How to handle with double integral with complicated expressions?

I want to do double integral with complicated expressions. simply , my code is that:

>> r=3;

>> t=2;

>> q= @(x,y) (x+(t./2)).^2+(x+(t./2)).*y.^2;

>> p = @(x,y) ((x+t)./2).^2+(x+(t./2)).*y.^2;

>> func = @(x,y) sqrt((q - x).^2 - r.^2) + asin(r.^2 - (p-x).^2);

>> y_min = @(x) sqrt(r.^2 - (x-t).^2);

>> y_max = @(x) sqrt(r.^2-(x-((t./2)-r)).^2);

after these codes, then I add this statement for double integral.

>> integral2(func, 0, t./2, y_min, y_max)

and I have this kinds of error messages.

함수 'minus'은(는) 'function_handle'형 입력 인수에 대해 정의되지 않았습니다.

오류 발생: @(x,y)sqrt((q-x).^2-r.^2)+asin(r.^2-(p-x).^2)

오류 발생: integral2Calc>integral2t/tensor (line 228) Z = FUN(X,Y); NFE = NFE + 1;

오류 발생: integral2Calc>integral2t (line 55) [Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);

오류 발생: integral2Calc (line 9) [q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);

오류 발생: integral2 (line 106) Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);

sorry for some Koreans.

how can I solve this problem?

Answer 1

[Orion]

Hi,

you defined q and p as function handles, and then you use them inside another function handle func.

in func you're asking to do a minus operation between x which will be a numerical parameter and q which is a function handle. this is a non-sense, that's the reason you get an error

Undefined function 'minus' for input arguments of type 'function_handle'.

just write func in one big line, replacing q and p with their expression.

func = @(x,y) sqrt(((x+(t./2)).^2+(x+(t./2)).*y.^2 - x).^2 - r.^2) + asin(r.^2 - (((x+t)./2).^2+(x+(t./2)).*y.^2-x).^2);

this way it will work.