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I need to plot g1, g2, g3 and g4 expressions. To do that, I am trying to evaluate the double defined integral in the three expressions, however I am having issues because one of the integration limits (alpha) is function of w. The resultante error says:

NIntegrate::nlim: α = α is not a valid limit of integration.

The question here is how can I compute a definite integral if I have a symbolic limit? Is it possible to do that?

Otherwise, is there anothet way to calculate and plot these expressions solving the indefinite integral?

Thanks in advance.

d = 0.0254; ad = 0.3; a = ad*d; young = 210 10^9; iner = (Pi/64)*d^4; ld = 24; L = ld*d; wcord = Sqrt[a*(d - a)]; α = a - (d/2) + Sqrt[((d/2)^2) - (w^2)]; αp = Sqrt[d^2 - (2*w)^2]; rα = α/αp; F = Sqrt[(2*αp)/(Pi*α)*Tan[(Pi*α)/(2*αp)]]*((0.923 + 0.199*(1 - Sin[(Pi*[Alpha])/(2*αp)])^4)/Cos[(Pi*α)/(2*αp)]); Fp = Sqrt[(2*αp)/(Pi*α)*Tan[(Pi*α)/(2*αp)]]*((0.752 + 2.02*rα + 0.37*(1 - Sin[(Pi*α)/(2*αp)])^3)/Cos[(Pi*α)/(2*αp)]); flxbInt = L^3/(48*young*iner) ; g1 = flxbInt + 2*(NIntegrate[((128*(L^2)*(αp^2))/(young*Pi*(d^8)))*α*F^2, {α, 0, α}, {w, 0, wcord}]) g4 = flxbInt + 2*(NIntegrate[((512*(L^2)*(w^2))/(young*Pi*(d^8)))*α*Fp^2, {α, 0, α}, {w, 0, wcord}]) g2 = g3 = 2*(NIntegrate[((256*(L^2)*αp*w)/(young*Pi*(d^8)))*α*F*Fp, {α, 0, α}, {w, 0, wcord}])

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  • $\begingroup$ There is a type in the definition of F; also, change the dummy integration variable. $\endgroup$ Commented Oct 28, 2016 at 7:24
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The most direct way of tackling this problem is defining a unnesting the integrals, converting to functions, and using NIntegrate

As before:

d = 0.0254; ad = 0.3; a = ad*d; young = 210 10^9; iner = (Pi/64)*d^4; ld = 24; L = ld*d; wcord = Sqrt[a*(d - a)]; \[Alpha] = a - (d/2) + Sqrt[((d/2)^2) - (w^2)]; \[Alpha]p = Sqrt[d^2 - (2*w)^2]; r\[Alpha] = \[Alpha]/\[Alpha]p; F = Sqrt[(2*\[Alpha]p)/(Pi*\[Alpha])*Tan[(Pi*\[Alpha])/(2*\[Alpha]p)]]*((0.923 + 0.199*(1 - Sin[(Pi*\[Alpha])/(2*\[Alpha]p)])^4)/Cos[(Pi*\[Alpha])/(2*\[Alpha]p)]); Fp = Sqrt[(2*\[Alpha]p)/(Pi*\[Alpha])*Tan[(Pi*\[Alpha])/(2*\[Alpha]p)]]*((0.752 + 2.02*r\[Alpha] + 0.37*(1 - Sin[(Pi*\[Alpha])/(2*\[Alpha]p)])^3)/Cos[(Pi*\[Alpha])/(2*\[Alpha]p)]); flxbInt = L^3/(48*young*iner);

Now define

i1[z_?NumericQ] := NIntegrate[k, {k, 0, z}];

Rewriting the integrals in terms of our new function i1

g1 = flxbInt + 2*(NIntegrate[((128*(L^2)*(\[Alpha]p^2))/(young*Pi*(d^8)))*i1[\[Alpha]]*F^2, {w, 0, wcord}]) g4 = flxbInt + 2*(NIntegrate[((512*(L^2)*(w^2))/(young*Pi*(d^8)))*i1[\[Alpha]]*Fp^2, {w, 0, wcord}]) g2 = g3 = 2*(NIntegrate[((256*(L^2)*\[Alpha]p*w)/(young*Pi*(d^8)))*i1[\[Alpha]]*F*Fp, {w, 0, wcord}])

Evaluating the integrals yields:

(*g1 = 1.20937*10^-6*) (*g4 = 1.13447*10^-6*) (*g2 = g3 = 4.88326*10^-8*)
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  • $\begingroup$ Thanks, your answer was very helpful. $\endgroup$ Commented Nov 15, 2016 at 17:29
  • $\begingroup$ In case I need a symbolic expression for each variable g. Do you think there is some way to get it? $\endgroup$ Commented Nov 17, 2016 at 18:45
  • $\begingroup$ I don't think you are going to be able to find an analytical expression for each integral. I would suggest sticking with numerics for this problem, though I don't want to close the door completely to the possibility that a solution exists. $\endgroup$ Commented Nov 17, 2016 at 19:54

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