Suppose $P(x,\,y,\,z)=Q(x,\,y,\,z)=R(x,\,y,\,z)=f((x^2+y^2)z)$ and $f$ has continuous derivative. Evaluate $$ \lim_{t\to0+}\frac{\iint_\Omega P(x,\,y,\,z)\,dydz+Q(x,\,y,\,z)\,dzdx+R(x,\,y,\,z)\,dxdy}{t^4}, $$ where $\Omega$ is the outside surface of $\{(x,\,y,\,z)\vert x^2+y^2\leq t^2,\,z\in[0,\,1]\}$.
I used Gauss Theorem to evaluate it, but could not get it.
Let's assume outside surface includes the planes on the ends. If not, you'll have to subtract off their contribution. By Divergence theorem you get that
$$\iint_\Omega (P,Q,R)\cdot \nu d\sigma = \iiint_{\text{Int}(\Omega)} 2xzf'((x^2+y^2)z)+2yzf'((x^2+y^2)z)+(x^2+y^2)f'((x^2+y^2)z) d^3 x$$
The first two terms are odd functions of $x$ and $y$, respectively, so their contributions to the final integral are $0$ by symmetry. Converting to cylindrical coordinates and computing the integral of the last term:
$$2\pi \int_0^t \int_0^1 r^3 f'(r^2z)dzdr = 2\pi\int_0^t rf(r^2)-rf(0)dr$$
Now plug this in to the integral and use L'Hopital
$$\lim_{t\to 0^+} \frac{2\pi\int_0^t rf(r^2)-rf(0)dr}{t^4} = \lim_{t\to 0^+} \frac{\pi t f(t^2) - \pi tf(0)}{2t^3} = \lim_{t\to 0^+} \frac{\pi f(t^2) - \pi f(0)}{2t^2}$$
Then L'Hopital again
$$= \lim_{t\to 0^+} \frac{\pi t f'(t^2) }{2t} = \frac{\pi}{2}f'(0)$$
