Compute $\int_M \omega$ 
Let $M=\{(x,y,z): z=x^2+y^2, z<1\}$ be a smooth 2-manifold in $\Bbb{R}^3$. Let $\omega=xdy\wedge dz+ydz\wedge dx+zdx\wedge dy\in \Omega^2(\Bbb{R}^3)$. Compute $$\int_M \omega.$$

I parametrised $M$ (up to a null-set) with $g(r,\theta)=(r\cos \theta, r\sin \theta,r)$ where $(r,\theta)\in (0,1)\times(0,2\pi)$, then computed $g^*\omega$ and found that it is zero, hence $$\int_M \omega = \int_{(0,1)\times(0,2\pi)} g^*\omega=0.$$
Is my result correct?
I feel like there might be an easier way to see that the integral is zero, avoiding finding an explicit parametrisation. Is that true? If so, how could one have argued?
 A: If $f(x,y)=(x,y,x^2+y^2)$ then $$ \int_M \omega = \int_{D} f^\ast \omega $$ where $D$ is a unit disk. Hence $$ \int_D (-x^2-y^2)dxdy = \int_D  (-r^2) rdrd\theta = 2\pi \frac{-r^4}{4}\bigg|_0^1 = -\frac{\pi}{2} $$
A: Use Stokes theorem:
$$ \int_{M} d\omega = \int_{\partial M} \omega$$
Find $\omega_1 $ such that $d \omega_1 = \omega$ and reduce integration to $\partial M = \{z=x^2+y^2 = 1\}$.
EDIT: This could work if $\omega$ would be exact but it is not so: $d \omega = 3\cdot dx \wedge dy \wedge dz \neq 0$.
A: Following the OP's notations, we write
$$g\left(r,\theta\right)=\left(r\cos\theta,r\sin\theta,r^2\right)$$
and then
$$g_r\left(r,\theta\right)=\left(\cos\theta,\sin\theta,2r\right),\quad g_\theta\left(r,\theta\right)=\left(-r\sin\theta,r\cos\theta,0\right).$$
Now we compute
$$\omega\left(g_r\left(r,\theta\right),g_\theta\left(r,\theta\right)\right)=-2r^2\cos^2\theta-2r^2\sin^2\theta+r^2=-r^2$$
whence
$$\int_M\omega:=\iint_{\left(r,\theta\right)\in\left(0,1\right)\times\left(0,2\pi\right)}\omega\left(g_r\left(r,\theta\right),g_\theta\left(r,\theta\right)\right)=\iint_{\left(r,\theta\right)\in\left(0,1\right)\times\left(0,2\pi\right)}\left(-r^2\right)\mathrm{d}x\mathrm{d}y$$
$$=-\iint_{\left(r,\theta\right)\in\left(0,1\right)\times\left(0,2\pi\right)}r^3\mathrm{d}r\mathrm{d}\theta=-\frac{\pi}{2}.$$
