm dv=0

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