Let $z=x^2+y^2$, and $0\leq z \leq 4$ and let a) $F=[x,y,2z]$ b) $F=[x,y,3z]$. Verifying Divergece theorem gives for the volum integral using a) $\nabla \cdot F=4$ and b) $\nabla \cdot F=5$ and using $\int_0^{2\pi} \int_0^2 \int_{r^2}^4 (\nabla \cdot F) r dz dr d\theta$ gives a) $32 \pi$ and b) $40 \pi$.
Surface integral for the top of paraboloide is for a) $32 \pi$ b) $48 \pi$ and for the side of the paraboliode gives for a) $0$ and b) $-8 \pi$ where we use normal vector for the side of paraboloid $(2r^2 \cos \theta , 2r^2 \sin \theta , -r)$ My question is if the direction of the unity vector is correct now?
This gives correct answer for part b) to get $48 \pi - 8 \pi = 40 \pi$ as well.
Remember the unity normal vector used for the top is [0,0,1] and for side is $\frac{[-2r^2 cos \theta, -2r^2\sin \theta, r]}{ r \sqrt{4r^2+1}} $ for this transformation the Jacobi is $ r \sqrt{4r^2+1} $ and Jacobi and the length of unity normal vector are simplified (deleted). Here we should choose the unity normal vector outward, so it would be: $\frac{[2r^2 cos \theta, 2r^2\sin \theta, -r]}{ r \sqrt{4r^2+1}} $