X=acos^3t,y=asin^3t 所 围成的平面图形的面积
问题描述：
X=acos^3t,y=asin^3t 所 围成的平面图形的面积
答
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
答
参数方程确定的函数的求导。 dy/dx=3asin^2tcost/[-3acos^2t sint]=-tanx d^2y/dx^2=-sec^2t/[-3acos^2t sint]=1/[3acos^4t sint].
答
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
答
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