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⟦d4e45d5a0⟧ TextFile

    Length: 1817 (0x719)
    Types: TextFile
    Names: »rado.tex«

Derivation

└─⟦060c9c824⟧ Bits:30007080 DKUUG TeX 2/12/89
    └─⟦this⟧ »./DVIware/lpr-viewers/crudetype/samples/rado.tex« 
└─⟦52210d11f⟧ Bits:30007239 EUUGD2: TeX 3 1992-12
    └─⟦af5ba6c8e⟧ »unix3.0/DVIWARE.tar.Z« 
        └─⟦ca79c7339⟧ 
            └─⟦this⟧ »DVIware/lpr-viewers/crudetype/samples/rado.tex« 

TextFile

%This is for the maths printer
\magnification = \magstephalf
\vsize=10 truein \hsize=170truemm
\voffset=0truein \hoffset=0truecm

\parskip = 1ex plus .5ex minus .1ex
\parindent 3em
\nopagenumbers

\font\Byg = cmr10 at 17.28 truept
\font\byg = cmr10 at 14.40 truept
\font\menu = cmti10 at 10.00 truept
\font\small = cmr10 at 10.00 truept
\font\bold = cmb10
\font\bcaps = cmssbx10
\font\pfont = cmu10
\font\tiny = cmr6 at 6 truept
\def\pound{{\pfont \$}}

\centerline{{\byg MA21}}
\vskip .75 truein
\centerline{{\bold Sheet Ten}}
\vskip 0.2 truein
{\menu (To be handed in to the box outside McCrea 232 on
the morning of the first lecturing day of next Term i.e.{\bold 
Wednesday 13th January}).}
\hfill\break
\vfill
\noindent
(1) \quad Evaluate
$$\int\limits_{y=0}^1dy\,\int\limits_{x=3y}^3 e^{x^2}dx.$$
(Hint: first find the region over which the integral is taken, and then 
reverse the order of integration.)
\hfill
\bigskip
\noindent
(2) \quad Show that
$$\int\!\!\!\int_Rxy^2 dx\,dy$$
where $R$ is the smaller segment of the circle $x^2+y^2=a^2$ cut off by the
line $x+y=a$ is equal to $\displaystyle{a^5\over 20}$.
\bigskip
\noindent
(3) \quad Evaluate
$$\int\!\!\!\int_Rxy dx\,dy$$
where $R$ is the quadrant of the circle $x^2+y^2=a^2$ for which $x>0$, $y>0$.
\hfill\break
(Hint: use polar coordinates.)
\hfill
\bigskip
\noindent
(4) \quad Evaluate
$$\int\!\!\!\int_Re^{-\left(x^2+y^2\right)}dx\,dy$$
where $R$ is the infinite quadrant given by $x>0$, $y>0$, by substituting
plane polar coordinates.
\par
Let $I=\int\limits _0^\infty e^{-x^2}dx$. Evaluate $I^2$. (Hint: remember
that $\int\limits _0^\infty e^{-x^2}dx$ is equal to 
$\int\limits _0^\infty e^{-y^2}dy$, and use the result of the first part.)
\par
Hence show that
$$\int_0^\infty e^{-x^2}dx={1\over 2}\sqrt\pi.$$
\vfill
\rightline{Peter Rado}
\end