Evaluate using Stokes' Theorem To evaluate $\oint_{C} -y^3dx+x^3dy+z^3dz,$ where $C$ is the intersection of cylinder $x^2 + y^2 =1$ and plane $x+y+z=1$. The orientation of $C$ is counter-clockwise motion in the $xy$ plane.
Now I have computed $\nabla \times\mathbf{F} = \left(0,0,3\left(x^2+y^2\right)\right).$ I am having difficulty finding out the curve $C$ of intersection and also I am confused about projection on the $xy$ plane. Should I take part inside $x+y=1 $ only or part between $x+y=1$ and $x^2+y^2=1?$
 A: Hint: Since $x^2+y^2=1$ you have that
$$x=\cos t\quad y=\sin t\quad 0\leq t\leq2\pi.$$
Thus, a parametrization for the curve $C$ is
$$x=\cos t\quad y=\sin t\quad z=1-\cos t-\sin t\quad 0\leq t\leq2\pi.$$
A: Let $\mathbf{F}=\left(-y^3,x^3,z^3\right).$
So we have
\begin{align*}
\oint_{C} -y^3dx+x^3dy+z^3dz&=\oint_C\mathbf{F}\cdot d\mathbf{r} \\
&=\iint_S(\nabla\times\mathbf{F})\cdot dS \\
&=\iint_S(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}\,dA \\
&=\sqrt{3}\iint_S\left(x^2+y^2\right)\,dA.
\end{align*}
My hunch is that we can consider the projection into the $xy$ plane of the surface $S$ for this problem, because the integrand $\nabla\times\mathbf{F}$ only has a $z$ component. If that is so, we will want to switch to polar coordinates:
\begin{align*}
\oint&=\sqrt{3}\int_0^{2\pi}\int_0^1\left(x^2+y^2\right)\,r\,dr\,d\theta \\
&=2\sqrt{3}\,\pi\int_0^1 r^3\,dr \\
&=\frac{\sqrt{3}\,\pi}{2}.
\end{align*}
As we can see from this wiki, to adjust for the projection, we need
$$A_{\text{proj}}=\cos(\beta) A, $$
since the angle is constant and pulls out of the integral. We have computed the projected area, so we must compensate by dividing by $\cos(\beta),$ we can calculate via the dot product formula:
$$\frac{1}{\sqrt{3}}(1,1,1)\cdot(0,0,1)=\cos(\beta). $$ 
This means the final result is
$$\frac{\sqrt{3}\,\pi}{2}\div\frac{1}{\sqrt{3}}=\frac{3\pi}{2}. $$
Now the question is, was the projection justified? Can we verify by, say, computing the original line integral? As suggested by DiegoMath in his answer, we can parametrize the curve $C$ as 
\begin{align*}
x&=\cos(t) \\
y&=\sin(t) \\
z&=1-\cos(t)-\sin(t),\\
0&\le t\le 2\pi.
\end{align*}
Then we have 
\begin{align*}
\mathbf{r}(t)&=(\cos(t), \sin(t), 1-\cos(t)-\sin(t))\\
\dot{\mathbf{r}}(t)&=(-\sin(t), \cos(t), \sin(t)-\cos(t)) \\
\mathbf{F}(t)&=(-\sin^3(t),\cos^3(t),(1-\cos(t)-\sin(t))^3) \\
\mathbf{F}\cdot\dot{\mathbf{r}}&=\sin^4(t)+\cos^4(t)+(\sin(t)-\cos(t))(1-\cos(t)-\sin(t))^3 \\
\oint_C\mathbf{F}\cdot d\mathbf{r}&=\oint_C\mathbf{F}\cdot \dot{\mathbf{r}}(t)\,dt \\
&=\int_0^{2\pi}\left[\sin^4(t)+\cos^4(t)+(\sin(t)-\cos(t))(1-\cos(t)-\sin(t))^3\right]dt \\
&=\frac{3\pi}{2},
\end{align*}
which is the answer we had above.
