3D plot in polar coordinates

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Hello! I made a code for solving the integral and it looks realistic in polar coordinates. But, how to present it in 3D as a figure in volume? Perhaps I should add an aditional rotation angle for this. Maybe it is easy for those who know ow to do it. If somebody knows, please, help me. Thank you.
s = 3;
n = 1;
t = 0.1;
r = 1;
a = 0:1:360;
a = a*pi/180;
b = sqrt(2*n*t);
L = sqrt((4*t+r^2)/3);
fun = @(k,u,c,a) ((k.^2).*exp(-1.5*k.^2)).*((u.^2).*(1-u.^2).*exp(-(b.*u).^2).*(cos(s.*k.*u.*cos(a)/L))).*(((cos(c)).^2).*(cos(s.*k.*sqrt(1-u.^2).*sin(a).*(cos(c)+sin(c))/(L*sqrt(2)))));
f3 = arrayfun(@(a)integral3(@(k,u,c)fun(k,u,c,a),0,Inf,-1,1,0,2*pi),a);
B = ((6*sqrt(6)*b^3)/(erf(b)*pi^2))*(1-(3/(2*b^2))*(1-((2*b*exp(-b^2))/(erf(b)*sqrt(pi))))).^(-1);
R = B*f3;
polar(a,R);

采纳的回答

Star Strider
Star Strider 2023-5-22
This took a while to get working correctly, and takes about 500 seconds to run, so I will post the code here slthough not the plot —
s = 3;
n = 1;
tv = 0:0.1:2;%3;
r = 1;
a = 0:2:360;
a = deg2rad(a);
tic
for k = 1:numel(tv)
t = tv(k)
b = sqrt(2*n*t);
L = sqrt((4*t+r^2)/3);
fun = @(k,u,c,a) ((k.^2).*exp(-1.5*k.^2)).*((u.^2).*(1-u.^2).*exp(-(b.*u).^2).*(cos(s.*k.*u.*cos(a)/L))).*(((cos(c)).^2).*(cos(s.*k.*sqrt(1-u.^2).*sin(a).*(cos(c)+sin(c))/(L*sqrt(2)))));
f3 = arrayfun(@(a)integral3(@(k,u,c)fun(k,u,c,a),0,Inf,-1,1,0,2*pi),a);
B = ((6*sqrt(6)*b^3)/(erf(b)*pi^2))*(1-(3/(2*b^2))*(1-((2*b*exp(-b^2))/(erf(b)*sqrt(pi))))).^(-1);
R = B*f3;
ta = t*ones(size(a));
[x,y,z] = pol2cart(a, R, ta);
X(k,:) = x;
Y(k,:) = y;
Z(k,:) = z;
toc % Total Time To This Point
LIT = toc/k % Mean Loop Iteration Time
end
toc
figure
surf(X, Y, Z)
colormap(turbo)
axis('equal')
axis('off')
hold on
xc = [0:0.25*r:r].'*cos(a);
yc = [0:0.25*r:r].'*sin(a);
xr = [0;r]*cos(0:pi/4:2*pi);
yr = [0;r]*sin(0:pi/4:2*pi);
plot3(xc.', yc.', zeros(size(xc)), '-k')
plot3(xr, yr, zeros(size(xr)), '-k')
zt = (0:1:max(tv)).';
zt1 = ones(size(zt));
surf(zt1*xc(end,:), zt1*yc(end,:), zt*ones(size(a)), 'FaceAlpha',0, 'MeshStyle','row')
hold off
text(xr(2,1:end-1)*1.1,yr(2,1:end-1)*1.1, zeros(1,size(xr,2)-1), compose('%3d°',(0:45:315)), 'Horiz','center','Vert','middle')
text(ones(size(zt))*xc(end,end-10), ones(size(zt))*yc(end,end-10), zt, compose('%.1f',zt))
toc
figure
surf(X, Y, Z, 'EdgeColor','none')
colormap(turbo)
axis('equal')
axis('off')
hold on
xc = [0:0.25*r:r].'*cos(a);
yc = [0:0.25*r:r].'*sin(a);
xr = [0;r]*cos(0:pi/4:2*pi);
yr = [0;r]*sin(0:pi/4:2*pi);
plot3(xc.', yc.', zeros(size(xc)), '-k')
plot3(xr, yr, zeros(size(xr)), '-k')
zt = (0:1:max(tv)).';
zt1 = ones(size(zt));
surf(zt1*xc(end,:), zt1*yc(end,:), zt*ones(size(a)), 'FaceAlpha',0, 'MeshStyle','row')
hold off
text(xr(2,1:end-1)*1.1,yr(2,1:end-1)*1.1, zeros(1,size(xr,2)-1), compose('%3d°',(0:45:315)), 'Horiz','center','Vert','middle')
text(ones(size(zt))*xc(end,end-10), ones(size(zt))*yc(end,end-10), zt, compose('%.1f',zt))
toc
TTmin = toc/60
% END
I only run it here from 0 to 2, because there is not much detail beyond that. The time dimension is the ‘Z’ dimension. Most of the detail is between 0 and 0.5.
.
  6 个评论
Hexe
Hexe 2023-5-23
编辑:Hexe 2023-5-23
Dear Star Strider.
Your previous answer is good and acceptable. I just wanted to clarify about another way of presentation the result, but probably it is not possible.
Thank you a lot.

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