Surface of a equation

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Daniel Jaló
Daniel Jaló 2012-9-15
评论： Walter Roberson 2020-11-11
Hi.
I want to plot the quadratic surface of a sphere:
x^2 + y^2 + z^2 = r , where r is equal to 1. Therefore:
x^2 + y^2 + z^2 = 1, where x, y and z are values between -1.5 and 1.5
Can anyone explain me how to do this? I've looked into mesh surfaces but I can only plot functions ( f(x,y) )..
Any help will be highly appreciated. Thanks.

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采纳的回答

Wayne King
Wayne King 2012-9-15
编辑：Wayne King 2012-9-15
I think you meant to square r in your equation, and you cannot have a value between -1.5 and 1.5 if the radius is 1. The radius has to be 1.5. Think about what happens if x=0,y=0,z=1.5 as you stated must be a point that satisfies the equation x^2+y^2+z^2 = r^2
You should probably do it with spherical coordinates:
n = 100;
r = 1.5;
theta = (-n:2:n)/n*pi;
phi = (-n:2:n)'/n*pi/2;
cosphi = cos(phi); cosphi(1) = 0; cosphi(n+1) = 0;
sintheta = sin(theta); sintheta(1) = 0; sintheta(n+1) = 0;
x = r*cosphi*cos(theta);
y = r*cosphi*sintheta;
z = r*sin(phi)*ones(1,n+1);
surf(x,y,z)
xlabel('X'); ylabel('Y'); zlabel('Z')
Note that MATLAB has a function for this with a unit sphere, sphere.m
  2 个评论
Varun Dogra
Varun Dogra 2020-11-11
Develop a sweep surface by sweeping a circle P (r, θ) of radius 7.5
units with center at (4,3) along the positive z-axis by 10 units
Please help
Walter Roberson
Walter Roberson 2020-11-11
Ran in:
Modifying some code from below, https://www.mathworks.com/matlabcentral/answers/48240-surface-of-a-equation#answer_243478
r = 7.5; xc = 4; yc = 3;
zl = 0; zh = 10;
fimplicit3(@(x,y,z) (x-xc).^2+(y-yc).^2-r^2,[xc-r xc+r yc-r yc+r zl zh], 'edgecolor', 'none')

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更多回答（3 个）

Jürgen
Jürgen 2012-9-15
[x,y,z] = sphere(); r = 5; figure,surf( r*x +cx, r*y+ cy, r*z +cz) wit (cx,,cy,cz) the center look at http://www.mathworks.com/matlabcentral/newsreader/view_thread/169373
Javier
Javier 2012-9-15
编辑：Walter Roberson 2020-11-11
Procedure done in Matlab R2012.
The problem that you want to solve gives complex solution for Z for arbitrary X and Y in [-1.5,1.5]. The square of X^2 + Y.^2 must be lower than 1, in other case,the solution for Z is a complex number (mesh function doesnt support complex data). To prove it, solve for Z. You get an square root of (1-(X.^2+Y.^2)). I show how to solve for arbitrary number X and Y lower than 0.70 (0.7071^2= 0.5).
Data=randn(10,10) % 10 is arbitrary. Matriz square.
u=find(Data>0.70);
d=find(Data<-0.70);
%Define limits of Data Matriz
Data(u)=0.70;
Data(d)=-0.70;
%Divide Data matriz in two
X=Data(:,1:5);
Y=Data(:,6:10);
%For arbitrary X and Y value Z must solve the equality
Z=feval(@(X,Y)[sqrt(1-(X.^2+Y.^2))],X,Y)
%Plot data
mesh(X,Y,Z)
If this solve your question please grade or make a comment to this answer. Best regards
Javier
auto2060
auto2060 2016-11-16
There is a new function in R2016b: fimplicit3
So
r = 1;
fimplicit3(@(x,y,z) x.^2+y.^2+z.^2-r^2,[-1.5 1.5 -1.5 1.5 -1.5 1.5])

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