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Hope you are doing great!

I am trying to solve a double Integral of a function let's say f(w,q). My upper and lower limits are

wmin = 0;
wmax[T_] = T - 5;
qmin[w_, T_] := Sqrt[T] - Sqrt[(T - w)];
qmax[w_, T_] := Sqrt[T] + Sqrt[(T - w)];

The integral I am trying to solve is:

Iw[q_,T_] := NIntegrate[f[w,q], {w,wmin,wmax[T]}]

Iq[w_,T_] := NIntegrate[q*Iw[q,T], {q,qmin[w, T], qmax[w,T]}] /;(wmin<w<=wmax[T])

Iq[w_,T_] := 0 /; (w < wmin or w >= wmax[T])

The problem is that if I give the integral let's say a value of Iq[1,25], it will go and replace w to NIntegrate[f[1,q],d1] which will give an error. I want Mathematica to first solve the I1 NIntegral with respect to w for a given T(must be same for all integrals) and then calculate the second integral which only has w-dependence in the limits of qmin, qmax and the condition.

Thanks in advance!

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  • $\begingroup$ Please provide your code (minimal working example) $\endgroup$ Mar 7 at 8:54

1 Answer 1

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Assuming

f = Function[{w, q}, 1];

we can integrate in one step as follows

int[ ww_?NumericQ, T_?NumericQ] := 
Boole[ (0 <= ww <= T - 5)] NIntegrate[q f[w , q], {q, Sqrt [T] - Sqrt[T - ww], Sqrt [T] + Sqrt[T - ww]} , {w, 0, T - 5}]

Plot3D[int[ww, T], {ww, 0, 10}  , {T, 0, 10},AxesLabel -> { ww, T, "int[ww,T]"} , MeshFunctions -> (#3 &), PlotPoints -> 50]

enter image description here

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