
\section{Integrating Differential Forms}\label{sec:lecture52}
Differential forms exist to be integrated on manifolds. In this section, we will talk about how this is done.

\subsection{Manifolds} For our purposes here, a {\bf smooth parametrized $k$-manifold} in $\rr^n$ will be given by a vector-valued
function ${\bf x}: U\to \rr^n$, where $U$ is a domain in $\rr^k$, ${\bf x}$ is $C^\infty$ and one-to-one on $U$ except possibly for
the boundary, and its derivative matrix
$$
D{\bf x}({\bf u})=
\left| \begin{matrix}
\frac{\partial x_1}{\partial u_1 } & \dots & \frac{\partial x_1}{\partial u_k }
\\
 \vdots &\ddots &\vdots \\
\frac{\partial x_n}{\partial u_1 } & \dots & \frac{\partial x_n}{\partial u_k }
\end{matrix} \right|
$$
has rank $k$ (maximal) at every point $\mathbf{u} \in D$.  This last condition (maximal rank) means that,  if we fix a point $\mathbf{u}_0 \in D$, and calculate the derivative matrix $D \mathbf{x} (\mathbf{u}_0)$ at $\mathbf{u}_0$, then the range of this matrix has dimension $k$. 

This is the set-up that we will use to define integration of differential forms. If $n=3$ and $k=1$, then we just 
have one parameter $t$ and a smooth parametrized curve ${\bf r}(t)$ with ${\bf r}'(t)\neq 0$. 
If $n=3$ and $k=2$, this is our previous definition of parametrized surfaces with two parameters $u,v$.


There are many other ways to define
a manifold. One is to use a system of equations $f_1({\bf x})=0, \dots, f_j({\bf x})=0$. Generically (in some neighbourhood of a point satisfying the given equations, with the precise conditions given by the Implicit
Function Theorem), each equation reduces the dimension of the manifold by 1. A useful term to describe this is the \textit{codimension}: each equation increases the codimension by 1, so that a manifold described by a nondegenerate system of $j$ equations in $\rr^n$ has codimension $j$ and dimension $n-j$.


Another way to define manifolds is more geometrical: essentially, a $k$-manifold without boundary is a subset of $\rr^n$ which ``locally" (in a neighbourhood of any fixed point) looks like $\rr^k$. A $k$-manifold with boundary is
similar, except at the boundary points the manifold locally looks like a half-space $\{{\bf x}\in\rr^k: x_k\geq 0\}$.
(The precise meaning of ``locally looks like" is provided by terms like {\it homeomorphism} or 
{\it diffeomorphism}. You don't have to know, in this class, what these are, but you will hear more about it if you take further courses in differential geometry.)

Each definition can be extended to include {\bf piecewise smooth} manifolds. This does not 
contribute anything conceptually new at this point, so we will skip it here.

\subsection{Tangent vectors and volume elements} Let ${\bf x}({\bf u})$ be a smooth parametrized manifold as above. 
Then we can find $k$ tangent vectors to the manifold by evaluating the derivatives:
$${\bf T}_i = \left\langle  \frac{\partial x_1}{\partial u_i } , \dots , \frac{\partial x_n}{\partial u_i }\right\rangle.$$
(For surfaces in $\rr^3$, these are the vectors ${\bf r}_u$ and ${\bf r}_v$, which we considered regularly throughout this course.) The vectors ${\bf T}_1,\dots,{\bf T}_k$ are the columns of the derivative matrix $D \mathbf{x}$.
Under the maximal rank condition, they are linearly independent at each point $P={\bf x}({\bf u})$ and span a $k$-dimensional affine space called the {\it tangent space} to the manifold at $P$.  

The \textbf{ $k$-volume element} on a manifold is the volume of the $k$-dimensional parallelogram spanned by
the tangent vectors. This is again a generalization of the work we did for surfaces in $\rr^3$. We will skip this part here, but you can check any reasonable textbook on the subject to learn more.

\subsection{Integrating Differential Forms} Recall that a differential $k$-form acts on $k$-tuples of vectors by
setting up a $k\times k$ determinant. So, there is a natural way to set up integration of $k$-forms on $k$-manifolds: just have the form act on the $k$ tangent vectors ${\bf T}_1,\dots,{\bf T}_k$ defined above.
Evaluating the determinant, we get a function of ${\bf u}$ with values in $\rr$. 
Then integrate that function on $U$. This is in fact what we will do. We will also look at how this recovers
some of the integrals we considered earlier, such as flux and circulation. 

\begin{remark}
For now, we will ignore issues related to orientation. This is, in fact, the entire focus of the next section.
\end{remark}

If our parametrized manifold $\mathcal{M}$ is given by a function ${\bf x}: U\to \rr^n$, where $U\subset \rr^k$, and 
if $\Phi$ is a differential $k$-form defined on $\mathcal{M}$, we define
$$
\int_\mathcal{M} \Phi = \int_U \Phi \left( {\bf T}_1,\dots, {\bf T}_k \right) du_1\dots du_k,
$$
where ${\bf T}_1,\dots, {\bf T}_k$ are the tangent vectors defined above. 

Let us see some examples and special cases.

\bigskip

\begin{enumerate}[(1)]

\item \textbf{Volume integration in $\rr^n$.} Let $\mathcal{M}=U$ be an open set in $\rr^n$, parametrized by ${\bf x}({\bf u})={\bf u}$. This is an $n$-manifold 
in $\rr^n$. The tangent vectors are ${\bf T}_i = \frac{\partial {\bf u}}{\partial u_i} = {\bf e}_i$. 

Let $\Phi = f(x_1,\dots,x_n)dx_1\wedge\dots\wedge dx_n $ be a differential $n$-form. Then
$$
\Phi \left( {\bf T}_1,\dots, {\bf T}_k \right) =  f(x_1,\dots,x_n)
\left| \begin{matrix}
1&0 &\ldots & 0\cr
            0 & 1 &\ldots &0\cr
            \vdots&\vdots&\ddots &\vdots\cr
           0 & 0 &\ldots & 1\cr
\end{matrix} \right|
=  f(u_1,\dots,u_n)
$$
so that
$$
\int_\mathcal{M} \Phi = \int_U   f(u_1,\dots,u_n)du_1\dots du_n
$$
is the usual $n$-dimensional integral of functions.

\bigskip

\item \textbf{Line integrals.} Let $\mathcal{M}$  be a smooth curve in $\rr^n$, parametrized by ${\bf r}(t)$ with $t\in[a,b]$. There is only one
tangent vector ${\bf r}'(t)$.

Let $\Phi=F_1 dx_1+ \dots + F_n dx_n$ be a 1-form. Then
\begin{align*}
\int_\mathcal{M} \Phi &= \int_a^b \left[ F_1 dx_1( {\bf r}'(t)) + \dots + F_n dx_n( {\bf r}'(t)) \right] dt
\\
&= \int_a^b \left[ F_1 x'_1(t) + \dots + F_n x'_n(t) \right] dt
\\
&= \int_a^b  F_1 dx_1 + \dots + F_n dx_n,
\end{align*}
generalizing vector line integrals to $n$ dimensions.


\bigskip

\item \textbf{Flux integrals.} Let $\calS$  be a 2-dimensional surface in $\rr^3$, parametrized by ${\bf r}(u,v)$ with $(u,v)\in D$. The
tangent vectors are ${\bf r}_u$ and ${\bf r}_v$.

For a vector field ${\bf F}$ in $\rr^3$, define the 2-form
$\Phi = F_1 dy\wedge dz + F_2 dz\wedge dx + F_3 dx\wedge dy$. Then
\begin{align*}
\int_\calS \Phi &= \int_\calS F_1 dy\wedge dz + F_2 dz\wedge dx + F_3 dx\wedge dy
\\
&= \iint_D F_1({\bf r}(u,v)) dy\wedge dz \left( {\bf r}_u,{\bf r}_v \right) + \dots
\end{align*}
Evaluating the 2-form, we get
$$
\int_\calS \Phi =  \iint_D \left[ F_1({\bf r}(u,v)) 
\left| \begin{matrix}
y_u &  z_u
\\
y_v &  z_v
\\
\end{matrix} \right| + \dots 
\right] du dv = 
\iint_D {\bf F}\cdot {\bf n} \,dudv,
$$
where
$$
{\bf n}= {\bf r}_u\times {\bf r}_v = \left| \begin{matrix}
\ii & \jj & \kk \\
x_u & y_u & z_u 
\\
x_v & y_v & z_v 
\\
\end{matrix} \right| 
$$
(Exercise: fill in the blanks and check that the calculation is correct!) This matches our formula for evaluating
$\iint_\calS {\bf F}\cdot d{\bf S}$, if ${\bf F}=F_1\ii + F_2\jj + F_3\kk$.

\bigskip

\item \textbf{Higher-dimensional flux integrals.} Let's try to extend the concept of flux integrals to $n$ dimensions. Let $\mathcal{M}$  be an $(n-1)$-dimensional surface in $\rr^n$, parametrized by ${\bf x}({\bf u})$ with ${\bf u}\in U$, where $U$ is a domain in $\rr^{n-1}$. The tangent vectors are
${\bf T}_i =   \frac{\partial {\bf x}}{\partial u_i } $. The vector
$$
{\bf n}= \left| \begin{matrix}
{\bf e}_1 & \dots & {\bf e}_n\\
\frac{\partial x_1}{\partial u_1 } & \dots & \frac{\partial x_n}{\partial u_{1} }
\\
 \vdots &\ddots &\vdots \\
\frac{\partial x_1}{\partial u_{n-1} } & \dots & \frac{\partial x_n}{\partial u_{n-1} }
\end{matrix} \right|
$$
is normal to $\mathcal{M}$. (Exercise: check that it is orthogonal to each ${\bf T}_i$.) Then we can 
set up integrals of $(n-1)$-forms $\Phi$ on $\mathcal{M}$ so that
$$
\int_\mathcal{M} \Phi = \int_U {\bf F}\cdot {\bf n} \,d{\bf u},
$$
where ${\bf F}$ is the vector field with the components $F_1,\dots,F_n$ equal to the corresponding
components of $\Phi$. 


\end{enumerate}



\section*{Worked Problems} 
In the hands-on examples below, we ignore orientation issues for now. If you use a different ordering of the tangent vectors in your solution, your answer may differ by a $\pm$ sign. (Exercise: complete the calculation of the integrals!)



\begin{enumerate}[1.]

\item{\it Integrate $\Phi=x_1x_3 dx_1\wedge dx_2 + x_3 dx_1 \wedge dx_4$ on
the 2-manifold ${\bf x}=\langle t,3s+t, s, s-t  \rangle$, $0\leq t \leq 1$, $0\leq s \leq 2$, in $\rr^4$.}

\medskip

The tangent vectors are $\mathbf{x}_t = \langle 1,1,0,-1 \rangle$ and $\mathbf{x}_s = \langle 0,3,1,1 \rangle$. Let us evaluate the 2-forms first:
$$
x_1 x_3 dx_1 \wedge dx_2 =
st
\begin{vmatrix}
1 & 1 \\
0 & 3
\end{vmatrix}
= 3st
$$
$$
x_3 d x_1 \wedge d x_4 =
s
\begin{vmatrix}
1 & -1 \\
0 & 1
\end{vmatrix}
= s.
$$
The integral is
$$
\int_0^1 \int_0^2 (3 st + s)ds dt = 5
$$
If you re-ordered your vectors $\mathbf{x}_t$ and $\mathbf{x}_s$ when evaluating the wedge product, you would have obtained an integral of $-5$, rather than $+5$.

\bigskip

\item{\it Integrate $\Psi=x_2x_3 dx_1\wedge dx_2  \wedge dx_4$ on the part of 
the paraboloid ${\bf x}=\langle s,t,u,s^2+t^2+u^2  \rangle$ given by $0\leq s \leq 1$, $0\leq t \leq 2$, 
$0\leq u \leq 3$ in $\rr^4$.}

\medskip

The tangent vectors are
$$
\mathbf{x}_s = \langle 1,0,0,2s \rangle
$$
$$
\mathbf{x}_t = \langle 0,1,0,2t \rangle
$$
$$
\mathbf{x}_u = \langle 0,0,1,2u \rangle
$$
so that
$$
x_2 x_3\, dx_1 \wedge dx_2 \wedge dx_4( \mathbf{x}_s,\mathbf{x}_t, \mathbf{x}_u)=
tu
\begin{vmatrix}
1 & 0 & 2s \\
0 & 1 & 2t \\
0 & 0 & 2u
\end{vmatrix}
=
tu
\begin{vmatrix}
1 & 2t \\
0 & 2u
\end{vmatrix}
=2 tu^2.
$$
Hence, we evaluate the definite integral
$$
\int_0^3 \int_0^2 \int_0^1 2 u^2 t \, ds dt du = 36.
$$
Again, note: if you re-ordered the vectors $\mathbf{x}_s, \mathbf{x}_t$ and $\mathbf{x}_u$ in your determinant, you may have obtained an answer of $-36$ instead.
\end{enumerate}

\section*{Practice Problems}

For now, ignore orientation issues. Your answer may differ by a $\pm$ sign, depending on how you order the tangent vectors.


\begin{enumerate}[1.]


\item
Compute $\int_M \Psi$, if $M$ is the parametrized manifold
$${\bf r}(u,v)=(u^3-v,v^3, uv^2 ,u^2v ),\ \ 0\leq u \leq 3,\ 0\leq v \leq 2,$$
and $\Psi=dx_1\wedge dx_2 + 2 dx_3 \wedge dx_4$. 

\medskip

\item{ Integrate $\Phi=4 dx_1\wedge dx_2 - x_2^2 dx_3 \wedge dx_4$ on the 2-manifold ${\bf x}=\langle t,s,t\cos s, t\sin s \rangle$, $0\leq t \leq 1$, $0\leq s \leq 2\pi$, in $\rr^4$. }
\medskip


\end{enumerate}








