Let $\mathcal{C}=\{(x,y,z)\in\mathbb{R}^3:y^2+z^2=1\}\cap\{(x,y,z)\in\mathbb{R}^3:x+y=1\}$ oriented such that the tangente vector at $(1,0,1)$ is $(1,-1,0)$.
(i) Find a regular parameterization for $\mathcal{C}$,
(ii) If $f:\mathbb{R}^3\to\mathbb{R}$ is a $C^1$ scalar field such that $\frac{\partial f}{\partial z}(x,y,z)=z$, evaluate $\int_{\mathcal{C}} f(x,y,z)dx + xydy + xzdz$.
We can parametrize $\mathcal{C}$ as the image of $\sigma:[0,2\pi]\to\mathcal{C}\subset\mathbb{R}^3,\sigma(\theta)=(1-\cos\theta,\cos\theta,\sin\theta)$. Hence,
$$\sigma'(\theta)=(\sin\theta,-\sin\theta,\cos\theta)\neq(0,0,0) \ \ \ \ \ \forall \theta\in[0,2\pi]$$
But I'm having trouble trying to evaluate the line integral. Since $\mathcal{C}$ is a smooth, simple, closed curve that delimits a region $S$, we could try to apply Stokes' Theorem. Let $F=(f(x,y,z),xy,xz)$, with $\text{curl } F=(0,0,y- f_x(x,y,z))$. Hence,
$$\oint_{\mathcal{C}}Fds \stackrel{\text{Stokes}}{=} \iint_S \text{curl }F dS = \iint_S (\text{curl }F) \cdot N \ dA$$
This would be solvable if the $z$-component of $N$ were $0$. Is this correct? How can I prove it?
Best Answer
If you know the generalized Stokes' Theorem, I think it may be easier to argue as follows: $S$ may be taken to be the part of the plane defined by $x+y=1$ cut off by the cylinder $y^2+z^2=1$. With $\varphi (x,z)=(x,x-1,z)$, Stokes' theorem gives
$\int_{\partial S}f(x,y,z)dx + xydy + xzdz=\int_{\varphi^{-1}(S)}\varphi^*d(f(x,y,z)dx + xydy + xzdz)=\int_{\varphi^{-1}(S)}\varphi^*(f_y(x,y,z)dy\wedge dx+f_z(x,y,z)dz\wedge dx + ydx\wedge dy + zdx\wedge dz)=\int_{\varphi^{-1}(S)}\varphi^*((y-f_y(x,y,z)dx\wedge dy)$
because $f_z(x,y,z)=z$ and because the wedges anticommute. Continuing, we get
$\int_{\varphi^{-1}(S)}\varphi^*((y-f_y(x,y,z)dx\wedge dy)=\int_{\varphi^{-1}(S)}(x-1-f_y(x,x-1,z)d\varphi^*x\wedge d\varphi^*y)=\int_{\varphi^{-1}(S)}(x-1-f_y(x,x-1,z)dx\wedge d(x-1)=\int_{\varphi^{-1}(S)}(x-1-f_y(x,x-1,z))dx\wedge dx=0$