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\begin{document}

\title{微分積分4 第2回課題解説}
\date{}
\maketitle

\section{課題}

\begin{itemize}
\item[1)] 次の三重積分を求めよ（a)b) 30点、c) 40点）。
\begin{itemize}
\item[a)] $\int_{x, y, z\geq 0, x+y+z\leq 3} xy(3-z) dxdydz$,
\item[b)] $\int_{1\leq z\leq y\leq x\leq e} \frac{1}{xyz} dxdydz$,
\item[c)] $\int_{x^2+y^2+z^2\leq 1} zdxdydz$.
\end{itemize}
\end{itemize}

\section{解}

\begin{itemize}
\item[1a)]
\[\begin{split}
\iiint\limits_{x, y, z\geq 0, x+y+z\leq 3} xy(3-z) & dzdydx
=\iint\limits_{x, y\geq 0, x+y\leq 3} xy \left(\int_0^{3-(x+y)}(3-z)dz\right) dydx \\
=& \iint\limits_{x, y\geq 0, x+y\leq 3} xy \frac{9-(x+y)^2}{2} dydx \\
=& \int_0^3 \int_0^{3-x} \frac{9xy-x^3 y-2x^2y^2-xy^3}{2} dydx \\
=& \int_0^3 \frac{9x(3-x)^2-x^3(3-x)^2}{4}-\frac{x^2(3-x)^3}{3}-\frac{x(3-x)^4}{8} dx
\end{split}\]
ここで
\[\begin{split}
& \frac{9x(3-x)^2-x^3(3-x)^2}{4}-\frac{x^2(3-x)^3}{3}-\frac{x(3-x)^4}{8} \\
=& \frac{9x^3-54x^2+81x -x^5+6x^4-9x^3}{4}+\frac{x^5-9x^4+27x^3-27x^2}{3} \\
& -\frac{x^5-12x^4+54x^3-108x^2+81x}{8} \\
=& -\frac{x^5}{24}+\frac{9}{4}x^3-9x^2+\frac{81}{8}x
\end{split}\]
なので
上記積分は
\[-\frac{3^6}{144}+\frac{9}{4}\times \frac{3^4}{4}-3^4+\frac{81}{8}\times\frac{3^2}{2}
=-\frac{81}{16}+\frac{729}{16}-81+\frac{729}{16}=\frac{81}{16}.
\]
あるいは
\[\begin{split}
\iiint\limits_{x, y, z\geq 0, x+y+z\leq 3} xy(3-z)& dxdydz \\
=& \iint\limits_{y, z\geq 0, y+z\leq 3-x} \frac{(3-y-z)^2y(3-z)}{2} dydz \\
=& \iint\limits_{y, z\geq 0, y+z\leq 3-x} \frac{(y(3-z)^3-2y^2(3-z)^2+y^3(3-z))}{2} dydz \\
=& \int_0^3 \frac{(3-z)^5}{4}-\frac{(3-z)^5}{3}+\frac{(3-z)^5}{8} dz \\
=& \int_0^3 \frac{(3-z)^5}{24} dz \\
=& \frac{3^6}{144}=\frac{81}{16}.
\end{split}\]
\item[1b)]
\[\begin{split}
& \iiint_{1\leq z\leq y\leq x\leq e} \frac{1}{xyz} dxdydz
=\iint_{1\leq y\leq x\leq e} \frac{1}{xy}\left(\int_1^y\frac{dz}{z}\right) dxdy \\
=& \iint_{1\leq y\leq x\leq e}\frac{\log y}{xy} dxdy
=\int_1^e \frac{1}{x}\left(\int_1^x\frac{\log y}{y} dy\right) dx \\
=& \int_1^e \frac{\log^2 x}{2x} dx
=\left[\frac{\log^3 x}{6x}\right]_1^e \\
=& \frac{1}{6}.
\end{split}\]
\item[1c)] 
\[
\iiint_{x^2+y^2+z^2\leq 1} zdxdydz
=\iint_{x^2+y^2\leq 1} \left(\int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}} z dz\right) dxdy
\]
だが
\[\int_{-a}^a z dz=\frac{a^2}{2}-\frac{a^2}{2}=0\]
だから
\[
\iiint_{x^2+y^2+z^2\leq 1} zdxdydz
=\iint_{x^2+y^2\leq 1} 0 dxdy=0.
\]
x, y, z は負の値もとることに注意する。

おまけ問題: $\iiint_{x^2+y^2\leq 1, 0\leq z\leq \sqrt{1-x^2+y^2}} zdxdydz$ を求めよ。
\end{itemize}


\end{document}
