I'm given the following vector field : $v=(yz,y^2z, yz^2)$ and I need to find its flux through the cylinder $x^2+y^2=1, 0 \leq z \leq 1$. I don't have solutions to this exercise, so I don't know if my results are correct.
The divergence of $v$ is simply $4yz$. So if we set up our integral in cylindrical coordinates, we get $$\int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{1}4zr\sin(\theta) r drd\theta dz$$. Now, if we evaluate this, we get $0$ because $\int_{0}^{2\pi}sin(\theta)=0$. So the result will simply be $0$ even if the original vector field could seem a bit intimidating. We could also use a symmetry argument : integrating the variable $y$, which is an odd function, over a symmetric interval, here a circle, will always yield $0$.
Also, the integrals over the two "surface circles", where $z=0$ and $z=1$, also yield $0$ because for the first one, we integrate over $0$ and for the second, we have $4y$ which is again odd over a symmetric interval.
Do you agree with this ? Thanks for your help !
Edit : as was pointed out in the comments, if we use the divergence theorem, we don't need to evaluate over surfaces, and thus the integrals over the circles where z=0 resp. z=1 are not required at all. This step would be required if we proceed without using the divergence theorem. In this case, we would also need to evaluate over the circular surface of the cylinder.
You can get the same answer even without the divergence theorem. The normal to the bottom surface is $(0,0,-1)$, so the flux is $$\Phi_B=\int_B(0,0,-1)\cdot(0,0,0)dS=0$$ The normal to the top ($z=1$) is $(0,0,1)$, so the flux is $$\Phi_T=\int_T(0,0,1)\cdot(y,y^2,y)dS=\int_TydS=0$$ In the last step I've relied on symmetry.
If you draw a picture of a circle with radius $1$, for any point $(x,y)$ on the circle, the normal to the circle is $(x,y)$, so in the case of the cylinder, the normal to the side is $(x,y,0)$. Then the flux through the side is $$\Phi_S=\int_S(x,y,0)\cdot(yz,y^2z,yz^2)dS=\int_S(xyz+y^3z)dS=\int_0^1zdz\int_0^{2\pi}d\theta(\sin\theta\cos\theta+\sin^3\theta)$$ It's trivial to show that the last integral is zero as well. So the total flux is zero, as you have found from the divergence theorem.