Use stokes' theorem to evaluate c f · dr where c is oriented counterclockwise as viewed from above. f(x, y, z = xyi + 5zj + 7yk, c is the curve of intersection of the plane x + z = 5 and the cylinder x2 + y2 = 36.

Respuesta :

The intersection can be parameterized by

[tex]C:=\mathbf r(t)=\begin{cases}x(t)=6\cos t\\y(t)=6\sin t\\z(t)=5-6\cos t\end{cases}[/tex]

with [tex]0\le t<2\pi[/tex].

By Stoke's theorem, the integral of [tex]\mathbf f(x,y,z)=xy\,\mathbf i+5z\,\mathbf j+7y\,\mathbf k[/tex] along [tex]C[/tex] is equivalent to

[tex]\displaystyle\int_C\mathbf f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r(t)=\iint_S\nabla\times\mathbf f\,\mathrm dS[/tex]

where [tex]S[/tex] is the region bounded by [tex]C[/tex]. The line integral reduces to

[tex]\displaystyle\int_0^{2\pi}(36\sin t\cos t\,\mathbf i+(25-30\cos t)\,\mathbf j+42\sin t\,\mathbf k)\cdot(-6\sin t\,\mathbf i+6\cos t\,\mathbf j+6\sin t\,\mathbf k)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{2\pi}(54(\cos3t-\cos t)-30(3\cos2t-5\cos t+3)+(126-126\cos2t)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{2\pi}(36+96\cos t-216\cos2t+54\cos3t)\,\mathrm dt[/tex]
[tex]=72\pi[/tex]
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