
\section{Orientation on Manifolds}

\begin{enumerate}[1.]


\item
{\it 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


We have ${\bf r}_u = (1,0,v,2u),\ \ {\bf r}_v = (0,1,u,-2v)$, so
$$dx_1\wedge dx_3 ({\bf r}_u, {\bf r}_v)
= 
\left|\begin{matrix}
1& v \\
0 & u\\
            \end{matrix}\right|
=u,$$
$$dx_2\wedge dx_4 ({\bf r}_u, {\bf r}_v)
= 
\left|\begin{matrix}
0& 2u \\
1& -2v \\
            \end{matrix}\right|
=-2 u$$
Therefore,
\begin{align*}
\iint_{M} \Psi
&= \int_0^1 \int_0^2 u^3 +2u^2 v \, dv du 
\\
&= \int_0^1   (u^3v+u^2v^2)\Big|_{v=0}^2  du
\\
&= \int_0^1 2u^3 +4u^2 \, du
\\
&= \frac{u^4}{2}+\frac{4u^3}{3} \Big|_0^1 = \frac{1}{2}+ \frac{4}{3} = \frac{11}{6}.
\end{align*}


\medskip

\item
{\it 
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

We have the tangent vectors ${\bf r}_u = (2u, 0, 1, 3u^2)$ and ${\bf r}_v = (0, 2v, 3v^2, -1)$. Now
\begin{align*}
\Psi({\bf r}_u, {\bf r}_v) &= 3u^2 dx_2\wedge dx_4 ({\bf r}_u, {\bf r}_v) + v^2 dx_3 \wedge dx_4 ({\bf r}_u, {\bf r}_v)\\
&= 3u^2
\begin{vmatrix}
0 & 2v\\
3u^2 & -1
\end{vmatrix}
+ v^2
\begin{vmatrix}
1 & 3v^2\\
3u^2 & -1
\end{vmatrix}
\\
&= 3u^2 (-6u^2v) + v^2(-1-9u^2v^2)\\
&= -18u^4v - v^2 - 9u^2v^4.
\end{align*}
Thus
\begin{align*}
\int_M \Psi &= \int_0^1 \int_0^1 \Psi({\bf r}_u, {\bf r}_v) \; du\,dv\\
&= \int_0^1 \int_0^1 \left(-18u^4v - v^2 - 9u^2v^4\right) \; du\,dv\\
&= \int_0^1 \left[-\frac{18}{5}u^5v - uv^2 - 3u^3v^4\right]_0^1 \; dv\\
&= \int_0^1 \left(-\frac{18}{5}v - v^2 - 3v^4\right) \; dv\\
&= \left[-\frac{9}{5}v^2 - \frac{1}{3}v^3 - \frac{3}{5}v^5\right]_0^1\\
&= -\frac{9}{5} - \frac{1}{3} - \frac{3}{5}\\
&= -\frac{41}{15}.
\end{align*}

\medskip

\item
{\it 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

We have ${\bf r}_u = (\cos t, \sin t, 1,0 )$, ${\bf r}_v = (0,0,1,1)$, ${\bf r}_t = (-u\sin t, u\cos t, 0,0)$,
$$dx_1\wedge dx_2 \wedge dx_4  ({\bf r}_u, {\bf r}_v, {\bf r}_t )
= 
\left|\begin{matrix}
\cos t & \sin t & 0 \\
0 & 0 & 1\\
-u \sin t & u \cos t & 0
            \end{matrix}\right|
= - \left|\begin{matrix}
\cos t & \sin t \\
-u \sin t & u \cos t 
            \end{matrix}\right| =-u
$$
\begin{align*}
\iint_{M} \Psi
&= \int_0^1 \int_0^2 \int_0^{2\pi} -2uv \, dt dv du 
\\
&= -2\pi \int_0^1 uv^2 \Big|_{v=0}^2 \, du
\\
&= -2\pi \int_0^1 4u \, du
\\
&= -4\pi u^2\Big|_0^1 = -4\pi.
\end{align*}

\bigskip



\item
{\it 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).}



\medskip


The tangent vectors are
\begin{align*}
{\bf r}_u&= (1,1,\cos v,\sin v),
\\
{\bf r}_v &= (1,-1,-u\sin v,u\cos v).
\end{align*}

(a) We have
\begin{align*}
\omega ({\bf r}_u, {\bf r}_v)
&= 
\left|\begin{matrix}
1&1 \\
1&-1 \\
            \end{matrix}\right| 
=-2
\end{align*}
This is always nonzero, so that the form does orient the manifold.            
            
\medskip

(b)
Since $\omega ({\bf r}_u, {\bf r}_v)<0$, we need to reverse the order of the tangent vectors in the computation:
\begin{align*}
\Phi ({\bf r}_v, {\bf r}_u)
&= 
\left|\begin{matrix}
-u\sin v& u\cos v \\
\cos v & \sin v \\
            \end{matrix}\right| 
=-u,
\end{align*}

\begin{align*}
\iint_{M} \Psi
&= \int_0^2 \int_0^\pi (u^2-v^2)(-u) dvdu = \int_0^2 \int_0^\pi (v^2u-u^3)  dvdu
\\[2ex]
&= \int_0^2 \left[  \frac{v^3u}{3}-u^3v  \right]_{v=0}^\pi  du
\\[2ex]
&=  \int_0^2 \left[  \frac{\pi^3u}{3}-\pi u^3  \right] du
\\[2ex]
&=  \left[  \frac{\pi^3u^2}{6}-\frac{\pi u^4}{4} \right]_{u=0}^2 = \frac{2\pi^3}{3}-4\pi.
\end{align*}









\end{enumerate}


