Variable in function as well as integral boundary

Question

Paul Heineman 2022-10-22

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1832833-variable-in-function-as-well-as-integral-boundary

评论： Paul Heineman 2022-10-22

采纳的回答： Torsten

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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);
Error using integral
Limits of integration must be double or single scalars.

Can someone please help with what I am doing wrong?

Thanks

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Answer 1

Torsten 2022-10-22

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https://ww2.mathworks.cn/matlabcentral/answers/1832833-variable-in-function-as-well-as-integral-boundary#answer_1081153

编辑：Torsten 2022-10-22

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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);

F_vdW1 = 1×90

1.0e-30 * 0 0.0461 0.0666 0.0793 0.0881 0.0948 0.1002 0.1046 0.1083 0.1115 0.1143 0.1167 0.1189 0.1209 0.1227 0.1243 0.1258 0.1272 0.1285 0.1297 0.1308 0.1319 0.1329 0.1338 0.1347 0.1355 0.1363 0.1371 0.1378 0.1385

plot(theta,F_vdW1)

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Torsten 2022-10-22

编辑：Torsten 2022-10-22

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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)

F_vdW1 = 1×360

1.0e-28 * 0.6801 0.5336 0.4630 0.4187 0.3872 0.3633 0.3442 0.3285 0.3152 0.3038 0.2938 0.2850 0.2771 0.2700 0.2636 0.2577 0.2523 0.2473 0.2427 0.2384 0.2343 0.2305 0.2270 0.2236 0.2204 0.2174 0.2146 0.2119 0.2093 0.2068

plot(theta,F_vdW1)

Paul Heineman 2022-10-22

This is what I needed, thanks!

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Answer 2

Bruno Luong 2022-10-22

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https://ww2.mathworks.cn/matlabcentral/answers/1832833-variable-in-function-as-well-as-integral-boundary#answer_1081158

编辑：Bruno Luong 2022-10-22

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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)
F_vdW1 = 1×90
1.0e-28 *

    0.6801    0.5336    0.4630    0.4187    0.3872    0.3633    0.3442    0.3285    0.3152    0.3038    0.2938    0.2850    0.2771    0.2700    0.2636    0.2577    0.2523    0.2473    0.2427    0.2384    0.2343    0.2305    0.2270    0.2236    0.2204    0.2174    0.2146    0.2119    0.2093    0.2068