[交流] 求助想用mathematica画一个缺一部分的实心球体 已有4人参与

clc
clear
close all
%两个球面
r1=1;
r2=2;
t=linspace(0,0.5*pi,25);
p=linspace(0,1.5*pi,25);
[theta,phi]=meshgrid(t,p);
x1=r1*sin(theta).*sin(phi);
y1=r1*sin(theta).*cos(phi);
z1=r1*cos(theta);
x2=r2*sin(theta).*sin(phi);
y2=r2*sin(theta).*cos(phi);
z2=r2*cos(theta); %
[m,n]=size(z1);
re=[0 0 1];%改数字 改变颜色、数字为0-1
colormap(re)
surf(x1,y1,z1);
hold on
surf(x2,y2,z2);
%底面
R=linspace(0,r2,25);
[r,ang]=meshgrid(R,p);
x3=r.*sin(ang);
y3=r.*cos(ang);
z3=zeros(25,25);
surf(x3,y3,z3);
%侧面
R1=linspace(0,r2,25);
[r,ang]=meshgrid(R1,t);
y4=r.*sin(ang);
z4=r.*cos(ang);
x4=zeros(25,25);
surf(x4,y4,z4);
R2=linspace(0,r2,25);
[r,ang]=meshgrid(R2,-t);
x5=r.*sin(ang);
z5=r.*cos(ang);
y5=zeros(25,25);
surf(x5,y5,z5);
shading interp
axis equal;
daspect([1,1,1])
view(3);
axis tight
camlight
grid on
lighting gouraud

3楼: Originally posted by zhangdegen at 2017-06-28 15:49:36
RegionPlot3D

Untitled-3.jpg

2 > Sqrt[x^2 + y^2 + z^2] > 1 && (3 \[Pi])/2 > ArcTan[z, Sqrt[x^2 + y^2]] > 0 && \[Pi]/2 > ArcTan[x, y] >  0

2楼: Originally posted by Yan_Jordan at 2017-06-28 15:13:35
我试着画了下，发现直接三维好像挺麻烦的，于是用了拼接面的方法，你可以参考下！
基本是画两个球面，一个底面加两个侧面，代码如下
clc
clear
close all
%两个球面
r1=1;
r2=2;
t=linspace(0,0.5*pi ...