# why the 0.1 is a high require in f2 line

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JICHAO ZHANG on 20 Jun 2023
Commented: JICHAO ZHANG on 20 Jun 2023
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 on 20 Jun 2023
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

Naman on 20 Jun 2023
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 on 20 Jun 2023
many thanks