X=acos^3t，y=asin^3t 所 围成的平面图形的面积

请把详细过程写下，谢谢!用定积分

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x=acos^3t,y=asin^3t是星形线，它的面仔肢积为
∫ydx=4*∫闭培asin^3t(acos^3t)'dt,t:π/2→0
=-3*a^2∫念态世sin^4t*cos^2tdt
=-3a^2∫(sin^4t-sin^6t)dt
=3/8*πa^2

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y=[a^(2/3) - x^(2/3)]^(3/2) （厅租设a>0)
S=4 定积分(0,a) ydx, 令x=a^(2/3) cos^2 t
S=4 定积念伏备分仔毁 -2/5 a^(2/3) sin^5 t
=8/5 a^(2/3)(1-a^(2/3) )^5

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S = |2∫ydx| (t=0到t=π) = |2∫asin^3t * a*3cos^2t*(-sint)dt| = 6aa∫旅或sin^4tcos^2tdt
= 6aa/5 ∫拆搏伍sin^6tdt
=3aa/20π
只能输银虚100字，没办法了

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