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
2025-01-21 04:48:46
推荐回答(3个)
回答1:

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

回答2:

y=[a^(2/3) - x^(2/3)]^(3/2) (设a>0)
S=4 定积分(0,a) ydx, 令x=a^(2/3) cos^2 t
S=4 定积分 -2/5 a^(2/3) sin^5 t
=8/5 a^(2/3)(1-a^(2/3) )^5

回答3:

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