$\displaystyle F = {x}^{2 }i + {y}^{2 }j+{z}^{ 2}k $ S the cylinder $\displaystyle 0 \leqslant {x}^{2 } + {y}^{2 } \leqslant 4, 0 \leqslant z \leqslant 4$ , including the top and base. So i have done $\displaystyle \nabla \cdot F = 2xi + 2yj + 2zk$ Now converting to cylindrical co-ordinates $\displaystyle x = 2 \cos \theta y = 2 \sin \theta z = z $ and $\displaystyle |r_\theta \times r_z| = 2$

Are the limits 2 and 0 for dz amd 2pi and 0 for dtheta?