curve integral - intersection between plane and sphere

Question

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.

Answer 1

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?

Answer 2

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

$$ \int_{\gamma}\left( z^4(\gamma)\frac{dx}{d\gamma}+x^2(\gamma)\frac{dy}{d\gamma}+y^8(\gamma)\frac{dz}{d\gamma}\right)d\gamma $$