I'm new to working with differential forms and integrating over manifolds. I think that I have the following problem solved, but I'm not all that confident in my work.
Let $D=\{(x,y,z)\in\mathbb{R}^{3}~:~y=x^{2}+z^{2},y\leq 4\}$, and let the region be oriented by the $2$-form $\sigma=dz\wedge dx$. Evaluate $\int_{D}\omega$, where $\omega=z~dx\wedge dy$.
Here's my work:
We parametrize the region with the map $\psi:(0,2]\times(-\pi,\pi)\to\mathbb{R}^{3}$ given by $$\psi(r,\theta)=(r\cos\theta,r^2,r\sin\theta)$$ (I think that the open intervals are okay, because we only need to cover the manifold modulo sets of measure zero). We now check orientations by computing the pullbacks $\psi^{*}\sigma$ and $\psi^{*}\omega$. We have
\begin{align} \psi^{*}\sigma&=\psi^{*}(dz\wedge dx)\\ &=d(r\sin\theta)\wedge d(r\cos\theta)\\ &=(\sin\theta~dr+r\cos\theta~d\theta)\wedge(\cos\theta~dr-r\sin\theta~d\theta)\\ &=-r\sin^{2}\theta~dr\wedge d\theta+r\cos^{2}\theta~d\theta\wedge dr\\ &=-r\sin^{2}\theta~dr\wedge d\theta-r\cos^{2}\theta~dr\wedge d\theta\\ &=-r~dr\wedge d\theta \end{align}
and we have
\begin{align} \psi^{*}\omega&=\psi^{*}(z~dx\wedge dy)\\ &=r\sin\theta~\left[d(r\cos\theta)\wedge d(r^{2})\right]\\ &=r\sin\theta~\left[(\cos\theta~dr-r\sin\theta~d\theta)\wedge(2r~dr)\right]\\ &=r\sin\theta~(-2r^{2}\sin\theta~d\theta\wedge dr)\\ &=(2r^{3}\sin^{2}\theta)~dr\wedge d\theta. \end{align}
Since $\psi^{*}\sigma$ is everywhere negative and $\psi^{*}\omega$ is everywhere positive, we have $\int_{D}\omega=-\int_{(0,2]\times(-\pi,\pi)}\psi^{*}\omega$. Hence,
\begin{align} \int_{D}\omega&=-\int_{-\pi}^{\pi}\int_{0}^{2}(2r^{3}\sin^{2}\theta)dr~d\theta\\ &=-8\int_{-\pi}^{\pi}\sin^{2}\theta~d\theta\\ &=-4\int_{-\pi}^{\pi}(1-\cos(2\theta))~d\theta\\ &=-4(\theta-\frac{1}{2}\sin(2\theta))\bigg|_{-\pi}^{\pi}\\ &=-4\pi-(-4)(-\pi)\\ &=-8\pi. \end{align}
Does my work look correct? I've tried this several times and I've gotten a couple of different answers, but I think this one is correct (I went through the steps methodically and tried not to make any stupid mistakes). Any help is appreciated.