4. The cross-sectional area shown is the shaded area between the function y=2 (h

Question

4. The cross-sectional area shown is the shaded area between the function y=2 (half an ellipse) andy=cos(.The area A of the cross section can be calculated by y.241-r :I ,dA The coordinate ofthe centroid ofthe area can be calculated by: Ve where M, is the moment of the area about the x axis, given by: y = cos(5 x -1 Use all methods that apply for the problem. You can use computer software as Matlab user-defined functions or MathCad explicit solution. Compare results and rank methods from best to worse

Explanation / Answer

matlab code

close all
clear
clc

funA = @(x) 2*sqrt(1-x.^2)-cos(pi*x/2);
A = integral(funA,-1,1)

funM = @(x) (2*sqrt(1-x.^2)+cos(pi*x/2)).*(2*sqrt(1-x.^2)-cos(pi*x/2));
M = 0.5*integral(funM,-1,1)

y = M/A

output

A =
1.8684
M =
2.1667
y =
1.1597

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