Integrate form $\omega=x_jdx_{1}\wedge\cdots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\cdots dx_n$ on $S=\{\textbf{x}\in\mathbb{R^n}:||\textbf{x}||=1\}$ Integrate form $\omega=x_jdx_{1}\wedge\cdots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\cdots dx_n$ on $S=\{\textbf{x}\in\mathbb{R^n}:||\textbf{x}||=1\}$. $i,j$ are constant.
Doing it from definition doesn't seem workable. Any hint would be greatly appreciated. Orientation is external.
I do not know Stoke's theorem yet. I know only basic facts connected with differential forms and some thorems about it working up to $\mathbb{R^3}$.
 A: Let's just do an example.  I am sure you can generalize.
How about we integrate $y \textrm{ d}x \wedge \textrm{d}z$ over the unit sphere in $\mathbb{R}^3$.
$
\begin{align} x&=  \sin\theta \, \cos\varphi \\ y&= \sin\theta \, \sin\varphi \\ z&=  \cos\theta \end{align}
$
The entire sphere is parameterized by $\theta \in [0,\pi]$ and $\phi \in [0,2\pi]$.  Let $R = [0,\pi] \times [0,2\pi]$ be that rectangle in the $\theta,r$ plane.
So we have
$$
\begin{align*}
\int_S y \textrm{ d}x \wedge \textrm{d}z &= \int_R \sin(\theta)\sin(\phi) \textrm{d}(\sin(\theta)\cos(\phi))\wedge\textrm{d}(\cos(\theta))\\
&=\int_R \sin(\theta)\sin(\phi)\left[-\sin(\theta)\sin(\phi) \textrm{d}\phi+\cos(\theta)\cos(\phi)\textrm{d}\theta\right]\wedge\left[-\sin(\theta)\textrm{d}\theta\right]\\
&=-\int_R \sin^3(\theta)\sin^2(\phi) \textrm{d}\theta \wedge \textrm{d}\phi\\
&= -\int_{\phi=0}^{\phi=2\pi} \int_{\theta=0}^{\theta=\pi} \sin^3(\theta)\sin^2(\phi) \textrm{d}\theta \textrm{d}\phi\\
&=-\left(\int_{\phi=0}^{\phi=2\pi} \sin^2(\phi)\textrm{d}\phi\right) \left(\int_{\theta=0}^{\theta=2\pi} \sin^3(\theta)\textrm{d}\theta\right)\\
&=-(\pi)(\frac{4}{3})\\
&=-\frac{4\pi}{3}
\end{align*}
$$
Alternatively, using Stoke's theorem, and letting $B$ be the unit ball, we have
$$
\begin{align*}
\int_S y \textrm{ d}x \wedge \textrm{d}z &= \int_B \textrm{d}(y \textrm{ d}x \wedge \textrm{d}z)\\
&=\int_B \textrm{d}y\wedge \textrm{ d}x \wedge \textrm{d}z\\
&=\int_B - \textrm{dx} \wedge \textrm{dy} \wedge \textrm{dz}\\
&=-\frac{4}{3} \pi
\end{align*}
$$
I obtained the last equality because I know the volume of the unit ball is $\frac{4}{3}\pi$.
If you need clarification of justification for any step, just let me know!
A: Hint:  Use Stoke's theorem to rewrite this as an integral over the interior of the ball.
