Is it possible to change solid fill to lines for patch and set limits to patch?

I am trying to fill the area between the z line and v line from from 0 until when they cross at 0.725 and cannot get the function to stop filling the area at that point. Does this require a loop or is there an easier way to do this? I also was wondering if it is possible to change the fill from being a solid fill to just a line pattern going diagonally through the area or if it has to be solid.

clear all
close all
clc

% V ideal
y = 0:10:35000;
x = 0:.01:0.9;
y = 14490 .* log( 1 ./ (1 - x));

% n max
z = 0:1:35000;
b = 0:0.01:0.9;
z = 14490 .* (log( 1 ./ (1 - b)) - ( 0.25 ) .* ( b ./ (1-b)));

% n = 1.1
v = 0:1:35000;
b = 0:0.01:0.9;
v = 14490 .* (log( 1 ./ (1 - b)) - (( 0.9091 ) .* b));

hold on
% All Functions Plotted
plot(x,y,'g');
plot(b,z,'r');
plot(b,v);

% Fill Area
patch([b fliplr(b)],[z fliplr(v)],'b')

% Labeling the Plot
title('\DeltaV vs. m_p/m_0')
xlabel('m_p/m_0')
ylabel('\DeltaV')
label1 = '\DeltaV Ideal';
label2 = '\eta_{max} = 4';
label3 = '\eta = 1.1';
legend(label1,label2,label3,'interpreter','tex')
hold off

Answer 1

[DGM]

You should be able to just limit the range used for drawing the patch object

% V ideal
y = 0:10:35000;
x = 0:.01:0.9;
y = 14490 .* log( 1 ./ (1 - x));

% n max
z = 0:1:35000;
b = 0:0.01:0.9;
z = 14490 .* (log( 1 ./ (1 - b)) - ( 0.25 ) .* ( b ./ (1-b)));

% n = 1.1
v = 0:1:35000;
b = 0:0.01:0.9;
v = 14490 .* (log( 1 ./ (1 - b)) - (( 0.9091 ) .* b));

% All Functions Plotted
hold on
plot(x,y,'g');
plot(b,z,'r');
plot(b,v);

% Fill Area
idx = find(b==0.72);
patch([b(1:idx) fliplr(b(1:idx))],[z(1:idx) fliplr(v(1:idx))],'b')

I haven't tried it, but this FEX submission appears to do hatch fills for patch objects:

https://www.mathworks.com/matlabcentral/fileexchange/53593-hatchfill2

[Student]

When using the code you provided, it does not shade anything. Any idea?

[DGM]

I chose the closest defined x-value. b has no value at 0.725. If you want to make sure the patch extends to the intersection, you'll have to insert an extra point (or make sure b and z actually have a point defined at that location).

% Fill Area
idx = find(b==0.72); % the closest x-value not beyond the intersection
xint = 0.724877; % this is still approximate
yint = interp1(b,z,xint);
patch([b(1:idx) xint fliplr(b(1:idx))],[z(1:idx) yint fliplr(v(1:idx))],'b')

I didn't expect the small gap to be an issue if you were to use a hatch fill anyway.