
\section{Integrating Differential Forms}

\begin{enumerate}[1.]


\item
{\it 
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


We have the tangent vectors ${\bf r}_u = (3u^2, 0, v^2, 2uv)$ and ${\bf r}_v = (-1, 3v^2, 2uv, u^2)$. Now
\begin{align*}
\Psi({\bf r}_u, {\bf r}_v) &=  dx_1\wedge dx_2 ({\bf r}_u, {\bf r}_v) + 2 dx_3 \wedge dx_4 ({\bf r}_u, {\bf r}_v)\\
&= 
\begin{vmatrix}
3 u^2  & -1\\
0 & 3v^2
\end{vmatrix}
+ 2
\begin{vmatrix}
v^2 & 2uv \\
2uv & u^2 
\end{vmatrix}
\\
&= (9u^2 v^2)  + 2 (-3 u^2v^2)\\
&= 3u^2v^2
\end{align*}
Thus
\begin{align*}
\int_M \Psi &= \int_0^2 \int_0^3 \Psi({\bf r}_u, {\bf r}_v) \; du\,dv\\
&= \int_0^2 \int_0^3 3u^2v^2 \; du\,dv\\
&= \int_0^2 \left[u^3v^2 \right]_{u=0}^3 \; dv\\
&= \int_0^2 27v^2 \; dv \\
&= \left[9v^3\right]_0^2\\
&= 72.
\end{align*}

\bigskip



\item{\it 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

The tangent vectors are ${\bf x}_t=\langle 1,0, \cos s, \sin s \rangle$ and ${\bf x}_s=\langle 0,1,-t\sin s, t\cos s \rangle$. We first evaluate the 2-forms:
$$
dx_1\wedge dx_2 ({\bf x}_t, {\bf x}_s)= 
\left| \begin{matrix}
1 &  0
\\
0 &  1
\\
\end{matrix} \right| =1,
$$
$$
x_2^2 dx_3\wedge dx_4 ({\bf x}_t, {\bf x}_s)= 
s^2 \left| \begin{matrix}
\cos s &  \sin s 
\\
-t \sin s & t \cos s 
\\
\end{matrix} \right| =s^2 t,
$$
so that the integral is
$$
\int_0^1 \int_0^{2\pi} (4-s^2 t) ds dt = 8\pi - \frac{4\pi^3}{3}.
$$











\end{enumerate}

