\documentclass[NoTitle]{math25}
% By Nils~R. Barth
% No title, as it's part of a set

\begin{document}
\begin{enumerate}
\item[(4)] As is standard, we denote \(\text{grad } f, \text{div } \vec{n}\)
by \(\nabla f, \nabla \cdot \vec{n}\) respectively. Beware of the notation \(dA\)! As for
``\(d\theta\),'' there will in general {\em not} be a 1-form \(A\) whose derivative
is \(dA\); if there were, then by Stoke's theorem, the surface area of surfaces without
boundary would be zero, which is not the case, as we see below. We drop the absolute
value bars in the definition of surface area, since all integrals evaluated are positive.

\begin{enumerate}
\item \(\nabla f = (2x, 2y, 2z), \lnorm \nabla f \rnorm = \lnorm (2x, 2y, 2z) \rnorm
        = 2 \lnorm (x, y, z) \rnorm = 2r\), so
        \(\vec{n}(x,y,z)= \nabla f /\lnorm \nabla f \rnorm = \frac{1}{r}(x,y,z)\).
        Also, \(\nabla \cdot \vec{n} = \frac{1}{r}(1+1+1)\,dx\,dy\,dz=\frac{3}{r}\,dx\,dy\,dz\).
        Thus, if we denote the ball centered at the origin of radius \(r\) by \(B_r\),
        and note that \(S_r = \partial B_r\) we obtain
\begin{equation*}
\int_{S_r} dA=\int_{\partial B_r} dA= \int_{B_r} \nabla \cdot dA
       = \int_{B_r} \frac{3}{r}\,dx\,dy\,dz = \frac{3}{r} \int_{B_r} dx\,dy\,dz
       = \frac{3}{r} \cdot \frac{4}{3}\pi r^3 = 4\pi r^2
\end{equation*}

\item Recall that \(h^*(\omega \wedge \eta) = (h^*\omega)\wedge(h^*\eta)\) and
                  \(h^*dx = d(h^*x)\).
\begin{align*}
h^*x& = r\cos \theta \sin \phi &
        h^*dx& = r(\cos \theta \cos \phi\,d\phi - \sin \theta \sin \phi\,d\theta) \\
h^*y& = r\sin \theta \sin \phi &
        h^*dy& = r(\cos \theta \sin \phi\,d\theta + \cos \phi \sin \theta\,d\phi)\\
h^*z& = r\cos \phi             &
        h^*dz& = -r\sin \phi\,d\phi
\end{align*}
We thus obtain, after some algebra, \(h^*dA = r^2\sin\phi\,d\phi\,d\theta\).
Thus,
\begin{equation*}
\int_{S_r} dA = \int_{[0, 2\pi] \times [0, \pi]} h^*dA
= \int_0^{2\pi}\!\!\int_0^\pi r^2\sin\phi\,d\phi\,d\theta
= 2\pi r^2 \int_0^\pi \sin\phi\,d\phi = 2\pi r^2 (1+1) = 4\pi r^2
\end{equation*}

\item Showing that \(d\omega = 0\) is a straight-forward calculation.
      Suppose there exists a 1-form \(\eta\) with \(d\eta = \omega\).
      Now, \(\omega\) restricted to the sphere of radius 1 is \(dA\),
      and the sphere has no boundary, so
\begin{equation*}
4\pi = \int_{S_1} dA = \int_{S_1} \omega = \int_{S_1} d\eta = \int_{\partial S_1} \eta = 0
\end{equation*}
      which is a contradiction.
\end{enumerate}
\end{enumerate}
\end{document}
