Multivariate Calculus and Surface Integral

Question

I am really confused with how to parameterize the piece S. Also, since it is pointing inward, do I fix at an axis and consider the orientation? Could I apply the right hand rule here?

Thanks.

Answer 1

You can parameterize $S$ by

$$\mathbf r(\theta,\zeta)=\langle x(\theta,\zeta),y(\theta,\zeta),z(\theta,\zeta)\rangle=\langle2\cos\theta,2\sin\theta,\zeta\rangle$$

where $0\le\theta\le2\pi$ and $0\le\zeta\le6-2\cos \theta$ (assuming the "$x=0$" is indeed a typo).

Meanwhile, an inward orientation would mean any unit vector normal to $S$ would point toward the interior of the cylinder so that the orientation would be considered negative by convention. Note that

$$\mathbf r_\theta\times\mathbf r_\zeta=\langle2\cos t,2\sin t,0\rangle$$ $$\mathbf r_\zeta\times\mathbf r_\theta=-\langle2\cos t,2\sin t,0\rangle$$

So given a vector field $\mathbf F(x,y,z)$, the negatively-oriented surface integral along $S$ would be given by

$$\iint_S\mathbf F(x,y,z)\,\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{\zeta=0}^{\zeta=6-2\cos\theta}\mathbf F(x(\theta,\zeta),y(\theta,\zeta),z(\theta,\zeta))\cdot(\mathbf r_\zeta\times\mathbf r_\theta)\,\mathrm d\zeta\,\mathrm d\theta$$