S is given by $z = \sqrt{x^2+ y^2} \ $ for $1 ≤ z ≤ 4$, then calculate $\nabla \cdot \vec{F} \ $ for the vector field $\vec{F} = (xz, −zy, x^3)$.
Evaluate $I = \iiint \ (\nabla \cdot \vec{F}) \ dV$ , where $V$ is the volume enclosed by the surface S and the two discs {(x, y, z) : x^2+y^2 ≤ 1 , z = 2} and {(x, y, z) : x^2+y^2 ≤ 9 , z = 3}$.
I got a negative answer it thet right?
On
Nothing forbids you from having a negative flux because it’s just an integral.
Through that surface you have portions where the field is outgoing, hence it has the same direction of the external local normal versor of the surface, and other portions where the field is incoming, hence it has the opposite direction of the external local normal versor.
The net balance is the value measured by the integral.
You are always calculating the flux as a outgoing flux, so if it’s positive it means that it’s mainly outgoing, while if it’s negative then it is mainly incoming.
There is no reason why the surface integral cannot be negative. Also the divergence of a vector field can be negative or negative in certain regions and positive in others.
As far as your integral, you have not completely written it so difficult to say where you went wrong but if the first limit mentioned is of $r$, it is not correct.
In your case, $\vec{F} = (z, -zy, x^2)$
The surface is cone $z = \sqrt{x^2+y^2} \ $ along with two discs $x^2+y^2 \leq 1$ at $z = 1$ and $x^2+y^2 \leq 9$ at $z = 3$. So we have a closed surface and applying divergence theorem,
$div{\vec{F}} = \nabla \cdot \vec{F} = -z$ as you mentioned. So your integral is,
$\displaystyle \int_0^{2\pi} \int_1^3 \int_0^z \ (-z) \ r \ dr \ dz \ d\theta = -20 \pi$