Circular arcs are drawn from a fixed point on the circumference of a unit circle.
What is the area bounded by the locus (shown in dotted blue and red) of the center of masses of the circular arcs?
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Shouldn't it be d x instead of d θ in the last few integrals?
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Yes, that was a typo. Thanks for letting me know.
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Let the arc subtend an angle of ϕ and arc length ϕ . Therefore the end points of the arc are ( 0 , 0 ) and ( cos ϕ , sin ϕ ) .
For points ( cos t , sin t ) along the arc, the average x-coordinate is given by ϕ 1 ∫ 0 ϕ cos t d t = ϕ sin ϕ while the average y-coordinate is given by ϕ 1 ∫ 0 ϕ sin t d t = ϕ 1 − cos ϕ Therefore the centre of mass is the point ( ϕ sin ϕ , ϕ 1 − cos ϕ ) . We will find the locus in polar coordinates: x = ϕ sin ϕ y = ϕ 1 − cos ϕ θ = tan − 1 ( x y ) = tan − 1 ( sin ϕ 1 − cos ϕ ) = tan − 1 ( tan 2 ϕ ) = 2 ϕ r ∴ r = x 2 + y 2 = ϕ 2 sin 2 ϕ + cos 2 ϕ − 2 cos ϕ + 1 = ϕ 2 − 2 cos ϕ = ϕ 2 − 2 + 4 sin 2 2 ϕ = ϕ 2 sin 2 ϕ = θ sin θ Now to find the area of this curve, we use the formula for area in polar coordinates: A = ∫ α β 2 1 r 2 d θ 2 ∫ 0 π 2 1 θ 2 sin 2 θ d θ ∫ 0 π θ 2 2 1 ( 1 − cos 2 θ ) d θ ∫ 0 2 π x 2 1 − cos x d x [ − x 1 − cos x ] 0 2 π − ∫ 0 2 π − x sin x d x = 0 + x → 0 lim x 1 − cos x + ∫ 0 2 π x sin x d x = 0 + x → 0 lim 1 sin x + ∫ 0 2 π x sin x d x = ∫ 0 2 π x sin x d x ( by letting x = 2 θ ) ( using integration by parts ) ( using L’Hospital’s rule )