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\begin{document}
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\section*{ 4. Integration of a function on $M$.}

�� �忡���� manifold�������� form�� ���лӸ��� �ƴ϶� �Լ���
������ �ٷ��� �Ѵ�.\\

{\bf 1. Volume element $\alp$ and $\int_Mf\alp$}\\

Let $\alp$ be a function given on $M^n$ such that $\forall p\in
M$,

$\alp(p)=|\om_p|$ for some $\om_p\in\Lam^n(T^*_pM)$

where
$|\om_p|(X_1,\cdot\cdot\cdot,X_n):=|\om_p(X_1,\cdot\cdot\cdot
X_n)|,\,\,X_i\in T_pM.$\\

Such a function $\alp$ is called a {\it volume element} - a way of
measuring n-dimensional volume(without sign).

$\alp$ is {\it smooth} if it can be written locally as
$\alp=g|dx_1\we\cdot\cdot\cdot\we dx_n|$ for $g\in\ccn,\,\,g\geq
0.$

Let $\alp$ be a smooth volume element on $M$.

$M$ �� $\rb^n$�� �� ��ǥ���� volume element�δ�
$|du_1\we\cdot\cdot\cdot\we du_n|$�� �ִ�.

( $\rb^n$������ $\int_{\rb^n}f|du_1\we\cdot\cdot\cdot\we
du_n|:=\int_{\rb^n}fdu_1\cdots du_n$ ����
���ǵȴ�.)\\

$<$ Change of variable formula. $>$


Domain $D\subset\rb^n$ �󿡼� diffeomorphism $\vphi:D\rightarrow
\vphi(D)$�� ����

$\int_D(f\circ\vphi)|det\vphi_*|dt_1\cdot\cdot\cdot
dt_n=\int_{\vphi(D)}fdu_1\cdot\cdot\cdot du_n$ (������ȯ����)

$\hs{2em}\Rightarrow\dis{\hs{2em}\int_D\vphi^*(f|du_1\we\cdot\cdot\cdot\we
du_n |)=\int_{\vphi(D)}f|du_1\we\cdot\cdot\cdot\we du_n|}$

(����) $\vphi^*|du_1\we\cdot\cdot\cdot du_n|(X_1,\cdot\cdot\cdot
X_n )=|du_1\we\cdot\cdot\we
du_n|(\vphi_*X_1,\cdot\cdot,\vphi_*X_n)$

$\hs{3em}=|du_1\we\cdot\cdot\cdot\we
du_n(\vphi_*X_1,\cdot\cdot\cdot,\vphi_*X_n)|$

$\hs{3em}=|\vphi^* du_1\we\cdot\cdot\cdot\we\vphi^*
du_n(X_1,\cdots,X_n)|$

$\hs{3em}=|d\vphi^*u_1\we\cdot\cdot\cdot\we
d\vphi^*u_n(X_1,\cdots,X_n)|$

$\hs{3em}=|d(u_1\circ\vphi)\we\cdots\we
d(u_n\circ\vphi)(X_1,\cdots,X_n)|$

$\hs{3em}=|det \vphi_*\cdot dt_1\we\cdots\we
dt_n(X_1,\cdots,X_n)|$

$\hs{3em}=|det\vphi_*||dt_1\we\cdots\we dt_n|(X_1,\cdots,X_n)$

$\therefore \vphi^*|du_1\we\cdots\we
du_n|=|det\vphi_*||dt_1\we\cdots\we dt_n|\hs{8em}\square$\\


For $f$ with $supp\,f\subset(U,x)$, let $\vphi=x\inv$,
$\widetilde{U}=x(U)$. Then

$\hs{3em}\dis{\int_Mf\alp:=\int_{\widetilde{U}}\vphi^*(f\alp)}$ is
well defined as before.\\

Note that if $M$ is orientable and $\om$ is a positive n-form,
then
$\int_Mf\om=\int_Mf|\om|$.\\

In general, for $f$ with compact support on $(M,\alp)$, choose a
finite cover $\{U_1,\cdots,U_k\}$ of $supp\,f$ and a partition of
unity subordinate to this cover and define
$\dis{\int_Mf\alp:=\sum_{i=1}^k\int_M\rho_if\alp}$. This is
well-defined as before independent of choice of a cover and
a partition of unity.\\


{\bf 2. Classical examples.}\\

(1) $n=1$ :

$\alp=ds=$length element�� $\alp=|"ds"|$ �� ���ǵȴ�. (���⼭ "ds"
�� Riemannian oriented volume form�̴�.) i.e., $ds(v_p)=|v_p|$.
�׷��� a parametrization $\vphi:[a,b]\rightarrow M\subset\rb^n$��
����

$\dis{\hs{2em}\int_Mfds=\int_a^bf(\vphi(t))\frac{ds}{dt}dt=\int_{[a,b]}\vphi^*(fds)}$ �� �ȴ�.\\

������ $\vphi^*(fds)=(\vphi^*f)(\vphi^*ds)$�̰� $\vphi^*ds$��
���ϱ� ���� $\vphi^*ds=g|dt|$�� ����. �׷���

$\vphi^*ds(\frac{d}{dt})=g|dt|(\frac{d}{dt})=g|dt(\frac{d}{dt})|=g|1|=g.$

$\therefore \vphi^*ds=\vphi^*ds(\frac{d}{dt})|dt|$

$\hs{3.2em}=ds(\vphi_*\frac{d}{dt})|dt|$

$\hs{3.2em}=ds(\frac{d\vphi}{dt})|dt|$

$\hs{3.2em}=|\vphi'(t)||dt|$\\

$\therefore \vphi^*(fds)=(f\circ\vphi)|\vphi'(t)||dt|=(f\circ\vphi)\frac{ds}{dt}|dt|$.\\


(2) $n=2$ :

$\alp=d\sigma=$area element=$|"d\sigma"|$,

$d\sigma(X_p,Y_p)=|"d\sigma"(X_p,Y_p)|$= $X_p$�� $Y_p$�� �̷��
����纯���� ���̰� �ȴ�. ������ \\

Choose $N=(a_1,a_2,a_3),$ a unit normal on $M$ such that
$[(N,X,Y)]=[(X,Y,N)]$ giving the standard orientation of $\rb^3$.
Then\\

$\hs{4em}$ area of parallelogram $(X,Y)$

$\hs{4.2em}$= volume of parallelepiped $(N,X,Y)\hs{2em}(\because
|N|=1)$

$\hs{4.2em}=|det(N,X,Y)|$

$\hs{4.2em}=|dx\we dy\we dz(N,X,Y)|\hs{1.5em}$

$\hs{4.2em}=|i_N\mu(X,Y)|\hs{1.5em}$ (let $ dx\we dy\we dz= \mu )$

$\hs{4.2em}=|i_N\mu|(X,Y)$

$\hs{4.2em}=|i_N(dx\we dy\we dz)|(X,Y)$

$\hs{4.2em}=|i_N(dx)\we dy\we dz-dx\we i_N(dy)\we dz+dx\we dy\we
i_N(dz) |(X,Y)\hs{1.5em}$

Since $a_1=dx(N),a_2=dy(N),a_3=dz(N),$

$\hs{4.2em}=|\underbrace{a_1 dy\we dz-a_2dx\we dz+a_3dx\we
dy}|(X,Y)$

$\hs{12em}"d\sigma"$

$\hs{4.2em}=d\sigma(X,Y)$\\

parametrization�� �־����� �� ���� �������� ����� ����.


$M$�� ���� parametrization �� $\vphi:D(\subset\rb^2)\rightarrow M$
�̶� �θ�

$\dis{\int_Mfd\sigma=\int_{D}\vphi^*(fd\sigma)=\int_{D}(f\circ\vphi)\vphi^*(d\sigma)}$
����

$\vphi^*(d\sigma)$�� ���ϱ� ���� �̸� $f|du\we dv|$�� ����.
($\vphi^*(d\sigma)$�� $D$������ measure�̹Ƿ� �̿� ���� �� ��
�ִ�. ) �׷��� $\vphi^*(d\sigma)(\frac{\pa}{\pa u},\frac{\pa}{\pa
v})=f$ �̰� ����

$\vphi^*(d\sigma)=\vphi^*(d\sigma)(\frac{\pa}{\pa
u},\frac{\pa}{\pa v})|du\we dv|$

$\hs{3.2em}=d\sigma(\vphi_*\frac{\pa}{\pa u},\vphi_*\frac{\pa}{\pa
v})|du\we dv|$

$\hs{3.2em}=d\sigma(\frac{\pa\vphi}{\pa u},\frac{\pa\vphi}{\pa
v})|du\we dv|$

$\hs{3.2em}=$area of $(\frac{\pa\vphi}{\pa u},\frac{\pa\vphi}{\pa
v})\cdot|du\we dv|$

$\hs{3.2em}=|\frac{\pa\vphi}{\pa u}\times\frac{\pa\vphi}{\pa
v}||du\we dv|$\\

$\hs{2em}\therefore\dis{\int_Mfd\sigma=\int_{D}(f\circ\vphi)|\,\frac{\pa\vphi}{\pa
u}\times\frac{\pa\vphi}{\pa v}\,|du\,dv}$\\\\

{\bf 3. Riemannian volume element.}\\

Let $g$ be a Riemannian metric and let $\{e_1,\cdots,e_n\}$ be an
orthonormal basis at $p\in M$ and $\{\eps_1,\cdots,\eps_n\}$ be
its dual basis. The {\it Rimannian volume element}, dvol at $p$ is
given by $|\eps_1\we\cdots\we\eps_n|$.
($\nu=\eps_1\we\cdots\we\eps_n$ is called a {\it oriented volume
form } if $(e_1,\cdots,e_n)$ is positive.)

Locally on $(U,x)$, let $g_{ij}=g(\paxi,\paxj)$ and $G=(g_{ij})$.
If we let $\paxj=\sum_{i=1}^na_{ij}e_i$, then
$\eps_i=\sum_ja_{ij}dx_j$.


$\therefore\eps_1\we \cdots\we\eps_n=(det\,A)dx_1\we\cdots\we
dx_n,\,\,where\,\,\,A=(a_{ij})$.

���� $g_{ij}=g(\paxi,\paxj)=g(\sum_ka_{ki}e_k,\sum_la_{lj}e_l)$

$\hs{3.6em}=\sum_{k,l}a_{ki}a_{lj}\delta_{kl}$

$\hs{3.6em}=\sum_{k}a_{ki}a_{kj}=\sum_k\,^ta_{ik}a_{kj}$

$\therefore G=\,^tAA$ and $det\,G=(det\,A)^2(>0)$.

$\therefore dvol=|\eps_1\we\cdots\we
\eps_n|=|det\,A||dx_1\we\cdots\we
dx_n|=\sqrt{det\,G}|dx_1\we\cdots\we dx_n|$\\

{\bf ��.} $\,\,M^{n-1}\subset\rb^n$ with induced Riemannian metric
�� ���� ���� $M$�� orientable�� ���, positive orthonormal frame
$e_2,\cdots,e_n$�� unit normal $N$�� $(N,e_2,\cdots,e_n)$��
$\rb^n$�� positive orthonormal frame�� �ǵ��� ��´�. �׸���
$e_1=N$�̶� �ΰ� $\{e_1,\cdots,e_n\}$�� dual basis
$\{\eps_1,\cdots,\eps_n\}$�� ���� $\nu$�� $M$�� Riemannian
oriented volume form�̶� ����. �׷���
$\nu=\eps_2\we\cdots\we\eps_n=i_N(\eps_1\we\cdots\we\eps_n)$ �̰�

($\,\,\because
\,\,\,i_N(\eps_1\we\cdots\we\eps_n)=i_N(\eps_1)\we\eps_2\we\cdots\we\eps_n-\eps_1\we
i_N(\eps_2)\we\cdots\we\eps_n+\cdots$ ���� ù���� ������ ��������
��� 0�� �ȴ�.) ���� $(\frac{\pa}{\pa x_1},\cdots,\frac{\pa}{\pa
x_n})$�� $\rb^n$�� orthonormal basis�̹Ƿ� $\eps_1\we\cdots\we
\eps_n=dx_1\we \cdots\we dx_n$���� �� �� �ִ�. ����

$\nu=\eps_2\we\cdots\we\eps_n=i_N(\eps_1\we\cdots\we\eps_n)=i_N(dx_1\we\cdots\we
dx_n )$.

$\therefore\,\,|\nu|=|\sum(-1)^{i-1}a_i
dx_1\we\cdots\we\overset{\we}{dx_i}\we\cdots\we dx_n
|\hs{1em}for\,\,\,N=(a_1,\cdots,a_n)$
\\

$M$�� orientable�� �ƴ� ��� ���� orientable�� ��츦 locally(in
fact pointwise)
�����ϸ� $|\nu|$�� orientation�� ������� �� ���ǵȴ�.\\

{\bf ��.} $M=\sbb^{n-1}_r(0)=$ sphere of radius $r$ �� ���
$N=\frac{1}{r}(x_1,\cdots,x_n)$�̰� ���� ���� Riemannian volume
form�� ������ ����.

$\nu=\frac{1}{r}\sum(-1)^{i-1}x_idx_i\we\cdots\we\overset{\we}{dx_i}\we\cdots\we
dx_n$.\\

$\rb^n$���� �߿��ϰ� ���̴� �Ǵٸ� form���δ� solid angle form
$"d\theta"$�� �ִ�. (��������� $d\theta$�� ����
$\rb^n\backslash\{0\}$�󿡼� exact form�� �ƴϴ�.)


$\dis{d\theta:=\frac{1}{r^{n-1}}\nu=\frac{1}{r^{n-1}}\sum(-1)^{i-1}x_i
dx_1\we\cdots\we
\overset{\we}{dx_i}\we\cdots\we dx_n}$\\

{\bf Exercise.} $d\theta$ �� closed���� ������.\\\\

{\bf Exercise.}(See. ��ȫ��,������, "�̺дپ�ü�� �Թ�")\\

1. (1) Let $\mu$ be an $m$-form on $M^m$ and $\nu$ be an $n$-form
on $N^n$.

Let $"\mu\we\nu"=\pi_1^*\mu\we\pi^*_2\nu$ where $\pi_i$ is the
canonical projection of $M\times N$ onto $M$ or $N$.
Then\\

$\hs{3em}\dis{\int_{M\times
N}f(x,y)\mu\we\nu=\int_M(\int_Nf(x,y)\nu)\mu}$.\\


(2) For $f\in \ccn_c(\rb^n) $, $\dis{\int_{\rb^n}fdx_1\we
\cdots\we
dx_n=\int_{\sbb^{n-1}}(\int_0^{\infty}f(r,\theta)r^{n-1}dr
)d\theta}$\\

2. Volume of $\mathbb{B}^n$ and $\sbb^n$:\\

$\hs{3em}$vol($\sbb^{n-1}$)=$\dis{\frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}}
\hs{1.5em}\hs{1em}where\,\,\,\Gamma(s)=\dis{\int_0^{\infty}e^{-t}t^{s-1}dt}\,\,\,$ and \\



$\hs{3em}$vol$(\mathbb{B}^n)=\dis{\frac{1}{n}}$vol$(\sbb^{n-1})\,\,\,$.\\

(proof of 2)
$\dis{\int_{\rb^n}e^{-r^2}=(\int_{\rb}e^{-x^2}dx)^n}=(\sqrt{\pi})^n$ ����\\

�� ���� �º��� ���������� ǥ���ϸ�\\

$\dis{\int_{\sbb^{n-1}}\int_0^{\infty}e^{-r^2}r^{n-1}drd\theta}=$vol$\dis{(\sbb^{n-1})\int_0^{\infty}e^{-r^2}r^{n-1}dr}\hs{3em}$


$\hs{9.2em}$=vol$(\sbb^{n-1})\dis{\int_0^{\infty}e^{-t}t^{\frac{n}{2}-1}\frac{1}{2}dt}$
$\hs{2em}$(Let $r^2=t$ then $2rdr=dt)$

$\hs{9.2em}=$vol$\dis{(\sbb^{n-1})\frac{1}{2}\Gamma(\frac{n}{2})}$\\

$\therefore\,\,\,\dis{(\sqrt{\pi})^n}=$vol$\dis{(\sbb^{n-1})\frac{1}{2}\Gamma(\frac{n}{2})}$.

vol$(\mathbb{B}^n)=\dis{\frac{1}{n}}$vol$(\sbb^{n-1})$ �� Stokes
������ ���� ���� �� �ִ�.



\end{document}
