Problem statement:
Calculate the integral of the vector field F (x, y, z) = ($ y^2 , z^2 , x^2 $) along the following surface: r(u, v) = ($u \cdot cos(v), u \cdot sin(v), u^2$ ) , u ∈ [0, 1], v ∈ [0, 2π].
I am fully aware that the surface along which the vector field is integrated is the boundary surface of a simple solid region (r is closed and piece-wise smooth) , hence Guass' divergence theorem can be applied here. \begin{equation} \iiint_V \nabla \cdot{\bf F } \: dV =\iint_S {\bf F} \cdot {\bf \hat{n}} \: {\bf dA} \end{equation}
It's easy to observe that $\nabla$$\cdot$ ${\bf F}$ = 0 so it follows that the flux of the vector field is zero as well, but evaluating the surface integral yields a different result.
"$\frac{\pi}{4}$".
Why does the divergence theorem fail here, any idea on what I'm doing wrong?
Your surface is the part of the paraboloid $z=x^2+y^2$ cut off below the plane $z=1$. It is not a closed surface, so the divergence theorem doesn't apply to it. However, if you add in a disk at the top, it becomes a closed surface. Since the integral over the whole surface is zero, the integral over the paraboloid part and the disk part must be opposites. The integral over the top disk is
$$ \iint_D x^2\,\mathrm{d}A = \int_0^{2\pi}\int_0^1 (r\cos\theta)^2\,\,r\mathrm{d}r\mathrm{d}\theta=\left(\int_0^{2\pi}\cos^2\theta\,\mathrm{d}\theta\right)\left(\int_0^1 r^3\,\mathrm{d}r\right)=(\pi)\left(\frac{1}{4}\right). $$
Note $\mathbf{F}\cdot\mathbf{N}=(y^2,z^2,x^2)\cdot(0,0,1)=x^2$ and $\int\cos^2\theta\,\mathrm{d}\theta$ is a standard by-parts integral and a substitution with polar coordinates gives $\mathrm{d}A=r\mathrm{d}r\mathrm{d}\theta$.