Here's the question: Evaluate the surface integral $\iint _S F\cdot n \space dA$ by the divergence theorem. $ \mathit F = [xy, yz, zx]$, S the surface of the cone $x^2 + y^2 \le 4z^2, \space \space 0 \le z \le 2 $
This is my working for this question. $$\nabla F = y + z + x $$
Parametric equation of S: $$ (r,v,u) = (r\cos(v), r\sin(v), u) $$
The limits for each variable: $$ 0\le v \le 2\pi \quad 0 \le u \le 2 \quad 0\le r \le 2u $$
Jacobian is $r$
The new divergence is $$\nabla F = r\cos (v) + r \sin (v) + u $$
Hence evaluating the triple integral yields $16\pi$
Now, my working out for the triple integral is correct but I am not sure about my parametric equation and limits. I do not know the correct answer to this question since my textbook only provides answers to odd-number problems. However, I came across this website http://www.slader.com/textbook/9780471488859-advanced-engineering-mathematics-9th-edition/463/problems/18/# and their answer is different to mine. Did I do something wrong?
Let's check. You are computing in cylindrical coordinates, which means you need to scale everything by the determinant of the transformation, which is $r$, so $$ \int_0^2\int_0^{2\pi}\int_0^{2z} r(r\cos\theta +r\sin\theta+z)\;\mathrm dr \;\mathrm d\theta\;\mathrm dz $$ The first two integrals in the sum are trivial, since we have an integral over a full period of sine and cosine, so we are left with $$ \int_0^2\int_0^{2\pi}\int_0^{2z} r(r\cos\theta +r\sin\theta+z)\;\mathrm dr \;\mathrm d\theta\;\mathrm dz\\ =\int_0^2\int_0^{2\pi}z\int_0^{2z} r\;\mathrm dr \;\mathrm d\theta\;\mathrm dz\\ =4\pi\int_0^2 z^3\mathrm dz\\ =16\pi $$ as you found.
The answer in the link provided doesn't make sense; the bounds on the radius clearly depend on the height of the cone.