2-form integration example I encountered the following example and I'm not getting $8\pi$ as the answer.  It seems to me that 
$\int_{D} (x^2+y^2) \mathrm{dx}\wedge \mathrm{dy} = \int\int r^3 \mathrm{dr}\mathrm{d\theta}$
with $0<r<1$ and $0 < \theta < 2\pi$.  This gives $\frac{\pi}{2}$.  Am I missing something?

 A: You are right, the correct value is $\frac{\pi}{2}$, not $8\pi$. The wrong value is however the only error, the argument for transforming the integral into
$$\int_0^{2\pi} \int_0^1 r^3\,dr\,d\theta$$
is correct.
Let's go into "just stick to the definitions and plug things in" mode to verify the value of $\frac{\pi}{2}$. We can parametrise the curve $C$ by $\gamma \colon [-\pi,\pi] \to \mathbb{R}^3$,
$$\gamma(t) = (\cos t, \sin t, \cos t + 1).$$
Then doggedly compute
\begin{align}
\int_C \mathbf{F}\cdot ds &= \int_{-\pi}^{\pi} \langle \mathbf{F}(\gamma(t)), \gamma'(t)\rangle\,dt \\
&= \int_{-\pi}^{\pi} (-\cos^2 t\sin t)\cdot (-\sin t) + (\cos t\sin^2 t)\cdot (\cos t) + (\cos t + 1)^3\cdot(-\sin t)\,dt \\
&= \int_{-\pi}^{\pi} 2\cos^2 t \sin^2 t - (\cos t + 1)^3\sin t\,dt \\
&= \frac{1}{2} \int_{-\pi}^{\pi} (2\sin t \cos t)^2\,dt - \int_{-\pi}^{\pi} (\cos t + 1)^3\sin t\,dt \\
&= \frac{1}{2} \int_{-\pi}^{\pi} \sin^2 (2t)\,dt \tag{1} \\
&= \frac{1}{4} \int_{-2\pi}^{2\pi} \sin^2 u\,du \\
&= \frac{1}{8} \int_{-2\pi}^{2\pi} 1 - \cos (2u)\,du \tag{2} \\
&= \frac{4\pi}{8} -\frac{1}{16}\sin (2u)\biggr\rvert_{-2\pi}^{2\pi} \\
&= \frac{\pi}{2}.
\end{align}
In line $(1)$, we have used the double-angle formula $\sin (2t) = 2\sin t\cos t$, and the fact that $(\cos t + 1)^3\sin t$ is an odd function, so its integral over a symmetric interval - in this case $[-\pi,\pi]$ - is $0$. In line $(2)$ we use the double-angle formula $\cos (2u) = \cos^2 u - \sin^2 u$ and $\cos^2 u + \sin^2 u = 1$ to rewrite $\sin^2 u = \frac{1}{2}(1-\cos (2u))$.
