\documentclass[b5paper,10pt]{jarticle}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\begin{document}
\begin{math}
~~~~ \displaystyle \\[0.2cm]
%\genfrac{}{}{0.5pt}{}{(x-1)(2x-3)}{x}\, dx\\[0.2cm]
B.433~~~z~軸を軸とすつ半径~1~の円柱の側面で、xy~平面上より上\\[0.2cm]
（z~軸の正の方向）にあり、平面~x-\sqrt 3 y +z=1~より下にある部分\\[0.2cm]
を~D~とする。D~の面積~S~を求めよ。\\[0.2cm]
解1~~~y~軸を縦軸と考え、z-x~平面に切り口の射影（積分範囲）を作り、\\[0.2cm]
面分の公式を用いて律義に求める。\\[0.2cm]
~~~~y=\genfrac{}{}{0.5pt}{}{x+z-1}{\sqrt 3}~~...~~(1)\\[0.2cm]
~~~~x^2+y^2=1~~~...~~(2)\\[0.2cm]
(2)~を~(1)~に代入。\\[0.2cm]
~~~~~x^2+\left(\genfrac{}{}{0.5pt}{}{x+z-1}{\sqrt 3}\right)^2=1\\[0.2cm]
~~~z^2+2(x-1)z+(4x^2-2x-2)=0\\[0.2cm]
~~~z=-x+1±\sqrt{3(1-x^2)}\\[0.2cm]
これは直線~z=-x+1~に楕円~z=±\sqrt{3(1-x^2)}~を足したグラフ。\\[0.2cm]
グラフの存在する範囲は~-1\leq x\leq 1~~~（点(-1,2)~と点(1,0)を通る。）\\[0.2cm]
また、~z=0~の時、x=1~or~-\genfrac{}{}{0.5pt}{}{1}{2}~~~~
\left(点(-\genfrac{}{}{0.5pt}{}{1}{2},0)を通る。\right)\\[0.2cm]
即ち、点(-1,2)~から点(1,0)~までを繋ぐ曲線~z=-x+1+\sqrt{3(1-x^2)}~\\[0.2cm]
は、y~軸の正の方向に存在する円柱の側面~S_1~に対応する積分範囲の限界を表し、\\[0.2cm]
点(-1,2)~から点(-\genfrac{}{}{0.5pt}{}{1}{2},0)~までを繋ぐ
曲線~z=-x+1-\sqrt{3(1-x^2)}~\\[0.2cm]
は、y~軸の負の方向に存在する円柱の側面~S_2~に対応する積分範囲の限界を表す。\\[0.2cm]
さて、面分~dS=\sqrt{\left(\genfrac{}{}{0.5pt}{}{\partial y}{\partial x}\right)^2
+\left(\genfrac{}{}{0.5pt}{}{\partial y}{\partial z}\right)^2
+1}~dzdxを求める。\\[0.2cm]
\genfrac{}{}{0.5pt}{}{\partial y}{\partial x}
=-\genfrac{}{}{0.5pt}{}{x}{\sqrt{1-x^2}},~~
\genfrac{}{}{0.5pt}{}{\partial y}{\partial z}=0~~であるから、\\[0.2cm]
S_1=\int _{-1}^1\left[\int _0^{-x+1+\sqrt{3(1-x^2)}}
\genfrac{}{}{0.5pt}{}{dz}{\sqrt{1-x^2}}\right]\,dx\\[0.2cm]
=\int _{-1}^1 \genfrac{}{}{0.5pt}{}{1}{\sqrt{1-x^2}}
\left[-x+1+\sqrt{3(1-x^2)} \right]dx\\[0.2cm]
=\int _{-1}^1 \left(\genfrac{}{}{0.5pt}{}{1}{\sqrt{1-x^2}}
-\genfrac{}{}{0.5pt}{}{x}{\sqrt{1-x^2}}
+\sqrt 3\right)\,dx\\[0.2cm]
=\left[\sin^{-1} x+\sqrt{1-x^2}+\sqrt 3 x\right]_{-1}^1\\[0.2cm]
=\pi +2\sqrt 3\\[0.2cm]
S_2=\int _{-1}^{-1/2}\left[\int _0^{-x+1-\sqrt{3(1-x^2)}}
\genfrac{}{}{0.5pt}{}{dz}{\sqrt{1-x^2}}\right]\,dx\\[0.2cm]
=\int _{-1}^{-1/2} \genfrac{}{}{0.5pt}{}{1}{\sqrt{1-x^2}}
\left[-x+1-\sqrt{3(1-x^2)} \right]dx\\[0.2cm]
=\int _{-1}^{-1/2} \left(\genfrac{}{}{0.5pt}{}{1}{\sqrt{1-x^2}}
-\genfrac{}{}{0.5pt}{}{x}{\sqrt{1-x^2}}
-\sqrt 3\right)\,dx\\[0.2cm]
=\left[\sin^{-1} x+\sqrt{1-x^2}-\sqrt 3 x\right]_{-1}^{-1/2}\\[0.2cm]
=-\genfrac{}{}{0.5pt}{}{\pi}{6}+\genfrac{}{}{0.5pt}{}{\pi}{2}
+\genfrac{}{}{0.5pt}{}{\sqrt 3}{2}-\genfrac{}{}{0.5pt}{}{\sqrt 3}{2}
=\genfrac{}{}{0.5pt}{}{\pi}{3}\\[0.2cm]
~~~S=S_1+S_2=\genfrac{}{}{0.5pt}{}{4\pi}{3}+2\sqrt 3~~~（答）\\[0.2cm]
解2~~~展開図を描くことにより求める方法。\\[0.2cm]
~~~~~~~~x=\cos \theta,~~y=\sin \theta ~~~
(0 \leq \theta \leq 2\pi)~~~...~~(1)\\[0.2cm]
~~~~~~~~~~~0 \leq z \leq 1-x+\sqrt 3y ~~~...~~(2)\\[0.2cm]
~~~(1)~を~(2)~に代入して、\\[0.2cm]
~~~~0 \leq z \leq 1-\cos \theta+\sqrt 3 \sin \theta
=1-2\cos \left(\theta +\genfrac{}{}{0.5pt}{}{\pi}{3}\right)~~~...~~(3)\\[0.2cm]
~~~(3)~の右辺と~z=0~との交点を求めて、展開図の存在する範囲を求める。\\[0.2cm]
~~~\cos \left(\theta +\genfrac{}{}{0.5pt}{}{\pi}{3}\right)
=\genfrac{}{}{0.5pt}{}{1}{2}\\[0.2cm]
~~~\theta=0~or~\genfrac{}{}{0.5pt}{}{4\pi}{3}\\[0.2cm]
S=\int _0^{4\pi/3} \left[1-2\cos \left(\theta+
\genfrac{}{}{0.5pt}{}{\pi}{3} \right)\right]\,d\theta\\[0.2cm]
=\left[\theta-2\sin \left(\theta+
\genfrac{}{}{0.5pt}{}{\pi}{3} \right)\right] _0^{4\pi/3}=
\genfrac{}{}{0.5pt}{}{4\pi}{3}+2\sqrt 3~~~（答）\\[0.2cm]
\end{math}
\end{document}