double_integral - et=abs(((a-I)/a)*100); %fprintf('(b) %1d,...

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function double_integral(d,u1,l1,u2,l2) %Problem 17.6 % % (a) Evaluate double integral analytically, note: imagine the integration % as two sweeping motions. Inner integration, a vertical line sweeps out % the area of a cross section. For the outer integration, the cross section % sweeps out the volume. if nargin~=5, error('Must specify a function to be integrated'); end syms x %Evaluates double integral with respect to x. yy1=int(d,x,l1,u1); syms y %Evaluates double integral with respect to y. yy2=int(yy1,y,l2,u2); a=eval(yy2); fprintf('\n%7.6f\n\n',a); % (b) Using the composite trapezoidal rule with n=2 n=2; h=(u1-l1)/n; % Need to first implement along the x-dimension for each y value l3=l2; i=d(l1,l2); j=d(h,l2); k=d(u1,l2); for s=1:3 y(s)=(u1-l1)*(i+2*j+k)/(2*n); l3=l3+h; i=d(l1,l3); j=d(h,l3); k=d(u1,l3); end I=(u2-l2)*(y(3)+2*y(2)+y(1))/(n*2);
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Unformatted text preview: et=abs(((a-I)/a)*100); %fprintf('(b) %1d, et=%2d',I,et); %??? Error using ==> fprintf disp('(b)'); %Function is not defined for 'sym' inputs. disp(I); disp('error'); disp(et); % (c) Using single applications of Simpson's Rule %Note: Simpson's 1/3 rule yields perfect results for cubic polynomials. n=2; h=(u1-l1)/n; % Need to first implement along the x-dimension for each y value l3=l2; i=d(l1,l2); j=d(h,l2); k=d(u1,l2); for s=1:3 y(s)=(u1-l1)*(i+4*j+k)/(3*n); l3=l3+h; i=d(l1,l3); j=d(h,l3); k=d(u1,l3); end I=(u2-l2)*(y(3)+4*y(2)+y(1))/(n*3); disp('(c)'); et=abs(((a-I)/a)*100); aa=eval(I); disp(aa); disp('error'); disp(et); %fprintf('(c) %7.6f, et=%1d\n\n',aa,et); %??? Error using ==> fprintf %Function is not defined for 'sym' %inputs....
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This note was uploaded on 09/27/2011 for the course EGM 3344 taught by Professor Raphaelhaftka during the Spring '09 term at University of Florida.

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double_integral - et=abs(((a-I)/a)*100); %fprintf('(b) %1d,...

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