Given the curve $C$ of equation $$\vec X=(3\cos t,3\sin t,6\cos t),\qquad0\leq t\leq2\pi$$ oriented according imposes this parameterization, find the circulation of $\vec f$ along $C$ if $\vec f\in\mathcal C^1$ in $\mathbb R^3$ and $$\text{rot }\vec f(x,y,z)=(0,2x-y,z),$$ where $\text{rot}$ means curl.
First I found the curve: $$\begin{cases}x=3\cos t\\y=3\sin t\\z=6\cos t\end{cases}\equiv\begin{cases}x^2+y^2=9\\z=6\dfrac x3\end{cases}\equiv\begin{cases}x^2+y^2=9\\z-2x=0,\end{cases}$$ so the normal becomes $$\vec N=(-2,0,1).$$ Now I have to use the Stokes theorem, that is
$$\begin{eqnarray*} &&\oint\limits_C{\vec f\;\text d\vec s}&(1)\\\\ &=&\iint\limits_S{\Big.\nabla\times\vec f\Big|_S\;\text d\vec\sigma}&(2)\\\\ &=&\iint\limits_{P_{xy}}{(0,2x-y,\underbrace z_{2x})\cdot(-2,0,1)\;\text dx\text dy}&(3)\\\\ &=&2\iint\limits_{P_{xy}}{x\;\text dx\text dy}&(4)\\\\ &\underset{\text{Using polar coordinates}}{=}&2\int_0^{2\pi}{\text d\theta}\int_0^3{\rho\cdot\rho\cos\theta\;\text d\rho}&(5)\\\\ &=&2\int_0^{2\pi}{\cos\theta\;\text d\theta}\int_0^3{\rho^2\;\text d\rho}&(6)\\\\ &=&2\int_0^{2\pi}{\cos\theta\;\text d\theta}\;\left.\left(\frac{\rho^3}3\right)\right|_0^3&(7)\\\\ &=&18\Big.\left(\sin\theta\right)\Big|_0^{2\pi}&(8)\\\\ \text{Circulation}&=&\boxed 0.&(9) \end{eqnarray*}$$
Is this correct?
Thank you!
You’re reached the correct answer, but I think you’ve done more work than you really needed to to get there.
First off, you can compute the surface normal directly from the parametric equation for the curve. Take two convenient points on it, say, at $t=0$ and $t=\pi/2$ and compute their cross product: $(3,0,6)\times(0,3,0)=(-18,0,9)$, which normalizes to $\vec N = \frac1{\sqrt5}(-2,0,1)$. I might also verify that $\vec N\cdot\vec X=0$ for all values of $t$ to be sure that I’m dealing with a planar curve.
You then have $\vec N\cdot\vec f = \frac1{\sqrt5}z=\frac6{\sqrt5}\cos t$, at which point you’re basically done. The region enclosed by the curve can be parameterized as $r\vec X(t)$ for $0\le r\le 1$, which makes the region of integration in the $r$-$t$ plane the rectangle $[0,1]\times[0,2\pi]$. The integrand will be some function of $r$ times $\cos t$, and its integral over a full cycle of $\cos t$ will vanish.