# How to calculate area between curve and horizontal line?

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Hamzah Mahmood
on 21 Aug 2020

Commented: esat gulhan
on 23 Aug 2020

Apologies if this has already been answered in another form elsewhere on the forums, however I couldn't find any answers that helped me directly with my code. I currently have code that maps the profile and area of a channel. I wish to have a horizontal y value across my channel width to simulate the fluctuating water level, the example line is shown in the image as the green line. So for example I would want to calculate and shade the area underneath the greenline only.

With the function would I be able to loop the same function for a range of values for the horizontal values, to provide an output I could then plot? A vector form would be ideal if possible.

I've already tried to use the area and fill commands unsuccesfully, or maybe I was not using the correct format for them.

I've also attached the relevant code below, any help would be much appreciated.

x = [ 0 1 2 3 4 5 6 7 8 9 10 11 12];

y = -[0 -0.5 -0.8 -0.8 -1 -1.1 -1.2 -1.2 -1.4 -1.2 -1.1 -1 0]; % y values in image were produced by rand function

subplot(2, 1, 1);

area(x,-y)

grid on;

hold on;

for k = 1 : length(x)

xline(x(k), 'Color', 'k');

end

yt = yticks;

for k = 1 : length(yt)

yline(yt(k), 'Color', 'k');

end

plot (x, -y, 'r.-', 'LineWidth', 3, 'MarkerSize', 30)

hold on

yline(-0.5,'g',11);

xlabel('Chainage (m)', 'FontSize', 11);

ylabel('Depth (m)', 'FontSize', 11);

An = (trapz(x,y)); % Overall area.

% Compute areas within each pair of x locations.

midpoints = (y(1:end-1) + y(2:end))/2;

deltax = diff(x);

Ani = deltax .* midpoints;

subplot(2, 1, 2);

bar(x(1:end-1)-x(1)+ 0.5, Ani);

title('Areas of Slices', 'FontSize', 14);

xlabel('x', 'FontSize', 11);

ylabel('Area (m^2)', 'FontSize', 11);

grid on;

##### 0 Comments

### Accepted Answer

esat gulhan
on 21 Aug 2020

clc;clear;

x = [ 0 1 2 3 4 5 6 7 8 9 10 11 12];

y = [0 -0.5 -0.8 -0.8 -1 -1.1 -1.2 -1.2 -1.4 -1.2 -1.1 -1 0];

h0= -0.5; %the axis in the graph

h=ones(size(x))*h0;

Int=pchip(x,y);

xx=linspace(x(1),x(end),500);

yy=y-h0;

yyy=pchip(x,yy,xx);

gt=yyy<0;

theArea =-trapz(xx(gt), (yyy(gt)))

hold off

area(xx, min(ppval(Int,xx), h(1)), h(1), 'EdgeColor', 'none', 'FaceColor', 'b'),grid,

hold on

plot(xx,ppval(Int,xx),'k',xx,h0,'Linewidth',2)

set(gca, 'XLim', [x(1) x(end)], 'YLim', [min(y) 0]);

title(strcat('BlueArea=', num2str(theArea)))

xlabel('Chainage (m)', 'FontSize', 11);

ylabel('Depth (m)', 'FontSize', 11);

You can change h0 which is horizontal line. when h0=-1

when h0=-0.5

when h0=0

I RESHAPED YOUR X AND Y DATAS 500 INTERVALS WITH PCHIP INTERPOLATION, IF NOT THE DATA IS NOT ENOUGH FOR GOOD SOLUTION.

##### 4 Comments

esat gulhan
on 23 Aug 2020

### More Answers (2)

esat gulhan
on 21 Aug 2020

Edited: esat gulhan
on 21 Aug 2020

x = [ 0 1 2 3 4 5 6 7 8 9 10 11 12]

y = -[0 -0.5 -0.8 -0.8 -1 -1.1 -1.2 -1.2 -1.4 -1.2 -1.1 -1 0]

Int=pchip(x,y)

plot(x,ppval(Int,x))

MIItemp = (fnint(Int))

Area = fnval(MIItemp,12)%Area between 0 12

If you want to find area between 0-5 , Area = fnval(MIItemp,5)

There are other ways but this is the easiest way .

Note:If your matlab is latetest version you can write makima(x,y),instead of phcip(x,y). If your lines are strict use makima

esat gulhan
on 22 Aug 2020

Did you see my second answer. I think it helps.

If you add pause to the code it will seem like an animation.

clc;clear;

x = [ 0 1 2 3 4 5 6 7 8 9 10 11 12];

y = [0 -0.5 -0.8 -0.8 -1 -1.1 -1.2 -1.2 -1.4 -1.2 -1.1 -1 0];

for h0= 0:-0.04:-2; %the axis in the graph

h=ones(size(x))*h0;Int=pchip(x,y);xx=linspace(x(1),x(end),500);

yy=y-h0;yyy=pchip(x,yy,xx);gt=yyy<0;theArea =-trapz(xx(gt), (yyy(gt)))

hold off

area(xx, min(ppval(Int,xx), h(1)), h(1), 'EdgeColor', 'none', 'FaceColor', 'b'),grid,

hold on

plot(xx,ppval(Int,xx),'k',xx,h0,'Linewidth',2)

set(gca, 'XLim', [x(1) x(end)], 'YLim', [min(y) 0]);

title(strcat('BlueArea=', num2str(theArea)))

xlabel('Chainage (m)', 'FontSize', 11);ylabel('Depth (m)', 'FontSize', 11);pause(0.00001)

end

##### 0 Comments

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