Double integral of a surface

Question

Andromeda 2021-11-11

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1584584-double-integral-of-a-surface

评论： Andromeda 2021-11-11

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)

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Andromeda 2021-11-11

Sure

my functon was supposed to be in the form of @(y,x). @Walter Roberson pointed that out

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Answer 1

Walter Roberson 2021-11-11

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1584584-double-integral-of-a-surface#answer_829609

在 MATLAB Online 中打开

https://www.mathworks.com/help/matlab/ref/integral2.html

q = integral2(fun,xmin,xmax,ymin,ymax)

xmin - real number

xmax - real number

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     