We have to verify above using stoke's theorem,

$\because \int\int\limits_{s}\bar{N}\cdot \bar{V} \times f)ds = \int\limits_{c}\bar{F}d\bar{r}$

Here c is the boundary of the upper half of the $(x^2 + y^2 + z^2) = 4$

$F dr = [(2x - y)\hat{i} - (yz^2)\hat{j} - (y^2 z)\hat{k}][dx\hat{i} + dy\hat{j} + dz\hat{k}] \\ Fdr = (2x - y)dx - (yz^2)dy - (y^2 z)dz$

Putting z = 0, dz = 0 $\therefore \bar{F}\cdot d\bar{r} = (2x - y)dx$

Now putting x = 2cos$\theta$ y = 2sin$\theta$

dx = -2sing$\theta$d$\theta$

$\therefore \int\limits_{c}F dr = \int\limits_{0}^{2\pi}(2 *2cos\theta - 2sin\theta)(-2sin\theta)d\theta \\ \therefore \int\limits_{c}F dr = \int\limits_{0}^{2\pi} -8sin\theta cos\theta + sin^2 \theta d\theta \\ \therefore \int\limits_{c}F dr = -4 \int\limits_{0}^{2\pi}sin2\theta + 4 \int\limits_{0}^{2\pi} \frac{1 - cos2\theta}{2} d\theta \\ \therefore \int\limits_{c}F dr = -4 \bigg[- \frac{cos2\theta}{2}\bigg]_{0}^{2\pi} + 4\bigg[\frac{\theta}{2} - \frac{sin2\theta}{4}\bigg]_{0}^{2\pi} \\ \therefore \int\limits_{c}F dr = -4\bigg[\frac{-1}{2} + \frac{1}{2}\bigg] + 4[\pi] \\ \therefore \int\limits_{c}F dr = 4\pi$