
\section{Orientation on Manifolds}

\subsection{Defining orientation on manifolds}


In dimensions $n=1,2,3$, we have always worked with the {\bf oriented} space $\rr^n$ in the sense that we 
assumed a certain ordering of the basis vectors (for example, the vectors $\ii$ and $\jj$ in $\rr^2$ are arranged
in the counterclockwise order). 
Switching to a different coordinate system preserves orientation if the new basis vectors are arranged the same way.
You can think about this in terms of manipulating the basis
vectors. If you want to preserve orientation, you can stretch the basis vectors or rotate them continuously, but with restrictions: you are not allowed to shrink any basis vector to zero length, and you are 
not allowed to rotate any one of the basis vectors so that it crosses the subspace spanned by the remaining vectors.



\begin{figure}[h!]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/53fig1.pdf}
\end{minipage}
\end{figure}






\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .60\linewidth]{figures/53fig2.png}
\caption{The original basis vectors $\mathbf{i}$ and $\mathbf{j}$,  as well as two linear transformations applied to them: one with positive determinant, the other, negative.}
\end{minipage}
\end{figure}




A more precise way to define this, and one that works in higher dimensions as well, is to look at the sign of the determinant of a linear mapping. Let's say that ${\bf u}_1,\dots,{\bf u}_n$ and ${\bf v}_1,\dots,{\bf v}_n$ are the basis vectors in two coordinate systems. Define a linear mapping $L:\rr^n\to\rr^n$ so that $L({\bf u}_j)={\bf v}_j$ for $j=1,\dots,n$. We will then say that the two coordinate systems 
define the same orientation if the
determinant of $L$ is positive, and that they define opposite orientations if the determinant is negative.



If the linear mapping $L$ is just a permutation of the basis vectors, then it preserves orientation if
the permutation involves an even number of transpositions of pairs of vectors, and reverses it if it
involves an odd number of transpositions. (Permutations can be written as superpositions of transpositions
of pairs in many different ways. A result in combinatorics says that for a given permutation, the number of
transpositions is either always odd or always even. This is called the {\bf sign} of the permutation: 
a permutation is odd if the number of
transpositions involved in it is always odd, and even if the number of transpositions is even.)


\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/53fig5.png}
\caption{Two different linear transformations. The first linear transformation ($2$-dimensional) does \textit{not} preserve orientation, since one transposition occurs. However, the second transformation ($3$-dimensional) does preserve orientation, since two transformations (an even number) of transpositions occured.}
\end{minipage}
\end{figure}



Note that the differential form $dx_1\wedge \dots \wedge dx_n$ captures the same information. We have
$$
dx_1\wedge \dots \wedge dx_n ({\bf e}_1,\dots,{\bf e}_n)=1.
$$
If we interchange two of the basis vectors, this reverses the orientation; it also changes the differential 
form evaluation to $-1$ (because two of the rows in the determinant get interchanged). If we interchange two pairs, we return to the original orientation, and the form now evaluates to $+1$; and so on.




That's how we get to differential forms defining orientation on manifolds. On a $k$-manifold, a small neighbourhood 
of any non-boundary point looks locally like $\rr^k$, so we can define orientation {\it on that neighbourhood}
like we would in $\rr^k$ -- by choosing a basis of tangent vectors in the tangent space at that point, 
and then choosing an ordering of them. However, this works only locally, and we want to orient the 
entire manifold so that the orientations on different small neighbourhoods match continuously. 
A $k$-differential form with possibly non-constant coefficients allows us to provide
such consistency on manifolds which are not flat. 




Let $\omega$ be a smooth differential $k$-form, defined everywhere on a $k$-manifold $\mathcal{M}$ in $\rr^n$. 
Assume that $\mathcal{M}$ has a smooth parametrization ${\bf x}: U\to \rr^n$, where $U\subset \rr^k$.
Fix some ${\bf x}={\bf x}({\bf u})\in \mathcal{M}$. We are interested in how $\omega$ acts on vectors tangent to $\mathcal{M}$ at that ${\bf x}$. 

Let ${\bf T}_1,\dots,{\bf T}_k$ be the tangent vectors to $\mathcal{M}$ at ${\bf x}$ that we defined above. From our definition of a smooth manifold, we know that they are linearly independent. 
Let $\mathcal{T}_{\bf x}$ be the linear space spanned by these vectors. (This is the $k$-dimensional space of all directions tangent to $\mathcal{M}$ at the given point.) 

We now want to zoom in on that space. First, we evaluate all coefficient functions of $\omega$ at ${\bf x}$. This gives us a $k$-form $\omega_{\bf x}$ with constant coefficients, still defined on $\rr^n$. Then, we restrict $\omega_{\bf x}$ to $\mathcal{T}_{\bf x}$. This is still linear in each variable and antisymmetric, hence a $k$-form with constant coefficients on a $k$-dimensional space. 


We can define a new coordinate system ${\bf y}=(y_1,\dots,y_k)$ on $\mathcal{T}_{\bf x}$ (for example, with
${\bf T}_1,\dots,{\bf T}_k$ as the basis vectors).
In these coordinates, $\Omega_{\bf x}$ has the form
\begin{equation}\label{53-eOmega}
\Omega_{\bf x}=a_{\bf x}\, dy_1\wedge\dots\wedge dy_k
\end{equation}
for some constant $a\in\rr$. We will say that $\Omega_{\bf x}$ is \textbf{non-degenerate} if $a_{\bf x}\neq 0$; since ${\bf T}_1,\dots,{\bf T}_k$ are linearly independent, this is true if and only if 
$$ %\begin{equation}\label{54-nondeg}
\Omega_{\bf x} ( {\bf T}_1,\dots, {\bf T}_k )\neq 0
$$ %\end{equation}
(see Example \ref{51-lin-indep}).
In that case, the sign of 
$\Omega_{\bf x}({\bf T}_1,\dots,{\bf T}_k)$ captures the ordering of the tangent vectors ${\bf T}_1,\dots,{\bf T}_k$ at ${\bf x}$. 


To do this consistently throughout the entire manifold, we need a ``global" version, as follows.

\begin{definition}
Let $\omega$ be a smooth differential $k$-form on a $k$-manifold $\mathcal{M}$ in $\rr^n$. We say that
 $\omega$ {\bf orients} $\mathcal{M}$ if the restricted $k$-form $\Omega_{\bf x}$, defined above, is non-degenerate for each ${\bf x}\in\mathcal{M}$. We say that $\mathcal{M}$ is {\bf orientable} if such a form $\omega$ exists.
\end{definition}
 
In this global picture, we go back from $\Omega_{\bf x}$ to $\omega$ as follows. First, since $\Omega_{\bf x}$ was just the restriction of $\omega_{\bf x}$ to tangent directions, we have $\Omega_{\bf x}({\bf T}_1,\dots,{\bf T}_k)=\omega_{\bf x}({\bf T}_1,\dots,{\bf T}_k)$. Second, we now allow ${\bf x}$ to vary, so that both the coefficients of $\omega$ and the tangent vectors ${\bf T}_j$ are variable as we move around the manifold. Thus, $\omega$ orients $\mathcal{M}$ if and only if 
$$
\omega ( {\bf T}_1,\dots, {\bf T}_k )\neq 0\hbox{ for all }{\bf u}\in U,
$$
and the given parametrization is \textbf{consistent} (or compatible) with the orientation given by $\omega$ if
the tangent vectors ${\bf T}_1,\dots, {\bf T}_k$ satisfy
\begin{equation}\label{53-eomega}
\omega ( {\bf T}_1,\dots, {\bf T}_k )>0\hbox{ for all }{\bf u}\in U.
\end{equation}
(If they don't, we need to reorder the variables so that the above condition is satisfied.)




\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .7\linewidth]{figures/53fig6.png}
\caption{The torus in $\mathbb{R}^3$ (pictured above) is an example of a ``non-flat'' manifold where we can define orientation and keep it consistent throughout the manifold.}
\end{minipage}
\end{figure}


\begin{example}
Let $\mathcal{M}=\{(x,y,z,g(x,y,z)):\ (x,y,z)\in U\}$ be the graph of a function of 3 variables in $\rr^4$. WIth $\mathcal{M}$ parametrized by the $(x,y,z)$ variables, the tangent vectors are
$$
{\bf T}_x=(1,0,0,g_x),\ {\bf T}_y=(1,0,0,g_y),\ {\bf T}_z=(1,0,0,g_z).
$$
We claim that the 3-form $\omega=dx\wedge dy\wedge dz$ orients $\mathcal{M}$. Indeed,
$$
\omega({\bf T}_x,{\bf T}_y,{\bf T}_z) =
\left| \begin{matrix}
1&0&0\\ 0&1&0\\ 0&0&1\\
\end{matrix} \right| =1,
$$
which is never zero. The same calculation shows that the parametrization above is consistent with the orientation given by $\omega$.
(Exercise: think about how this would extend to functions of $n$ variables!)



\end{example}



\begin{example}\label{53-ex-cylinder}
To orient manifolds that are not graphs of functions, we usually need forms $\omega$ with variable coefficients. For instance, let 
$\bfr(t,z)=\cos t\ii +\sin t\jj+z\kk$ be the parametrization of a cylinder in $\rr^3$. The two tangent vectors are
$$
\bfr_t = -\sin t \ii +\cos t \jj ,\ \ \bfr_z =\kk.
$$
If we try to use the elementary 2-forms for orientation, we get
$$
dx\wedge dy(\bfr_t,\bfr_z)=0, \ dx\wedge dz(\bfr_t,\bfr_z)=-\sin t, \ dy\wedge dz(\bfr_t,\bfr_z)=\cos t,
$$
so that none of them work. However, if $\omega=x dy\wedge dz- y dx\wedge dz$, then
$$
\omega(\bfr_t,\bfr_z)=\cos^2 t +\sin^2 t =1,
$$
so that this form does provide a valid orientation. A geometric interpretation is as follows. 
Let ${\bf N}=\cos t \ii + \sin t \jj$ be the normal vector to the cylinder. Then the form $\omega$ above can also be written in the 
determinant form
$$
\omega({\bf v}_1,{\bf v}_2) = \left| \begin{matrix}
\cos t & v_{11} & v_{21} \\
\sin t & v_{12} & v_{22} \\
0 & v_{13} & v_{23} \\
\end{matrix} \right| ,
$$
with columns ${\bf N}$, ${\bf v}_1$, ${\bf v}_2$. In particular,
$$\omega(\bfr_t,\bfr_z)=  \left| \begin{matrix}
\cos t & -\sin t & 0\\
\sin t & \cos t & 0 \\
0 & 0& 1 \\
\end{matrix} \right| =1,
$$
corresponding to the fact that ${\bf N}, \bfr_t,\bfr_z$ are three orthonormal vectors oriented the same way as the $\ii,\jj,\kk$ triple.

\end{example}






%\vfill
%
%\pagebreak

Some important special cases we should keep in mind are the following.

\begin{enumerate}[$\bullet$]

\item{\bf Orientation on $\rr^n$:} We will always use the $n$-form $dx_1\wedge\dots\wedge dx_n$ to orient solid $n$-dimensional domains in $\rr^n$ including the entire space. In $\rr^2$ and $\rr^3$, this means $dx\wedge dy$ and $dx\wedge dy\wedge dz$, respectively. This corresponds to the standard (counterclockwise for $n=2,3$) orientation we have been using all along.

\medskip


\item{\bf Orientation on curves:} if ${\bf r}(t)$ is a smooth parametrized curve, then ${\bf r}'(t)$ provides
orientation. In the language of differential forms, the 1-form $\omega_{\bf r}({\bf v}) = {\bf r}'\cdot {\bf v}$
orients the curve. Recall that, for a smooth parametrized curve, we assumed that ${\bf r}'(t)\neq {\bf 0}$ for all $t$,
so that ${\bf r}'\cdot {\bf v}\neq 0$ for any vector ${\bf v}$ tangent to the curve at ${\bf r}(t)$.


\medskip

\item {\bf Orientation on surfaces in $\rr^3$:} this is a generalization of Example \ref{53-ex-cylinder}. If ${\bf n({\bf x})}$ is the normal vector to $\calS$
at ${\bf x}$, let 
\begin{equation}\label{53eq10}
\omega_{\bf x}({\bf v}_1,{\bf v}_2) = \left| \begin{matrix}
{\bf n({\bf x})} & {\bf v}_1 & {\bf v}_2
\end{matrix} \right| 
\end{equation}
(the determinant of the matrix with columns ${\bf n({\bf x})}$, ${\bf v}_1$, ${\bf v}_2$). If $\calS$ is orientable,
then ${\bf n({\bf x})}$ can be chosen consistently on $\calS$ so that it is never zero--and therefore (\ref{53eq10}) defines a nonvanishing differential form. The determinant in (\ref{53eq10}) is then positive if and only if the vectors ${\bf n({\bf x})}$, ${\bf v}_1$, $ {\bf v}_2$ are oriented the same way as the vectors $\ii,\jj,\kk$ in $\rr^3$, meaning that ${\bf v}_1,{\bf v}_2$ are oriented counterclockwise when viewed from the direction of ${\bf n({\bf x})}$. A parametrization ${\bf r}(t,s)$ is consistent with this orientation if the same is true of the ${\bf T}_t,{\bf T}_s$ vectors (i.e. if they are oriented counterclockwise when viewed from the direction of ${\bf n({\bf x})}$). 

\medskip

\item {\bf Orientation on hypersurfaces of dimension $n-1$ in $\rr^n$:} the last definition generalizes to
all $n\geq 2$. Suppose that $\mathcal{M}$ is a manifold of dimension $n-1$ in $\rr^n$, and that 
we can choose a non-zero normal vector ${\bf n}$ continuously throughout $\mathcal{M}$. Then
orientation on $\mathcal{M}$ can be defined using the form
\begin{equation}\label{53eq11}
\omega_{\bf x}({\bf v}_1,{\bf v}_2, \dots, {\bf v}_{n-1}) = \left| \begin{matrix}
{\bf n}({\bf x}) & {\bf v}_1 & {\bf v}_2 & \dots & {\bf v}_{n-1}
\end{matrix} \right| 
\end{equation}
In many situations, there is a natural choice of ${\bf n}$. For example, if $\mathcal{M}$ is given by an equation
$g(x_1,\dots,g_n)=0$ with $\nabla g\neq 0$ on $\mathcal{M}$, then we can choose ${\bf n}=\nabla g$. If $\mathcal{M}$ is the boundary of a solid region $D\subset\rr^n$, we may choose ${\bf n}$ to point away from $D$. As in $\rr^3$, the determinant in (\ref{53eq11}) is positive if and only if the ordering of ${\bf n}, {\bf v}_1,\dots ,{\bf v}_{n-1}$ matches the standard orientation in $\rr^{n}$.

\end{enumerate}

\medskip

\subsection{Inherited orientation on boundary} 
Suppose that $\mathcal{M}$ is a manifold of dimension $k$ in $\rr^n$, oriented by a $k$-form $\omega$, and
that its boundary $\partial \mathcal{M}$ is a $(k-1)$-manifold. Then the {\it inherited orientation} on $\partial \mathcal{M}$ is given by the form
\begin{equation}\label{53eq12}
\partial \omega_{\bf x}({\bf v}_1,{\bf v}_2, \dots, {\bf v}_{k-1}) :=
\omega_{\bf x}(      {\bf n}({\bf x}),      {\bf v}_1,{\bf v}_2, \dots, {\bf v}_{k-1}) ,
\end{equation}
where $ {\bf n}({\bf x})$ is the vector ``tangent to $\mathcal{M}$" (more precisely, it would be tangent to $\mathcal{M}$ if
$\mathcal{M}$ were extended past its boundary), normal to $\partial\mathcal{M}$, and pointing away from $\mathcal{M}$.  

\medskip

\begin{enumerate}[$\bullet$]

\item Suppose that $\mathcal{M}=D$ is a solid region in $\rr^n$. We use our standard orientation in $\rr^n$, given by the form $\omega=dx_1\wedge \dots\wedge dx_n$. Then the formula (\ref{53eq12}) defining $\partial\omega$ on $\partial\mathcal{M}$ matches (\ref{53eq11}) if we let $ {\bf n}({\bf x}) $ point away from $D$. This is consistent with our earlier convention for surfaces bounding solid regions in $\rr^3$.

\medskip

\item Let $\mathcal{M}=\calS$ be a 2-dimensional surface in $\rr^3$, bounded by a curve $\mathcal{C}$. Assume that $\calS$ is orientable according to our definition in Chapter 3, so that it has a unit normal vector ${\bf N}$ varying smoothly throughout the manifold. We then have the 2-form
\begin{equation}\label{53eq13}
\omega_{\bf x}({\bf v}_1,{\bf v}_2) = \left| \begin{matrix}
{\bf N} & {\bf v}_1 & {\bf v}_2
\end{matrix} \right| 
\end{equation}
as in (\ref{53eq10}),
providing counterclockwise orientation on $\calS$ when viewed from the direction of ${\bf N}$. We then imagine $\calS$ extending beyond its boundary, and orient the pair  $({\bf n}, {\bf T})$ consistently with this orientation, where ${\bf n}$ is as described above and ${\bf T}$ indicates the orientation of the boundary curve. You can check that this matches our earlier convention.

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/54fig1.png}
\caption{Inherited orientation on the boundary of the surface.}
\end{minipage}
\end{figure}


\end{enumerate}

\medskip

\subsection{Integration on oriented manifolds} Our integration formula from Section \ref{sec:lecture52} was as follows: if $\mathcal{M}$ is parametrized by a function ${\bf x}: U\to \rr^n$, where $U\subset \rr^k$, and 
if $\Phi$ is a differential $k$-form defined on $\mathcal{M}$, then
$$
\int_\mathcal{M} \Phi = \int_U \Phi \left( {\bf T}_1,\dots, {\bf T}_k \right) du_1\dots du_k,
$$
where ${\bf T}_1,\dots, {\bf T}_k$ are the tangent vectors defined above. 

But the thing is, how do we know the ordering of the ${\bf T}_1,\dots, {\bf T}_k$ vectors in $\Phi$? For example, if a 2-manifold is parametrized by $(s,t)$, we could use $(s,t)$ in that order, or we could switch them and use $(t,s)$ instead. Such a change in parametrization would affect the order of two of the tangent vectors, therefore the sign of $\Phi ( {\bf T}_1,\dots, {\bf T}_k )$ and, ultimately, the sign of the integral.

When we introduce orientation on the manifold, {\it we fix the order in which we should write the tangent vectors.} Specifically, if $\omega$ is the form providing orientation on $\mathcal{M}$, we want to label our parameters $u_1,\dots,u_k$ (therefore the tangent vectors ${\bf T}_1,\dots, {\bf T}_k$) so that $\omega ( {\bf T}_1,\dots, {\bf T}_k )>0$. 
If this holds, then the parametrization is {consistent} with the orientation given by $\omega$. This is the order of tangent vectors we need to use in our integration formula. We illustrate this in the Worked Problems section.

\begin{remark}
We assumed earlier that the tangent vectors ${\bf T}_1,\dots, {\bf T}_k$ are linearly independent everywhere on $\mathcal{M}$ except possibly on its boundary. We also assumed that $\omega$ is nonzero on $\mathcal{M}$. Therefore $\omega ( {\bf T}_1,\dots, {\bf T}_k )$ cannot change sign on $\mathcal{M}$, so that it must be either always positive or always negative on $\mathcal{M}$ except possibly for boundary points. We will see this in the examples below.
\end{remark}




\section*{Worked Problems}
In the Worked Problems in Section \ref{sec:lecture52}, our solutions ignored orientation issues. We now revisit the same questions with attention paid to the orientation of the underlying manifolds.

\begin{enumerate}[1.]
\item{\it
Check that the 2-manifold ${\bf x}=\langle t,3s+t, s, s-t  \rangle$, $0\leq t \leq 1$, $0\leq s \leq 2$, in $\rr^4$, is oriented by the 2-form $\omega=dx_1 \wedge dx_2$. Integrate $\Phi=x_1x_3 dx_1\wedge dx_2 + x_3 dx_1 \wedge dx_4$ on
this manifold, with this choice of orientation.}

\medskip

The tangent vectors are ${\bf x}_t=\langle 1,1, 0, -1 \rangle$ and ${\bf x}_s=\langle 0,3, 1, 1 \rangle$. Since
$$
dx_1\wedge dx_2 ({\bf x}_t, {\bf x}_s)= 
 \left| \begin{matrix}
1 &  1
\\
0 &  3
\\
\end{matrix} \right| =3
$$
is never 0, the form orients the manifold; moreover, since it evaluates to a positive value on $({\bf x}_t, {\bf x}_s)$ taken in this order, this is the order of the tangent vectors we want to use. 




The rest of the solution is as before. We evaluate the 2-forms:
$$
x_1x_3 dx_1\wedge dx_2 ({\bf x}_t, {\bf x}_s)= 
ts \left| \begin{matrix}
1 &  1
\\
0 &  3
\\
\end{matrix} \right| =3ts,
$$
$$
x_3 dx_1\wedge dx_4 ({\bf x}_t, {\bf x}_s)= 
s \left| \begin{matrix}
1& -1 
\\
0 & 1
\\
\end{matrix} \right| =s,
$$
so that the integral is
$$
\int_0^1 \int_0^{2} (3ts+s)  ds dt = 5.
$$



(Additional note: if we wanted to orient
$\mathcal{M}$ by $-\omega$ instead, we would have to reverse the order of the $t$ and $s$ variables.)

\medskip

\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 $\rr^4$, with orientation given by the form $\omega = dx_1 \wedge dx_2 \wedge dx_3$.}

\medskip

The tangent vectors are ${\bf x}_s=\langle 1,0, 0, 2s \rangle$, ${\bf x}_t=\langle 0,1,0,2t \rangle$,
 and ${\bf x}_u=\langle 0,0,1,2u \rangle$. Then
 $$
dx_1\wedge dx_2 \wedge dx_3 ({\bf x}_s, {\bf x}_t. {\bf x}_u )= 
 \left| \begin{matrix}
1 &  0 & 0
\\
0 &  1 & 0 \\
0& 0& 1 
\\
\end{matrix} \right| =1>0,
$$
so that the above order of the $s,t,u$ variables works for us. Now
$$
x_2 x_3 dx_1\wedge dx_2 \wedge dx_4 ({\bf x}_s, {\bf x}_t. {\bf x}_u )= 
tu \left| \begin{matrix}
1 &  0 & 2s
\\
0 &  1 & 2t \\
0& 0& 2u 
\\
\end{matrix} \right| = tu (2u) =2tu^2 
$$
so that the integral is
$$
\int_0^1 \int_0^{2} \int_0^3 2tu^2 \, du dtds = 36.
$$
Calculation omitted here (but not in your own work)!

\medskip

\noindent
(Additional note: the above calculation shows that the form $\omega'=dx_1\wedge dx_2\wedge dx_4$ could also be used to orient the manifold, with $\omega'({\bf x}_s, {\bf x}_t, {\bf x}_u )=2u>0$ except for the part of boundary where $u=0$.
This would still work for the purposes of integration here, but there are times (as in the Generalized 
Stokes's Theorem, to come next time) when we want a form that does not vanish on the boundary.)










\end{enumerate}

\section*{Practice Problems}
\begin{enumerate}[1.]

\item
{Compute $\int_M \Psi$, if $\Psi=x_1^2 dx_1\wedge dx_3- x_3 dx_2\wedge dx_4$ and
$M$ is the parametrized manifold
${\bf r}(u,v)=(u,v, uv,u^2-v^2  )$, $0\leq u \leq 1$, $0\leq v \leq 2$, 
oriented consistently with this parametrization.}

\medskip

\item
{
Compute $\int_M \Psi$, if $M$ is the parametrized manifold
${\bf r}(u,v)=(u^2,v^2, u+v^3 ,u^3-v  )$, $0\leq u \leq 1$, $0\leq v \leq 1$, 
oriented consistently with this parametrization, and $\Psi=3x_1dx_2\wedge dx_4 + x_2 dx_3 \wedge dx_4$.}

\medskip

\item
{Compute $\int_M \Psi$, if $\Psi = 2 x_4 dx_1\wedge dx_2  \wedge dx_4$
and
$M$ is the parametrized manifold
${\bf x}(u,v,t)=( u\cos t ,u\sin t , u+ v, v )$, $0\leq u \leq 1$, $0\leq v \leq 2$, $0\leq t\leq 2\pi$,
oriented consistently with this parametrization.}

\medskip

\item
Let $M$ be the manifold parametrized by
$${\bf r}(u,v)=( u+v,u-v,u\cos v,u\sin v),$$
with $0\leq u \leq 2$, $0\leq v\leq \pi$. 
 

(a) Prove that the differential form $\omega= dx_1\wedge dx_2$ orients $M$.


(b) Compute $\int_M \Phi$, if $\Phi=x_1x_2  dx_3\wedge dx_4$ and $M$ is oriented by the form $\omega$ from (a).





\end{enumerate}





