这是一份 Imperial帝国理工大学 MATH 192作业代写的成功案例
This looks like Greens Theorem with only the dimension of $U$ changed from 2 to 3 . This is right. The proof is the same too.
Recall that we had for
$$
\begin{aligned}
g: & I^{2} \longrightarrow U \subset \mathbb{R}^{2} \
& {\left[\begin{array}{l}
u \
v
\end{array}\right] \rightsquigarrow\left[\begin{array}{l}
x \
y
\end{array}\right] }
\end{aligned}
$$
and if $\omega$ is a $1-$ form on $U, \omega=P d x+Q d y, g^{} \omega$ was the $1-$ form on $I^{2}$ defined by $$ \left(g^{} \omega\right)\left[\begin{array}{l}
u \
v
\end{array}\right]=P\left[g\left[\begin{array}{l}
u \
v
\end{array}\right]\right] d x+Q\left[g\left[\begin{array}{l}
u \
v
\end{array}\right]\right] d y
$$
and
$$
\left[\begin{array}{l}
d x \
d y
\end{array}\right]=g^{\prime}\left[\begin{array}{l}
d u \
d v
\end{array}\right]=\left[\begin{array}{ll}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \
\frac{\partial y}{\partial u} & \frac{\partial u}{\partial v}
\end{array}\right]\left[\begin{array}{l}
d u \
d v
\end{array}\right]
$$
MATH 192 COURSE NOTES ：
If $\omega$ is a smooth differential 1-form on $U \subseteq \mathbb{R}^{3}$ and $g: I^{2} \longrightarrow U$ is a smooth embedding a.e.,
$$
\int_{\partial g\left(I^{2}\right)} \omega=\int_{\partial I^{2}} g^{} \omega $$ Proof We have for each edge of $\left(I^{2}\right), g$ defines a parametric curve in $U \subseteq \mathbb{R}^{3}$ which is part of the boundary of $g\left(I^{2}\right)$ and $g^{} \omega$ is constructed to take care of the length stretch automatically:
$$
\begin{aligned}
\int_{g(I)} \omega=\int_{0}^{1}\left(P \frac{d x}{d t}+Q \frac{d y}{d t}+R \frac{d z}{d t}\right) d t \
&=\int_{I} g^{*} \omega
\end{aligned}
$$