这是一份uwa西澳大学MATH1011的成功案例

If $g: I \longrightarrow U \subseteq \mathbb{R}^{2}$ is a smooth embedding and if $\omega$ is a differential 1-form on $U, g^{} \omega$ is the differential 1-form on $I$ defined by $$ g^{} \omega \triangleq P \circ g \frac{d x}{d t} d t+Q \circ g \frac{d y}{d t} d t
$$
where
$$
\omega \triangleq P d x+Q d y
$$
and
$$
g(t)=\left[\begin{array}{l}
x(t) \
y(t)
\end{array}\right]
$$

MATH1011COURSE NOTES ：

$$
\begin{aligned}
\omega & \triangleq P d x \wedge d y \
g^{} \omega & \triangleq(P \circ g) d x \wedge d y \end{aligned} $$ and $$ \left[\begin{array}{l} d x \ d y \end{array}\right]=\left[\begin{array}{ll} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t} \end{array}\right]\left[\begin{array}{l} d s \ d t \end{array}\right] $$ where allows us to calculate $d x \wedge d y$ in terms of $d s \wedge d t$. This gives: $$ \begin{aligned} d x \wedge d y &=\left(\frac{\partial x}{\partial s} d s+\frac{\partial x}{\partial t} d t\right) \wedge\left(\frac{\partial y}{\partial s} d s+\frac{\partial y}{\partial t} d t\right) \ &=\left(\frac{\partial x}{\partial s} \frac{\partial y}{\partial t}-\frac{\partial x}{\partial t} \frac{\partial y}{\partial s}\right) d s \wedge d t \end{aligned} $$ So $$ g^{}(\omega)=P \circ g\left(\frac{\partial x}{\partial s} \frac{\partial y}{\partial t}-\frac{\partial x}{\partial t} \frac{\partial y}{\partial s}\right) d s \wedge d t
$$