getting error as integralParseArgs

Question

nikhil 2020-5-16

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/525937-getting-error-as-integralparseargs

评论： nikhil 2020-5-16

采纳的回答： Ameer Hamza

在 MATLAB Online 中打开

Hi all,

i am trying to solve the above equation in matlab, but when i am trying to substitute the alp values and x1 values in code and i am getting an error as matrix dimensions must agree while solving the code

any suggestions and thanks in advance

clc;
clear;
format short
ro= 17.5; R = 65; %polar boss opening radius and radius of cylinder
wa= asind(ro/R); b= 1.225*ro; % winding angle
x = b:5:R;
y = ro:0.45:b;
x1 = x/R;
y1 = y/R;
alp = asind(ro./x); alp1 = asind(ro./y);
i = zeros(size(alp));
j= zeros(size(x1));
for c = 1:numel(alp,x1)
func1= @(x)((cosd(wa).*(x.^3))./((sqrt(1-(x.^3))).*((cosd(wa)*cosd(wa)).*((x.^2).*(1+(x.^2)))-(sind(alp).*sind(alp)))));
i(c) = integral(func1,1,alp(c),x1(c));
end

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Answer 1

Ameer Hamza 2020-5-16

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/525937-getting-error-as-integralparseargs#answer_432831

在 MATLAB Online 中打开

Which variable represent ρ is not clear. Also, the syntax for calling integral is incorrect. Try this code

clc;
clear;
format short
ro= 17.5; R = 65; %polar boss opening radius and radius of cylinder
wa= asind(ro/R); b= 1.225*ro; % winding angle
x = b:5:R;
y = ro:0.45:b;
x1 = x/R;
y1 = y/R;
alp = asind(ro./x); alp1 = asind(ro./y);
i = zeros(size(alp));
j= zeros(size(x1));
for c = 1:numel(alp)
    func1= @(x)((cosd(wa).*(x.^3))./((sqrt(1-(x.^3))).*((cosd(wa)*cosd(wa)).*((x.^2).*(1+(x.^2)))-(sind(alp(c)).*sind(alp(c))))));
    i(c) = integral(func1,1,x(c));
end

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Ameer Hamza 2020-5-16

在 MATLAB Online 中打开

There seems to be some problem with the limits of integration or the function itself. If you plot the denominator in the limit of integration, you get the following curve

figure
den = @(y) ((sqrt(1-(y.^3))).*((cosd(wa).*cosd(wa)).*((y.^2).*(1+(y.^2)))-(sind(alp1(d)).*sind(alp1(d)))));
a1 = linspace(y(1)/65, 1, 1000);
a2 = den(a1);
plot(a1, a2);

As you can see, the curve passes through y=0, which means that it has a singularity at that point, which might make the integral diverge. You need to figure out how to avoid this singularity.

nikhil 2020-5-16