3D plot in polar coordinates

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);

Answer 1

[Star Strider]

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.

.

[Hexe]

Dear Star Strider!

Thanks a lot for your help. This looks cool and similar to surfaces I obtained using pol2cart. But I also wanted to ask - is it possible to make a 3D plot (R,a,a) with fixed t (for example, t=0.1) so that we get a pattern in polar coordinates in the form of volumetric drops?

Thank you.

[Star Strider]

I do not understand.

Perhaps:

[x,y,z] = pol2cart(a, R, a);

would work, with a fixed value for ‘t’.

[Hexe]

I tried to do it, but this is not what I mean. I ask about possibility to draw the 3D plot of the attached 2D at fixed time, like k-axis rotation. Maybe it can be done trhrough polar3D?

[Star Strider]

I am lost. I have no idea what you want to do.

I will delete my answers in a few minutes.

[Hexe]

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.

[Star Strider]

As always, my pleasure!