绘图如何实现立体效果？

仍然不太优雅（本可以避免一些dummy的语句的）... 就当抛砖引玉了... 一定有更符合tikz-style更优雅的办法：

基本思路：

• 先画右圆(实线)
• 再画左圆把右圆的一部分盖住
• 最后画右圆(虚线)

\documentclass[border=8pt,tikz]{standalone}
\usepackage{amssymb}
\usetikzlibrary{patterns}
\begin{document}
    \begin{tikzpicture}
        \path [draw,very thick,->] node at (-.3,-.3) {$O$} (-0.5,0) -- (8,0) node[anchor=north] {$x$};
        \path [draw,very thick,->] (0,-0.5) -- (0,6.1) node[anchor=east] {$y$};
        \node at (7.5,5.5) {$\mathbb{R}^{2}$};
        \draw[very thick,blue,dashed,-] (0.5,0.5) -- (6.5,0.5) -- (6.5,5) --(0.5,5)--(0.5,0.5);
        \node at (6.2,1) {$\Lambda$};
        \node at (5.5,3) {$S$};
        \draw (5,3) circle (1.2);
        \draw[fill=white] (3,3) circle (1.5);
        \draw[dashed] (5,3) circle (1.2);
        \draw[pattern=north west lines,draw=red]  (3,3) circle (1.5);
        \node[fill=white,rounded corners,below of=3] at (2.5,4) {$B$};
        \node at (2,1.3) {$ {\color{red}\partial B=B_{0}}$};
    \end{tikzpicture}  
\end{document}

思路如下：先找到两圆的交点，再分别画右边小圆的两段弧，左边一段弧画成虚线，右边一段弧正常即可。
注：1.使用let ... in ... 操作是为了找到两段弧的开始角度与结束角度。
2.使用atan反正切函数，注意此函数求出的角的范围是(-90,90)度。
3.使用库：intersections求交点坐标,calc实现坐标计算功能。

\usepackage{amssymb}
\usetikzlibrary{patterns,intersections,calc}
\begin{document} 

    \begin{tikzpicture}
        \path [draw, very thick, ->] node at (-.3,-.3) {$O$} (-0.5,0) -- (8,0) node[anchor=north] {$x$};
        \path [draw,very thick, ->](0,-0.5) -- (0,6.1)  node[anchor=east]  {$y$};
        \node  at (7.5,5.5) {$\mathbb{R}^{2}$};
        \draw[very thick,blue,dashed,-] (0.5,0.5) -- (6.5,0.5) -- (6.5,5) --(0.5,5)--(0.5,0.5);
        \draw[pattern=north west lines,name path=circle1,draw=red] (3,3) circle (1.5);
        \node [fill=white,rounded corners,below of=3] at (2.5,4) {$B$};
        \node at (2,1.3) {$ {\color{red}\partial B=B_{0}}$};
        \path[name path=circle2] (5,3) circle (1.2);
        %左边一段弧，dashed
        \draw[dashed,name intersections={of=circle1 and circle2,name=i}]
             let \p1=($(5,3)-(i-1)$), \p2=($(5,3)-(i-2)$), \n1={atan(\y1/\x1)}, \n2={atan(\y2/\x2)}
             in (i-1) arc({\n1+180}:{\n2+180}:1.2);
        %右边一段弧
        \draw let \p1=($(5,3)-(i-1)$), \p2=($(5,3)-(i-2)$), \n1={atan(\y1/\x1)}, \n2={atan(\y2/\x2)}
                 in     (i-1) arc({\n1+180}:{\n2-180}:1.2);
        \node at (6.2,1) {$\Lambda$};
        \node at (5.5,3) {$S$};
    \end{tikzpicture}  

认为有回答符合需求，请点个采纳，谢谢！
第一步，问AI，让它计算出角度。

第二步，引用解析式结果，画图：

\documentclass[tikz,border=2pt]{standalone}
\usetikzlibrary{calc,patterns}
\usepackage{amssymb}
\begin{document}
\newcommand{\MyAngle}{{acos(319/480)-180}}% about 48.4-180=-131.6
\newcommand{\MyAngleNeg}{{180-acos(319/480)}}
\begin{tikzpicture}

    \draw [->,very thick] (-.5,0) -- (8,0) node [below] {$x$};% x轴
    \draw [->,very thick] (0,-.5) -- (0,6) node [left]  {$y$};% y轴

    \draw [blue,very thick,dashed]            (.5,.5) rectangle (6.5,5);% 蓝框
    \draw [draw=red,pattern=north west lines] (3,3)   circle    (1.5);% 左圆

    \draw [dashed] ($(5,3)+(\MyAngleNeg:1.2)$) arc (-\MyAngle:\MyAngle   :1.2);% 虚线弧
    \draw          ($(5,3)+(   \MyAngle:1.2)$) arc ( \MyAngle:\MyAngleNeg:1.2);% 实线弧
    
    \node at (0,0)    [below left]                          {$O$};
    \node at (3,3)    [left=9pt,fill=white,rounded corners] {$B$};
    \node at (5,3)    [right=9pt]                           {$S$};
    \node at (6.5,.5) [above left=3pt]                      {$\Lambda$};
    \node at (.5,.5)  [above right=15pt,text=red]           {$\partial B=B_0$};
    \node at (6.5,5)  [above right=9pt]                     {$\mathbb{R}^2$};

\end{tikzpicture}
\end{document}