Double integral of a surface

5 views (last 30 days)
Andromeda
Andromeda on 11 Nov 2021
Commented: Andromeda on 11 Nov 2021
The correct answer of the double integral of the surface sqrt(x)-y^2 is 1/7 but the program output is 301/1430. Where have I gonne wrong? See code below
syms x y
format rational
Function = @(x,y) x.^(1/2)-y.^2;
xmin = @(y) y.^4;
xmax = @(y) y.^(1/2);
Double_integral = integral2(Function,0,1,xmin,xmax);
disp(Double_integral)
  1 Comment
Andromeda
Andromeda on 11 Nov 2021
Sure
my functon was supposed to be in the form of @(y,x). @Walter Roberson pointed that out

Sign in to comment.

Accepted Answer

Walter Roberson
Walter Roberson on 11 Nov 2021
xmin - real number
xmax - real number
ymin - real number | file handle
ymin - real number | file handle
Notice that your function handles are named xmin and xmax suggesting that they are limits on x rather than limits on y. But integral2() requires that the x limits be placed before the y limits, and does not permit function handles for the x limits.
What is the solution? This: exchange your x and y in your function.
format rational
Function = @(y,x) x.^(1/2)-y.^2;
xmin = @(y) y.^4;
xmax = @(y) y.^(1/2);
Double_integral = integral2(Function,0,1,xmin,xmax);
disp(Double_integral)
1/7
  1 Comment
Andromeda
Andromeda on 11 Nov 2021
Hahaha, yes!! Thank you.

Sign in to comment.

More Answers (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by