$\displaystyle S:\ \ x^2+y^2+z^2=1\\ is\ the\ surface\ of\ the\ sphere\ with\ centre\ (0,0,0)\ and\\radius\ =\ 14. \\[2ex] \displaystyle F=yzi+zxj+xyk\ and\ dr=dxi+dyj+dzk \\[2ex] \displaystyle \therefore{}F\bullet{}dr=yzdx+xzdy+xydz \\[2ex] $

$\displaystyle Part\ I: \\[2ex] \displaystyle Consider,\ \int F\bullet{}dr=\ \int yz dx+xzdy+xydz\ \\[2ex] \displaystyle In\ the\ plane\ z=0,\ equation\ of\ sphere\ from\ (1)\ is\ x^2+y^2=1 \\[2ex] \displaystyle It\ is\ a\ circle\ with\ centre\ (0,0)\ and\ radius\ =1\ ...(2)\\[2ex] Using\ parametric\ form, \\[2ex] \displaystyle X=a\cos\theta{}=1\cos\theta{} \\[2ex] \displaystyle \therefore{}dx=-\sin\theta{}d\theta{} \\[2ex] \displaystyle Y=a\sin\theta{}\ =\ 1\sin\theta{} \\[2ex] \displaystyle \therefore{}dy=\cos\theta{}d\theta{} \\[2ex] \displaystyle Z=0\\[2ex] \therefore{}dz=0 \\[2ex] $

$\displaystyle For\ complete\ circle\ limits\ of\ \theta{}\ are\ 0\ to\ 2\pi{}. \\[2ex] \displaystyle \therefore{}\int F\bullet{}dr=\ \int_0^{2\pi{}}0+0+0=0\ ...(3)$

$Part\ 2: \displaystyle F=yzi+zxj+xyk \\[2ex] \displaystyle \therefore{}\nabla{}\times{}F=\left\vert{}\begin{array}{ ccc} i & j & k \\ \frac{\partial{}}{\partial{}x} & \frac{\partial{}}{\partial{}y} & \frac{\partial{}}{\partial{}z} \\ yz & zx & xy \end{array}\right\vert{} \\[2ex] \displaystyle =\ -i(x-x)-j(y-y)-k(z-z) \\[2ex] \displaystyle =\ 0 \\[2ex] \displaystyle In\ xy\ plane,\ Unit\ normal\ is\ N=k\ and\ ds=dxdy \\[2ex] \displaystyle \therefore{}\ N\bullet{}\left(\ \nabla{}\times{}F\right)=k\bullet{}0=0 \\[2ex] \displaystyle \therefore{}\iint N\bullet{}\left(\ \nabla{}\times{}F\right)ds=0\ ..\left(4\right) \\[2ex] \begin{equation} From\ (3)\ and\ (4) %eq12 \end{equation} \displaystyle \int F\bullet{}dr\ \ =\iint N\bullet{}\left(\ \nabla{}\times{}F\right)ds \\[2ex] \displaystyle \therefore{}\ Stokes\ theorem\ is\ verified. \\[2ex]$