Variable in function as well as integral boundary
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Paul Heineman
le 22 Oct 2022
Commenté : Paul Heineman
le 22 Oct 2022
Hi,
I have an integral I want to compute and plot for different angles theta. 0-90 degrees.
For every angle I want to compute the integral of the function from 0 up to the angle theta.
theta = 1:90;
A = 0.4*10^-19;
R = 4215 * theta.^(-1.35)*10^-9;
D_0 = 0.3;
D_c = R.*(1-cosd(theta));
D = (D_0 + D_c)*10^-9;
b = 0.2*10^-6;
Fun = @(theta) (A ./ (6 .* pi .* (D_0 + R .* ( 1-cosd(theta))).^3)) .* b .* R;
F_vdW1 = integral(Fun, 0, theta);
Can someone please help with what I am doing wrong?
Thanks
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Torsten
le 22 Oct 2022
Modifié(e) : Torsten
le 22 Oct 2022
Is it this what you want ?
If not, please clarify how the theta in the integrand and in the upper bound of the integral should be interpreted. Mathematically, your notation to use the same name for both is wrong.
theta = 1:90;
A = 0.4*10^-19;
R = 4215 * theta.^(-1.35)*10^-9;
D_0 = 0.3;
D_c = R.*(1-cosd(theta));
D = (D_0 + D_c)*10^-9;
b = 0.2*10^-6;
Fun = (A ./ (6 .* pi .* (D_0 + R .* ( 1-cosd(theta))).^3)) .* b .* R;
F_vdW1 = cumtrapz(theta,Fun);
plot(theta,F_vdW1)
3 commentaires
Torsten
le 22 Oct 2022
Modifié(e) : Torsten
le 22 Oct 2022
Are you sure that "phi" should enter the equation for R in degrees and not in radians ?
And according to your graphics, only the 6 and not the complete expression 6*pi*[D0+R*(1-cos(phi))]^3 would appear in the denominator of the integrand.
A = 0.4*10^-19;
R = @(phi) 4215 * phi.^(-1.35)*10^-9;
D_0 = 0.3;
D_c = @(phi) R(phi).*(1-cosd(phi));
D = @(phi) (D_0 + D_c(phi))*10^-9;
b = 0.2*10^-6;
Fun = @(phi)(A ./ (6 .* pi .* (D_0 + R(phi) .* ( 1-cosd(phi))).^3)) .* b .* R(phi);
theta = 1:360;
F_vdW1 = arrayfun(@(theta)integral(Fun,0,theta),theta)
plot(theta,F_vdW1)
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Bruno Luong
le 22 Oct 2022
Modifié(e) : Bruno Luong
le 22 Oct 2022
I have no idea if the code correspondons to the formula; I just modify your code to make it work on array
theta = 1:90;
A = 0.4*10^-19;
R = @(theta) 4215 * theta.^(-1.35)*10^-9;
D_0 = 0.3;
D_c = @(theta) R.*(1-cosd(theta));
b = 0.2*10^-6;
Fun = @(theta) (A ./ (6 .* pi .* (D_0 + R(theta) .* ( 1-cosd(theta))).^3)) .* b .* R(theta);
F_vdW1 = arrayfun(@(onetheta) integral(Fun, 0, onetheta), theta)
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