This time my question is based on this example Divergence theorem
I wanted to change the solution proposed by Omnomnomnom to cylindrical coordinates. $$ \iiint_R \nabla \cdot F(x,y,z)\,dz\,dy\,dx = \iiint_R 3x^2+3y^2+3z^2\,dz\,dy\,dx = $$ $$ \int_{0}^1 \int_{0}^{2\pi}\int_0^2 3r+3z^2r\,dz\,d \varphi \,dr =\int_{0}^{1} \int_{0}^{2\pi} 6r+8r\,d \varphi \,dr =\int_{0}^{1} 28 \pi r \,dr = 14 \pi$$ Which is a different result than the one obtained from calculating the triple integral proposed by Omnomnomnom (with the help of wolframalpha Triple Integral Calculator) - it gives $34.5575$
What have I done wrong?
EDIT: As suggested by DeltaG: $$ \int_{0}^1 \int_{0}^{2\pi}\int_0^2 3r^2+3z^2r\,dz\,d \varphi \,dr =\int_{0}^{1} \int_{0}^{2\pi} 6r^2+8r\,d \varphi \,dr =\int_{0}^{1} 12 \pi r^2+16 \pi r \,dr = 12 \pi$$
Which unfortunately still is not equal to $34.5575$
You've lost an $r$: $3x^2+3y^2 = 3r^2$, rather than $3r$. The $r$ just before the differentials should be distributed to both term in the integrand as well.
Response to edit: the $z$ is fine now, but that $r$ at the end isn't making it into the first term, I think. $$ \int_{0}^1 \int_{0}^{2\pi}\int_0^2 (3r^2+3z^2)r\,dz\,d \varphi \,dr =\int_{0}^{1} \int_{0}^{2\pi} (6r^2+8)r\,d \varphi \,dr =\int_{0}^{1} (12 \pi r^3+16 \pi r) \,dr = 3\pi+8\pi=11 \pi$$