Cfrgtkky

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 $D={(x,y,z)inmathbb{R}^{3}~:~y=x^{2}+z^{2},yleq 4}$, and let the region be oriented by the $2$-form $sigma=dzwedge dx$. Evaluate $int_{D}omega$, where $omega=z~dxwedge dy$.

Here's my work:

We parametrize the region with the map $psi:(0,2]times(-pi,pi)tomathbb{R}^{3}$ given by $$psi(r,theta)=(rcostheta,r^2,rsintheta)$$ (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^{*}(dzwedge dx)\
&=d(rsintheta)wedge d(rcostheta)\
&=(sintheta~dr+rcostheta~dtheta)wedge(costheta~dr-rsintheta~dtheta)\
&=-rsin^{2}theta~drwedge dtheta+rcos^{2}theta~dthetawedge dr\
&=-rsin^{2}theta~drwedge dtheta-rcos^{2}theta~drwedge dtheta\
&=-r~drwedge dtheta
end{align}

and we have

begin{align}
psi^{*}omega&=psi^{*}(z~dxwedge dy)\
&=rsintheta~left[d(rcostheta)wedge d(r^{2})right]\
&=rsintheta~left[(costheta~dr-rsintheta~dtheta)wedge(2r~dr)right]\
&=rsintheta~(-2r^{2}sintheta~dthetawedge dr)\
&=(2r^{3}sin^{2}theta)~drwedge dtheta.
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~dtheta\
&=-8int_{-pi}^{pi}sin^{2}theta~dtheta\
&=-4int_{-pi}^{pi}(1-cos(2theta))~dtheta\
&=-4(theta-frac{1}{2}sin(2theta))bigg|_{-pi}^{pi}\
&=-4pi-(-4)(-pi)\
&=-8pi.
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.