curve integral - intersection between plane and sphere I am going to calculate the line integral
$$ \int_\gamma z^4dx+x^2dy+y^8dz,$$
where $\gamma$ is the intersection  of the plane $y+z=1$ with the sphere $x^2+y^2+z^2=1$, $x \geq 0$, with the orientation given by increasing  $y$.
Since $\gamma$ is an intersection curve, I decided to use Stoke's theorem,  applied to  the vector field $(z^4,x^2,y^8)$ and an oriented surface $Y$ with  boundary $\gamma$. But how am I going to parametrize the surface so I can use it with Stoke's thoerem?
If I parametrize the surface by $(x(s,t),y(s,t),z(s,t))=(0,t,1-t),$ $x^2+y^2+z^2\leq 1$, I will get the normal vector $(0,0,0)$, but the normal vector is going to point upwards, I think.
 A: The intersection curve (an ellipse) is given by $x^2+2(y-\frac{1}{2})^2=\frac{1}{2}$. It can be parametrized as $x=\frac{\sqrt2}{2}\cos t$, $y=\frac{1}{2}\sin t+\frac{1}{2}$, where $t\in[0,2\pi]$ (also don't forget z=1-y). Could you check all this and try to carry out the integration, and maybe show your work below?
A: Given the plane $\Pi\to(p-p_0)\cdot\vec n = 0$ and the sphere $S\to\|p\|=r$ the distance from $\Pi$ to the origin is calculated as $p_i$ where $p_i$ is the intersection point between $\Pi$ and the line $L\to p = \lambda \frac{\vec n}{\|\vec n\|}$ or $d=\lambda^* = \frac{p_0\cdot\vec n}{\|\vec n\|}$. The intersection $S\cap\Pi$ produces a circumference with radius $r_c = \sqrt{r^2-d^2}$. Now we can build a parameterized circumference representing the intersection $S\cap\Pi$ as
$$
C(\gamma) = p_i + \left(\vec s\sin\gamma +\vec t\cos\gamma\right)r_c
$$
with $\{\vec s, \vec t\}$ such that
$
\cases{
\vec t\cdot\vec n = 0\\
\vec t\times\frac{\vec n}{\|n\|}=\vec s\\
\|\vec t\|=1
}
$
NOTE
Here $p=(x,y,z),\ \vec n = (0,1,1)$ and $p_0=(0,0,1)$.
With $C(\gamma)$ we have $\{x(\gamma),y(\gamma),z(\gamma)\}$ and finalizing with
$$
 \int_{\gamma}\left( z^4(\gamma)\frac{dx}{d\gamma}+x^2(\gamma)\frac{dy}{d\gamma}+y^8(\gamma)\frac{dz}{d\gamma}\right)d\gamma
$$
