Finding radii of a droplet for each y

Imagine that a droplet is a stack of discs, each one having it's own radius, which you have to find. I have the contour of the droplet, that is array C and I have tip position (position of min y), that is i. Is it possible to write a code with the following meaning: for each y, subsctract x values of C's opposite parts?

[Image Analyst]

Yes. So is each disc a different z? Like this is a confocal microscope image or some other kind of volumetric image? And x and y are with respect to each 2-D image from the "stack"? What exactly is a "tip"? What exactly is (the badly-named) "i"? Is it

i = min(y); % Get tip position.

Are your images binary or gray scale? If gray scale, can you binarize them to produce a good binary image?

Can you provide a few images at different z levels, after reading this of course:

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[Nijat Azimzade]

Sorry for lack of details. Here you can see my code. I use online MATLAB. In attachment I give the original image of the droplet (I am not sure how to download post-processed image, but if you run this code you'll get exactly same result as me). In short, I want to get the volume of the droplet but I do not have 3D image. Idea is to imagine this droplet as a stack of discs (one on another), each with own diameter, find these discs' areas and then add them to get volume. So basically to approach ice droplet as just a sliced potato. On image these dics diameters are straight lines connecting right side and left side of a contour.

clc; clear all;
I = imread('Droplet 10m40s.png');
x=im2gray(I);
%binarization
v=im2bw(x,0.2);
v1=imfill(v,'holes'); % to fill the holes
imshow(v1)
bw= bwareaopen(v1,3000);% threshold in order to remove the undesired parts
imshow(bw);
[L n]=bwlabel(bw);
r=regionprops(L, "BoundingBox");
% rectangle('Position',r.BoundingBox, 'EdgeColor','g');
% height = r.BoundingBox(4)
% width=r.BoundingBox(3)
B=bwboundaries(bw)

C=B{1}; %first contur in the file B
hold on
plot(C(:,2),C(:,1),'b')
%this plot function must be memorized. For some reason y is first position
%and x is second. 1 and 2 always stay like this
A=min(C(:,1)) %is to take the minimum y value of contur
i=find(C(:,1)==min(C(:,1))); %it is to determine the position of minimum y in the list

Cc=C(i-10:i+10,:);  %to define tip angle with n values
hold on
plot(Cc(:,2),Cc(:,1),'r')

%we divide our angle into two lines and get equation of each line with
%polyfit. Then we find angle 1, 2 between line 1, 2 and imginary vertical under
%the tip (dividing line). Then we add angles 1 and 2
Ccc1=C(i-15:i,:);
yy1=Ccc1(:,2)
xx1=Ccc1(:,1);
p  = polyfit(xx1,yy1,1)
angle1=atan(p(1))
% pp= poly2sym(p)
Ccc2=C(i:i+15,:)
yy2=Ccc2(:,2)
xx2=Ccc2(:,1)
o = polyfit(xx2,yy2,1)
angle2=atan(o(1))
oo=poly2sym(o)
angle=-angle1+angle2
teta=rad2deg(angle)
x=max(C)-min(C)

sum(bw(:) == 0) %black
sum(bw(:) == 1) %white            %to get number of white and black pixels