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Find the perimeter of the ellipse: (x/3)^2+(y/2)^2=1.?

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  • 1 year ago

    My solution: Let x=3cosA,y=2sinA, A is the parameter.

    ds=dx^2+dy^2=>ds=sqr[9-5cos^2(A)]dA=3sqr[1-5cos^2(A)/9]dA=>

    .....2pi

    s=3Ssqr[1-0.55556cos^2(A)]dA=3(5.286621)=15.859863,

    ......0

    by my numerical method of polygon with n=3500 points; each tiny

    interval=0.001795709 (totally 3499 intervals).

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