Properties of integrals of differential forms

Independence of orientation

Lemma: Given $C$ oriented curve in ${\bf R}^2$ parametrized by $p \colon [a,b] \rightarrow {\bf R}^2$, with $p=(p_1,p_2)$. Then $$\displaystyle\int_C A dx = \displaystyle\int_a^b A(p(t))p'_1(t)dt$$ and $$\displaystyle\int_C B dy = \displaystyle\int B(p(t))p'_2(t)dt.$$

Theorem: Same conditions; then $$\displaystyle\int_C A dx + B dy = \displaystyle\int_a^b \left( A(p(t))p'_1(t) + B(p(t))p'_2(t) \right) dt.\blacksquare $$ $\blacksquare$

These two are easy. The next one is more challenging.

Theorem: Same conditions, then $$\displaystyle\int_{-C} \varphi = -\displaystyle\int_C \varphi . $$

Proof: Let $\varphi = A dx + B dy$. Suffices to prove for $\varphi = Adx$ (then prove for $\varphi = Bdy$, then add the two).

Note: We reply on the fact that there are just $2$ orientations.

So, given

• $p \colon [a,b] \rightarrow {\bf R}^2$, $p=(p_1,p_2)$.

We reverse it

• $q \colon [a,b] \rightarrow {\bf R}^2$, $q = (q_1,q_2)$.

How do we get $q$ from $p$?

Consider the diagram:

$$ \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & & C & & \\ & ^{p=p(t)} \nearrow & & \nwarrow ^{q=q(u)} & \\ & [a,b] & \la{T=?} & [a,b] \end{array} $$

Here:

• $pT=q$, so that
• $p_1T=q_1$, and
• $p_2T=q_2$.

The arrows are functions and you can discover another function, $T$, that completes the diagram. Then $T$ should be understood as a change of variables.

We also want: $T(a)=b$, $T(b)=a$. So choose $T(s)=-s + a + b$.

Now,

• $\displaystyle\int_C Adx = \displaystyle\int_a^b A(p(t))p'_1(t) dt,$ (1)

and

• $\displaystyle\int_{-C} A dx = \displaystyle\int_a^b A(q(s))q'_1(s)ds.$ (2)

Now we need to get from (1) to (2) somehow.

Recall the key step of integration by substitution: given a continuous $F$ we have

• $\displaystyle\int_c^d F(T(s))\frac{dT}{ds} ds = \displaystyle\int_{T(c)}^{T(d)} F(t)dt.$ (3)

For (3) we choose $$F(t)=A(p(t))p'_1(t)$$ and $c=b,d=a$.

Now we turn to our integral: $$(1) = \displaystyle\int_a^b A(p(t))p'_1(t) dt$$ ... almost the right-hand side of (3) except for the limits, then $$= -\displaystyle\int_a^b F(T(s))\frac{dT}{ds}ds$$ ...the left-hand side of (3) except for "$-$".... $$= -\displaystyle\int_a^b A(p(T(s)))p'_1(T(s))\frac{dT}{ds}ds$$ ... some algebra now... $$=-\displaystyle\int_a^b A(q(s))p'_1(T(s))\frac{dT}{ds} ds.$$ How do we deal with $p'_1(T(s))$ here? To find the missing part, differentiate this: $$q_1(s)=p_1(T(s))$$ to get $$q'_1(s)=p'_1(T(s))\cdot T'(s) = -p'_1(T(s)).$$ Now substitute. Then our integral is:

$\begin{align*} &=-\displaystyle\int_a^b A(q(s))(-q'_1(s))(-1)ds \\ &=-\displaystyle\int_a^b A(q(s))q'_1(s) ds \\ &= -\displaystyle\int_{-C} Adx. \end{align*}$

$\blacksquare$

Independence of change of variables

Next, what if $q$ was given? We can still find a change of variables $T$.

Consider: $$ \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & & C & & \\ & ^p \nearrow & & \nwarrow ^q & \\ & [a,b] & \la{T=?} & [a,b] \end{array} $$

Let's prove this:

Theorem (Change of variables). $$\displaystyle\int_{C_p} \varphi = \displaystyle\int_{C_q} \varphi,$$ where $C_p$ is $C$ parametrized by $p$ and $C_q$ is $C$ parametrized by $q$, but both $p$ and $q$ have the same orientation.

Proof: The idea of proof is: a change of variables, again.

First, what are the requirements on $p$, $q$?

• They are differentiable.
• They are onto, and 1-1 except possibly for the end-points $p(a)$,$p(b)$,$q(a)$,$q(b)$.
• Their derivatives are non-zero, $p' \neq 0$, $q' \neq 0$. (Exercise: why do we need this?)

We need to find a change of variables, $T$. Simple.

We construct it via compositions. We need:

• (1) $pT=q$.

We deal with the same diagram as above: $$ \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & & C & & \\ & ^p \nearrow & & \nwarrow ^q & \\ & [a,b] & \la{T=?} & [a,b] \end{array} $$ and it becomes clear that we should choose $$T=p^{-1}q.$$ This is OK, since both are invertible.

Observe that to apply a change of variables, $T$ has to be differentiable, and so does $p^{-1}$. That's why we need $p' \neq 0$.

Next, we choose:

• (2) $T(c)=a$, $T(d)=b$,

since $p$,$q$ have the same orientation.

Now,

• $\displaystyle\int_{C_p} A dx = \displaystyle\int_a^b A(p(t))p'_1(t)dt$ (1)

and

• $\displaystyle\int_{C_q} A dx = \displaystyle\int_c^d A(q(s))q'_1(s)ds$ (2).

We now differentiate

• $T = p^{-1}q,$

or

• $pT=q.$

Then

• $p' \frac{dT}{ds} = q',$

so

• $\frac{dT}{ds}=\frac{q'}{p'}.$ (3)

Note: When $\frac{dT}{ds} > 0$, there is no change. We multiply by the sign of $\frac{dT}{ds}$, etc.

Recall, integration by substitution, again.

Given a continuous $F$,

• $\displaystyle\int_c^d F(T(s))\frac{dT}{ds} ds = \displaystyle\int_{T(c)}^{T(d)} F(t) dt.$ (4)

Now, let's get from (1) to (2).

Pick

• $F(t) = A(p(t))p'_1(t).$ (5)

Next we follow the same pattern as in the above theorem.

Our integral is: $$(1) = \displaystyle\int_{T(c)}^{T(d)} A(p(t))p'_1(t)dt,$$ ...almost the right hand side of $(4)$ except for the limits... $$= \displaystyle\int_c^d F(T(s))\frac{dT}{ds}ds,$$ ...left hand side of $(4)$, except for "$-$... $$= \displaystyle\int_c^d F(T(s))\frac{q'}{p'}ds.$$ Substitute $F=(s)$, then... $$=\displaystyle\int_c^d A(p(T(s))p'_1(T(s))\frac{q'_1}{p'_1} ds.$$ Some algebra... $$=\displaystyle\int_c^d A(q(s))q'(s) ds$$ Differentiate: $$q_1(s)=p_1(T(s))$$ to get $$q'_1(s)=p'_1(T(s))T'(s) = p'_1(T(s))\frac{q'_1}{p'_1},$$ That's (2), done. $\blacksquare$

Now $\displaystyle\int_{C} \varphi$ makes sense as it's independent from parametrization!