Integration of forms on \(\mathbb {R}^n\) and on manifolds.

A reference for this material is Chapter 16 of John M. Lee. Introduction to smooth manifolds. Second edition. Grad. Texts in Math., 218. Springer, New York, 2013. xvi+708pp. ISBN: 978-1-4419-9981-8.

Solution:

For convenience, we write \((x^1, \ldots , x^n)\) for the coordinates in \(U\) and \((y^1, \ldots , y^n)\) for the coordinates in \(V\). Write

\[ \omega = f \, dy^1 \wedge \cdots \wedge dy^n . \]
Notice that
\[ \varphi ^{\ast } dy ^j = \sum _{i =1 } ^n \frac {\partial y ^j }{\partial x^i } dx ^i \]
Then
\begin{align*} \varphi ^{\ast } \omega & = ( f \circ \varphi ) \, \varphi ^{\ast }dy^1 \wedge \cdots \wedge \varphi ^{\ast } dy^n \\ & = (f \circ \varphi ) \sum _{i_1, \ldots , i_n = 1 } ^n \frac {\partial y ^1}{\partial x ^{i_1}} \cdots \frac {\partial y ^n}{\partial x ^{i_n}} dx^{i_1} \wedge \cdots \wedge dx ^{i_n} \\ & = (f \circ \varphi ) \sum _{\sigma \in S_n} \text {sign}(\sigma )\, \frac {\partial y ^{ 1}}{\partial x ^{\sigma (1) }} \cdots \frac {\partial y ^{n}}{\partial x ^{\sigma (n) }} dx^1 \wedge \cdots \wedge dx ^n \\ & = (f \circ \varphi ) \text {det} (d\varphi ) dx^1 \wedge \cdots \wedge dx ^n \\ & = (f \circ \varphi ) \vert \text {det} (d\varphi ) \vert dx^1 \wedge \cdots \wedge dx ^n, \end{align*}

where the absolute value appeared because \(\varphi \) is orientation-preserving. Integrating,

\begin{align*} \int _U \varphi ^{\ast } \omega & = \int _{U} (f \circ \varphi ) \vert \text {det} (d\varphi ) \vert \\ & = \int _V f \\ & = \int _V \omega . \end{align*}
Solution:

Exercise. Use the change of variables theorem.