Compute $\int_M \omega$

Question

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?

Answer 1

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 $$\int_M\omega:=\iint_{\left(r,\theta\right)\in\left(0,1\right)\times\left(0,2\pi\right)}\omega\left(g_r\left(r,\theta\right),g_\theta\left(r,\theta\right)\right)=\iint_{\left(r,\theta\right)\in\left(0,1\right)\times\left(0,2\pi\right)}\left(-r^2\right)\mathrm{d}x\mathrm{d}y$$ $$=-\iint_{\left(r,\theta\right)\in\left(0,1\right)\times\left(0,2\pi\right)}r^3\mathrm{d}r\mathrm{d}\theta=-\frac{\pi}{2}.$$

Answer 2

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: $d \omega = 3\cdot dx \wedge dy \wedge dz \neq 0$.

Answer 3

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 $$ \int_D (-x^2-y^2)dxdy = \int_D (-r^2) rdrd\theta = 2\pi \frac{-r^4}{4}\bigg|_0^1 = -\frac{\pi}{2} $$