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\noindent
\parbox{2.25in}{\bf Math 3510}
\hfill {\Large \bf Homework \#2} \hfill
\parbox{2.25in}{\bf \hfill Due: Mon., Sept. $17^\mathrm{th}$, 2018}

\vspace{0.2in}

%\noindent {\bf\large Name:} \underline{\qquad\sc\large This is my fabulous name! \qquad}
%
%\vspace{0.2in}

\begin{description}
\item[\#1 Orienting Ourselves:] Let $C$ be the circle: $x^2+z^2=1$ and $y=0$. 
%
%Also, let $S_1$ be the ellipsoid: $\dfrac{(x+1)^2}{4}+\dfrac{(y-2)^2}{9}+(z-5)^2=1$.
\begin{enumerate}[(a)]
\item Can $\Omega = xz^2\,dx+y^3e^z\,dy+xyz\,dz$ be an orientation for $C$?
\item Give a smooth parameterization for $C$ to show it is a $1$-manifold in $\mathbb{R}^3$. Make sure you check that all of the various criterion for a smooth parameterization are met.
%\item Let $\B X:[0,2\pi] \times [0,\pi] \to \mathbb{R}^3$ be defined by $\B X(\theta,\phi) = (2\cos(\theta)\sin(\phi)-1, 3\sin(\theta)\sin(\phi)+2, \cos(\phi)+5)$. 
%         Given that $\B X$ is a smooth parameterization for $S_1$\dots
%         
%         Is $\B X$ compatible with $\Omega = (x+1)\,dy \wedge dz+ (y-2)\,dz \wedge dx+z\,dx \wedge dy$?
%  trouble at boundary
\end{enumerate}

\item[\#2 Exterior Derivatives:] Some quick computations.

\begin{enumerate}[(a)]
\item Let $f(x,y)$ be a smooth function and $\omega = P\,dx+Q\,dy$ both defined on $\mathbb{R}^2$. Compute $df$, $d\omega$, and explain why $d^2=0$ for forms on $\mathbb{R}^2$.
\item Let $\omega = (x^4+y^5z^3)\,dy\wedge dz+ xyz\,dz \wedge dx + (x^2+y^2+z^2)\,dx\wedge dy$. Compute $d\omega$.
\end{enumerate}

\item[\#3 Into the Fourth Dimension!] Consider the following ``innocent'' coordinate transforms\dots

\[ 
\begin{array}{ccc} & \mbox{\bf (1)} & \\  x & =  & r\cos(\theta) \\ y & = & r\sin(\theta) \\ z & = & z \\ t & = & t \end{array}
\qquad\quad
\begin{array}{ccc} & \mbox{\bf (2)} & \\  z & = & \rho\cos(\varphi) \\ r & =  & \rho\sin(\varphi) \\ t & = & t \\ \theta & = & \theta \end{array}
\qquad\quad
\begin{array}{ccc} & \mbox{\bf (3)} & \\  t & = & \beta\cos(\alpha) \\ \rho & =  & \beta\sin(\alpha) \\ \theta & = & \theta \\ \varphi & = & \varphi \end{array}
\]
\[
\begin{array}{ccc} & \mbox{\bf (1)} \circ \mbox{\bf (2)} & \\  x & =  & \rho\sin(\varphi)\cos(\theta) \\ y & = & \rho\sin(\varphi)\sin(\theta) \\ z & = & \rho\cos(\varphi) \\ t & = & t \end{array}
\qquad\qquad
\begin{array}{ccc} & \mbox{\bf (1)} \circ \mbox{\bf (2)} \circ \mbox{\bf (3)} & \\  x & =  & \beta\sin(\alpha)\sin(\varphi)\cos(\theta) \\ y & = & \beta\sin(\alpha)\sin(\varphi)\sin(\theta) \\ z & = & \beta\sin(\alpha)\cos(\varphi) \\ t & = & \beta\cos(\alpha) \end{array}
\]

Notice that each transform is converting a pair of variables to a {\it polar plane}. For transform (1), we have $x^2+y^2=r^2$, $r \geq 0$, and $0 \leq \theta \leq 2\pi$ (or any $2\pi$-period). For transform (2), we have $r^2+z^2=\rho^2$ and $\rho \geq 0$. But now $0 \leq \varphi \leq \pi$. Why only $\pi$ and not $2\pi$? Because $r \geq 0$ so $r=\rho\sin(\varphi) \geq 0$ so we need $0 \leq \varphi \leq \pi$ to keep things positive. Finally, for transform (3), $\rho^2+t^2 = \beta^2$, $\beta \geq 0$, and $0 \leq \alpha \leq \pi$. 

Putting things together, $x^2+y^2+z^2+t^2=r^2+z^2+t^2=\rho^2+t^2=\beta^2$. Notice that (1) is a 4D cylindrical coordinate system, (2) is spherical coordinates in leaving the 4-th coordinate (i.e., $t$) alone, and (3) is a kind of 4D version of spherical coordinates.

The Jacobian determinant of the (1) transform is $J_1 = r$. Likewise, $J_2=\rho$ and $J_3=\beta$. Next, $J_{F \circ G} = J_F J_G$ (chain rule + property of determinants), so $J_{1 \circ 2} = r\rho = \rho^2\sin(\varphi)$ and $J_{1 \circ 2 \circ 3} = r\rho\beta = \rho^2\sin(\varphi)\beta = \beta^3\sin^2(\alpha)\sin(\varphi)$.

\begin{enumerate}[(a)]
\item Let $B_R = \{ (x,y,z,t) \in \mathbb{R}^4 \;|\; x^2+y^2+z^2+t^2 \leq R^2 \}$ (for some fixed $R>0$). 

Find the 4-volume of $B_R$: \qquad $\iiiint\limits_{B_R} 1\,dx\,dy\,dz\,dt$.

\item Consider the unit sphere in $\mathbb{R}^4$: $S^3 = \{ (x,y,z,t) \;|\; x^2+y^2+z^2+t^2 = 1\}$. Use the 4D spherical coordinates to parameterize $S^3$. Compute $\iiint\limits_{S^3} t \,dx\wedge dy \wedge dz$ using your parameterization (i.e., assume that $S^3$ is oriented consistently with your parameterization).
\end{enumerate}

\item[\#4 Verify Stokes' Theorem:] Let $M = \{ (x,y,z) \;|\; x^2+y^2+z^2=4 \mbox{ and } z \geq 0 \}$ be the $2$-manifold with orientation $\Omega = -x\,dy \wedge dz -y\,dz \wedge dx -z\,dx\wedge dy$.

\begin{enumerate}[(a)]
\item Identify $\partial M$. Then find the induced orientation on $\partial M$.
\item Parameterize $\partial M$. Then check if your parameterization is compatible with the induced orientation found in part (a).
\item Let $\omega = (y^2+z^2)\,dx+x^2\,dz$. Verify the generalized Stokes' theorem for this $\omega$ and $M$. 
\end{enumerate}

\end{description}

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