I've been working on this code for some hours and it seems I'm doing something wrong which I just can't figure out. I have a function which I integrate and then plot it's values. But the graph I get looks strange. I expected some form of an exponential function but the one I get is far away from my anticipations, and when I try changing some of my constants, it does still behave in a strange manner. This is how the code looks like:

y = 0:0.5:D;
p = 65.5;
k = 1;
D =10:0.5:100; %Create a vector for D

M =zeros(1,181); %an array
for j = k:181 %For loop for all water values

  M(k) = integral(@(y) Func5(p,D,k,y),0,D(k));

title('Hydrostatic pressure in bars');
xlabel('water level in fot');
ylabel('pressure i bars');

And this is the code for my function:

function FD = Func5(p,D,k,y)
 w = 40-20*exp(-(0.01*y))*2; %Width of the dam
 FD = p.*(D(k)-y).*w; %Claculate the pressure

Thanks in advance for any help.

  • 1
    $\begingroup$ it's graph/it's value --> its graph, its value $\endgroup$ Commented Nov 21, 2015 at 12:14
  • $\begingroup$ Hey Wolfgang! Thanks for your help but I don't think I'm getting what you are conveying to me. Could you help explaining a little more? $\endgroup$ Commented Nov 21, 2015 at 13:35
  • $\begingroup$ Can you indicate which ODE you are solving? There are some issues with your code: 1. you need D for vector y but don't specify it, later you then redefine D which may introduce bugs if you are not careful. 2. Why do you manually increase j using k+1 in the loop? it seems to me simply using j=1:181 and M(j) works just fine. 3. It would be best not to use a for-loop and vectorize the call to integral but since i am not sure what you are solving, it is difficult to say if that is possible. 4. It is unclear what the different variables (D, k, etc) are... $\endgroup$
    – nluigi
    Commented Nov 21, 2015 at 14:27
  • $\begingroup$ @LloydKizito -- I'm simply pointing out a persistent grammar error in your question. $\endgroup$ Commented Nov 22, 2015 at 1:04

1 Answer 1


I think the main problem you encounter is that your for-loop probably doesn't do what you expect.

k = 1;
M =zeros(1,181);
for j = k:181 
  M(k) = integral(@(y) Func5(p,D,k,y),0,D(k));
  k+1;  # does not increase k

The problem here is that the statement k+1 is useless and does not increase k which i think is what you are after. The problem with this is that j is still increased each iteration but the values of the integral are assigned to M(k) which remain M(1) rather than increase with the loop. Also the argument D(k) remains unchanged which i guess should instead change with the iteration aswell.

try this instead:

M =zeros(1,181);
for j = 1:181
  M(j) = integral(@(y) Func5(p,D,j,y),0,D(j));

this give me atleast a graph which resembles some sort of exponential function.

Some general advice about your code: try using comments which add useful information to improve readability. M =zeros(1,181); %an array is not a useful comment because i can already see this is 'an array', what i cant see is what M means; a better comment would be % preallocate the results vector , the same applies for D =10:0.5:100; %Create a vector for D; what is D? and for that matter all other variables? Often it is also better to use meaningful variable names like results or pressure rather than D and M. Furthermore, in a function like Func5 it is useful to include a larger comment block at the top to say what the function calculates (which equation) and what its in/outputs are.

  • $\begingroup$ You are awesome nluigi! I worked just great!! $\endgroup$ Commented Nov 21, 2015 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.