> $x\geqq 0,y\geqq 0,z\geqq 0$, 平面$z=x+y$および柱面$x^2+y^2=2^2$で
> 囲まれた立体を描きたいのですが，上手い方法はないでしょうか？

Perl と for 文を用いたゴリ押しです。
x, y を非負で制限するときは \開始角のコメントアウトを入れ替えてください。

\documentclass[a4j]{jarticle}%
\usepackage{emathP}%
\usepackage{emathPp}%
\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%
\def\分割数{40}
%\def\開始角{0}
\def\開始角{-$pi/4}%<= 昔、基礎解析でよく見かけた立体
%%%%%%%%%%%%%%%%%%%%%%%%
\calcval{(($pi/4-(\開始角)))/\分割数}\Step%
\calcval{$pi/2-(\開始角)}\終了角%
\rtenretu*{Ex(1,-15);Ey(1,45);Ez(1,90)}%
\unitlength=8mm%
\begin{Zahyou*}[\Ex][\Ey][\Ez](0,4)(0,4)(0,5)
\iiitenretu*{O(0,0,0);A(2,0,2);B(0,2,2);Ax(2,0,0);Bx(0,2,0)}
\def\Cx{2*cos(X)}
\def\Sx{2*sin(X)}
\Ifor\ICnt{0}\分割数\Do{%
\calcval{($pi/4)+(\ICnt)*(\Step)}\X%
\funcval\Cx\X\Px%
\funcval\Sx\X\Py%
\calcval{(\Px)+(\Py)}\Pz%
\def\PL{(\Px,\Py,0)}%
\def\PH{(\Px,\Py,\Pz)}%
\iiiNuritubusi[0]{\O\PL\PH\O}%
\begin{color}{green}
\iiiDrawlines{\O\PH}%
\end{color}%
%\iiiDrawlines{\PL\PH}%
}%
\Ifor\ICnt{0}\分割数\Do{%
\calcval{($pi/4)-(\ICnt)*(\Step)}\X%
\funcval\Cx\X\Px%
\funcval\Sx\X\Py%
\calcval{(\Px)+(\Py)}\Pz%
\def\PL{(\Px,\Py,0)}%
\def\PH{(\Px,\Py,\Pz)}%
\iiiNuritubusi[0]{\O\PL\PH\O}%
\begin{color}{green}
\iiiDrawlines{\O\PH}%
\end{color}%
\iiiDrawlines{\PL\PH}%
}%
%%%%%%%%%%% お色直し %%%%%%%%%%%%%
\iiiPut\Ax[s]{2}
\iiiPut\Bx[nw]{2}
\iiiPut\O[s]{O}
\iiiArrowLine\O{(\Xmax,0,0)}%
\iiiPut{(\Xmax,0,0)}[e]{$x$}%
\iiiArrowLine\O{(0,\Ymax,0)}%
\iiiPut{(0,\Ymax,0)}[e]{$y$}%
\iiiArrowLine\O{(0,0,\Zmax)}%
\iiiPut{(0,0,\Zmax)}[e]{$z$}%
  \def\C{2*cos(T)}
  \def\S{2*sin(T)}
  \iiiBGurafu\C\S{0}{\開始角}{\終了角}
\end{Zahyou*}
\end{document}

▼関連発言
│
└◆4818:空間図形 [genio] 06/13 20:36
　└◆4819:Re:空間図形 [田中徹] 06/14 00:13
　　└◆4820:Re[2]:空間図形 [genio] 06/14 08:35
　　　└◆4821:Re[3]:空間図形 [wizard] 06/17 13:11<-last