How do I integrate this expression?

Question

Yagnaseni Roy 2012-11-15

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I have an expression for force here:

f =1.3e7*((500*y).*(0.46775*cos(60*t+0.2094395101999999e6*x)+0.37843*cos(2*(60*t+0.2094395101999999e6*x))+0.25228*cos(3*(60*t+0.2094395101999999e6*x))+0.1169*cos(4*(60*t+0.2094395101999999e6*x)))/1);

I need to integrate this expression w.r.t y and x (double integral), keeping t fixed.

Any idea how I could do this?

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Answer 1

Walter Roberson 2012-11-15

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在 MATLAB Online 中打开

That expression is numeric GIGO (Garbage In, Garbage Out). You have 1.3e7 which is only two decimal places of precision, but you also have 0.2094395101999999e6 which is 16 decimal places ending in 999999 that looks like blindly copied floating point numbers that should have been rounded.

But anyhow. Letting the limits be y0 and y1, and x0 and x1, the integral is

(5021948750/3848451) * (y0+y1) * ((140325/25228) * sin(60*t + (10262536/49) * x0) + (113529/50456) * sin(120*t + (20525072/49) * x0) + sin(180*t + (30787608/49) * x0) + (2505/7208) * sin(240*t + (41050144/49) * x0) - (140325/25228) * sin(60*t + (10262536/49) * x1) - (113529/50456) * sin(120*t + (20525072/49) * x1) - sin(180*t + (30787608/49) * x1) - (2505/7208) * sin(240*t + (41050144/49) * x1)) * (y0-y1)

Choose the number of digits of precision wisely when you convert to floating point.

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Yagnaseni Roy 2012-11-16

Pls forgive me if this is a very obvious question, but how did you get this formula for integration? I mean just by looking at it, the coefficients are not the same as in the expression I put up. Anyway, I got it to work.