Hello!
I'm getting a problem to calculate the Area and Volume (solid of revolution) from a curve. I've generated this curve using the Curve Fitting App from my data. This data consist in two vectors with 75 elements, one from measures of x, and another to measures of y, both related (like x(1) is related to y(1)).
The problem is: When i integrate the fittedmodel i don't get the expected result to the integer, the result is very far from the expected. Below is my data and the results:
%xi and yi is the data to generate the curve
xi = [8 11 14 17 20 ... 23 26 29 32 35 ... 38 41 44 47 50 ... 53 56 59 62 65 ... 68 71 74 77 80 ... 83 86 89 92 95 ... 98 101 104 107 ... 110 113 116 119 ... 122 125 128 131 ... 134 137 140 143 ... 146 149 152 155 ... 158 161 164 167 ... 170 173 176 179 ... 182 185 188 191 ... 194 197 200 203 ... 206 209 212 215 ... 218 221 224 227 230]'; yi = [48 48.75 48.85 48.85 48.95 48.9 ... 48.875 48.725 48.7 48.65 ... 48.425 48.40 48.15 48.05 ... 48 47.8 47.625 47.35 47.05 ... 46.85 46.5 46.25 46.1 45.65 ... 45.2 44.825 44.5 44.15 ... 43.7 43.2 43 42.55 42 ... 41.4 41 40.4 39.825 39.4 ... 38.975 37.625 36.9 36.05 ... 35.515 34.935 34.01 33.06 ... 32.49 31.425 30.74 29.905 ... 29.335 28.435 27.865 27.59 ... 27.2 26.73 26.475 26.33 ... 26.07 25.985 25.95 25.97 ... 26.125 26.26 26.5 26.9 27.19 ... 27.685 28.07 28.705 29.46 ... 30.035 30.96 31.425 31.5]';This is the Curve Fitting App results:
The results from Curve Fitting Tool:Linear model Poly9: f(x) = p1*x^9 + p2*x^8 + p3*x^7 + p4*x^6 + p5*x^5 + p6*x^4 + p7*x^3 + p8*x^2 + p9*x + p10 where x is normalized by mean 119 and std 65.38Coefficients (with 95% confidence bounds):
p1 = 0.1351 (0.01098, 0.2593) p2 = 0.1677 (0.06146, 0.2739) p3 = -0.9629 (-1.743, -0.1826) p4 = -1.728 (-2.319, -1.138) p5 = 2.014 (0.3559, 3.672) p6 = 6.094 (5.048, 7.139) p7 = 2.268 (0.9082, 3.628) p8 = -6.89 (-7.526, -6.254) p9 = -14.36 (-14.71, -14) p10 = 39.11 (39.01, 39.2)Goodness of fit: SSE: 1.826 R-square: 0.9997 Adjusted R-square: 0.9996 RMSE: 0.1676The expected Volume is: 1e+06 aprox.
Things that I've tried:
1:
Yfun = @(p) interp1(xi, yi, p); vol = 2*pi*integral(@(r) Yfun(r).^2, 8, 230);figure(2)fplot(Yfun,[0 240]);The result: vol =
2.117974233163996e+06
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2: After using the Curve Fitting App to generate the curve, and saved the data of fit in the Workspace as "fit":
plot(fit); %i don't know how to integrate the fit, I've tried this:
inte = integrate(fit,[xi],8); %this returns a vector of values of the integral in each point, but i need the area inside the curve, and not the vector of values.
disp(inte);The Results:
1.0e+04 * -5.298410022994969 -5.207195743886399 -5.115538164258902 -5.023588889525035 -4.931468089706592 -4.839271915381278 -4.747078332694966 -4.654951650401904 -4.562945978684072 -4.471107828746543 -4.379478033805406 -4.288093145994929 -4.196986439838003 -4.106188631165216 -4.015728400649765 -3.925632793364640 -3.835927549881680 -3.746637410336017 -3.657786420490639 -3.569398258071206 -3.481496588417348 -3.394105450730244 -3.307249669804079 -3.220955283027533 -3.135249968547648 -3.050163457718799 -2.965727913230844 -2.881978253539507 -2.798952404325326 -2.716691458601843 -2.635239728695733 -2.554644675548018 -2.474956703553063 -2.396228812377401 -2.318516100800259 -2.241875121508695 -2.166363089879153 -2.092036954000738 -2.018952337460231 -1.947162370631597 -1.876716430310121 -1.807658811419972 -1.740027358120832 -1.673852084860661 -1.609153820684662 -1.545942912331533 -1.484218023244017 -1.423965066508125 -1.365156309831062 -1.307749689888364 -1.251688371632910 -1.196900585378846 -1.143299770568881 -1.090785050020480 -1.039242052041939 -0.988544090029842 -0.938553699921691 -0.889124525098248 -0.840103525926018 -0.791333477018046 -0.742655699387482 -0.693912956889896 -0.644952426613754 -0.595628631100551 -0.545806196373474 -0.495362273642866 -0.444188434154865 -0.392191815873104 -0.339295267448934 -0.285436199160039 -0.230563812097151 -0.174634335770652 -0.117603860409718 -0.059418303453184 0
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The second test is far better in precision, but i don't get the result that i wanted, the integral under the curve (area) to calculate the volume (solid of revolution).
Can someone help me or give me a light to what to do?
If yes, please describe using a code example, cause i'm new using Matlab.
Thanks!
Best Answer