why the 0.1 is a high require in f2 line

f = @(r,theta,phi,xi) r.^3 .* sin(theta).^2 .* sin(phi);
f1=@(theta,phi,xi)integral(@(r)f(r,theta,phi,xi),0,2,'ArrayValued',1);
df2=@(theta,phi,xi)cell2mat(arrayfun(@(theta,phi,xi)f1(theta,phi,xi),theta,phi,xi,'UniformOutput',0));
f2=integral3(@(theta,phi,xi)f(theta,phi,xi),0,pi,0,pi,0,2*pi,'AbsTol', 0,'RelTol',0.1);
ret=f2

[Walter Roberson]

f = @(r,theta,phi,xi) r.^3 .* sin(theta).^2 .* sin(phi);

So f expects 4 input parameters

f2=integral3(@(theta,phi,xi)f(theta,phi,xi),0,pi,0,pi,0,2*pi,'AbsTol', 0,'RelTol',0.1);

but here f is invoked with three input parameters

Answer 1

[Naman]

Hi Jichao Zhang,

The 'RelTol' parameter in the integral3 function sets the relative error tolerance for the numerical integration.

In the given code, 'RelTol' is set to 0.1, which allows for a maximum relative error of 10% in the numerical integration. This value is relatively high and may not be appropriate for some applications that require high accuracy. However, depending on the specific requirements of the application, a 'RelTol' of 0.1 may be sufficient to achieve reasonable accuracy while still maintaining computational efficiency.

Hope it helps.

[JICHAO ZHANG]

many thanks