A learning question about volume forms I am learning differential manifold and got a question.
How do we calculate the surface area? Or how to calculate the volume of a submanifold?
Like for the surface area of $S^n$, if $\phi$ is the embedding map, then it seems that
$S=\int\phi^*(\sum_{j=1}^{n+1}(-1)^{j-1}x_j dx_1\wedge dx_2...dx_{j-1}\wedge dx_{j+1}...\wedge dx_{n+1})$ according to some webpage I found. But where did that volume form come from? For a general case, if $(N,\phi)$ is a n-dimension submanifold embedding in a m-dimension manifold M, what is the n-form in $A(M)$ that should be pulled back and integrate on $N$?
Thank you for your patience.
 A: I think that in general the best approach is the following: For all this discussion, we start with a Riemannian metric $ds^2$ on $M$, and we look at the induced Riemannian metric $i^*ds^2$ on $N$. We write 
$$i^*ds^2 = \sum_{j=1}^n \omega^j\otimes\omega^j$$
for a suitable collection of $1$-forms $\omega^j$. Then the induced volume ("area") form on $N$ will be $\omega^1\wedge\dots\wedge\omega^n$. 
For example, consider $S^2\hookrightarrow \mathbb R^3$. Considering spherical coordinates, $i(\phi,\theta) = (\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi)$, we have 
\begin{align*}
i^*ds^2_{\mathbb R^3} &= i^*\big(dx^1\otimes dx^1+ dx^2\otimes dx^2+dx^3\otimes dx^3\big) \\
&= d\phi\otimes d\phi + \sin^2\phi\, d\theta\otimes d\theta \\
&= \omega^1\otimes\omega^1 + \omega^2\otimes\omega^2\,,
\end{align*}
where $\omega^1 = d\phi$ and $\omega^2 = \sin\phi\,d\theta$. [We order these to give the orientation we want on the submanifold.] Then our area form on $S^2$ is
$$\omega^1\wedge\omega^2 = \sin\phi\,d\phi\wedge d\theta\,.$$
A: Suppose you have an orientable manifold $M$ and a volume form $\operatorname{vol}_g$ on $M$. Then to get an induced form on a submanifold $N$ you just need to choose an outward pointing unit vector field $X$ on $N$ (this can always be done) and then do $\iota^\ast(N \lrcorner \operatorname{vol}_g)$, where $\iota : N \hookrightarrow M$ is inclusion and the upper star indicates the pullback. This also gives an induced orientation on $N$. In your case above, all we do is take the standard volume form $dx^1 \wedge \ldots \wedge dx^n$ on $\Bbb{R}^n$ and contract it with the Euler vector field
$$X = x^i \frac{\partial}{\partial x^i}$$
that is outward pointing on $S^n$. For example when $n = 3$ assuming I have calculated this correctly you should get the induced form to be $x dy \wedge dz - y dx \wedge dz + z dx \wedge dy$. Then if you integate this using spherical coordinates you should get the area of $S^2$ (remember $``$volume on $S^2"$  now is area.)
