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OleMash [197]
3 years ago
7

Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the

positive x direction, and f=⟨x^5,0,y^1⟩. begin by writing down the "standard" parametrization of ∂m as a function of the angle θ (denoted by "t" in your answer)
Mathematics
1 answer:
ludmilkaskok [199]3 years ago
7 0
By Stokes' theorem,

\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S

where \mathcal C is the circular boundary of the hemisphere \mathcal M in the y-z plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle

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

\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt
=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi

We can check this result by evaluating the equivalent surface integral. We have

\nabla\times\mathbf f=\langle1,0,0\rangle

and we can parameterize \mathcal M by

\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle

so that

\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv

where 0\le v\le\dfrac\pi2 and 0\le u\le2\pi. Then,

\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi

as expected.
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Answer:

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