Calculate the line integral $$\int_{\gamma}(y-x)dx+(z-x)dy+(x-y)dz$$ where $\gamma$ is a circunference obtained from the intersection of the sphere $$S=\lbrace (x,y,z)\in \mathbb{R}^{3}:x^2+y^2+z^2=a^2 \rbrace$$ and the plane $S_{1}$ given by the ecuation $y=x\tan \alpha$ and orientation counterclockwise from positive axis $x$ Is my approach below correct or i forget some detail, any feedback or recomendation was very helful and useful. Thanks so much for read me
Notice that the surface $S_{1}$ give recorres all $\mathbb{R}^3$ of form that the plane recorres all $\alpha \in [0,\pi]$ where $\alpha \neq \pi/2$, too can assume that $a\in \mathbb{R}$ with $a\neq 0$.
the intersection of $S \cap S_{1}$ is when both equations are satisface , therefore $y=x\tan \alpha$ in $x^2+y^2+z^2=a^2$ we get $x^2+x^2\tan^2 \alpha+z^2=a^2$ that is equivalent to $x^2(1+\tan^2\alpha)+z^2=a^2$ Which is a circle in the plane $XZ$ Of form that we can see the surface like a \begin{equation} x^2\sec^2 \alpha+z^2=a^2 \end{equation}
For $\alpha \in [0, \pi]$ y $a\in \mathbb{R},a \neq 0$ with $\alpha \neq \pi/2$. Dividing $(1)$ by $a^2$ we get \begin{equation} \frac{x^2}{\frac{a^2}{\sec^2 \alpha}}+\frac{z^2}{a^2}=a^2 \end{equation} \begin{equation} \frac{x^2}{a^2\cos^2 \alpha}+\frac{z^2}{a^2}=a^2 \end{equation} Notice $(3)$ is a easy equation of parametrize.
Let $t\in [0,\pi/2]$ y $\alpha \in [0, \pi]$ with $\alpha \neq \pi/2,a \neq 0$ then \begin{equation} x=a\cos \alpha \sin t \end{equation} \begin{equation} y=0 \end{equation} \begin{equation} z=a\cos t \end{equation} Is one parametrization of the intersection $S \cap S_{1}$ of form that found $\frac{dx}{dt}=a\cos \alpha \cos t$, $\frac{dy}{dt}=0$, $\frac{dz}{dt}=-a\sin t$ and remplazion in the integral
$$\int_{\gamma}(y-x)dx+(z-x)dy+(x-y)dz=4 \int_{0}^{\pi/2}{(-a^2\cos^2 \alpha \sin t \cos t-a^2\cos \alpha \sin^2 t)dt}$$ since $$4\int_{0}^{\pi/2}{(-a^2\cos^2 \alpha \sin t \cos t-a^2\cos \alpha \sin^2 t)dt}=-a^2\cos \alpha(2\cos \alpha+ \pi)$$ then $$\int_{\gamma}(y-x)dx+(z-x)dy+(x-y)dz=-a^2\cos \alpha(2\cos \alpha+ \pi)$$