Determine the integral of the vector field $F(x,y,z)=(y,z,x)$ across the intersection of the unit sphere $x^2+y^2+z^2=1$ and the plane $Ax+By+Cz=0$ with an arbitrary orientation.
My attempt:
My idea was to apply Stokes theorem because $\operatorname{curl}F=(\partial_yF_3-\partial_zF_2,\partial_zF_1-\partial_xF_3,\partial_xF_2-\partial_yF_1)=(-1,-1,-1).$ The intersection of the unit sphere and the plane $Ax+By+Cz=0$ is a circle $\gamma,$ which encloses a disc $D$ in the given plane so, if $\gamma$ is positively oriented wrt $D,$ by Stokes, $$\int_\gamma Fd\gamma=\int_D\operatorname{curl}FdD.$$ Suppose $D'$ is the orthogonal projection of the disc $D$ onto the $xy$ plane and that $D=\Phi(D'), \Phi(x,y)=\left(x,y,-\frac{A}Cx-\frac{B}Cy\right).$ Then, $$\vec N(x,y)=\partial_x\Phi\times\partial_y\phi=\begin{vmatrix}\vec i&\vec j&\vec k\\1&0&-\frac{A}C\\0&1&-\frac{B}C\end{vmatrix}=\left(\frac{A}C,\frac{B}C,1\right).$$ Therefore, $$\begin{aligned}\int_D FdD&=\int_{D'}F\circ\Phi(x,y)\cdot N(x,y)dxdy\\&=\int_{D'}(-1,-1,-1)\cdot\left(\frac{A}C,\frac{B}C,1\right)dxdy\\&=-\frac{A+B+C}C\int_{D'}dxdy\\&=-\frac{A+B+C}{C\sqrt{\frac{A^2}{C^2}+\frac{B^2}{C^2}+1}}\int_{D'}\sqrt{\frac{A^2}{C^2}+\frac{B^2}{C^2}+1}dxdy\\&=-\frac{A+B+C}{\sqrt{A^2+B^2+C^2}}\operatorname{area}(D)\\&=-\frac{A+B+C}{\sqrt{A^2+B^2+C^2}}\pi,\end{aligned}$$ so, depending on the orientation of the circle, $\int_\gamma Fd\gamma=\mp\frac{A+B+C}{\sqrt{A^2+B^2+C^2}}\pi.$
Is my answer valid?