Question

найти площадь фигуры ограниченной астроидой x=a cos(t)^3 y= b sin(3t)^3

Answer 1

S = ∫[a,b] y dx,

x = a cos(t)^3,

y = b sin(3t)^3,

S = ∫[t1,t2] x dy,

S = ∫[0,2π] a cos(t)^3 * 3b sin(3t)^2 cos(t) dt

S = 3ab∫[0,2π] cos^4(t) sin^2(3t) dt

S = 3ab∫[0,2π] (1-sin^2(t))^2 sin^2(3t) dt

S = 3ab∫[0,2π] (sin^6(t) - 2sin^4(t) + sin^2(t)) sin^2(3t) dt

S = 3ab ( [sin^7(t)/7 - 2sin^5(t)/5 + sin^3(t)/3]sin^3(3t)/9 |0 to 2π )

S = 81abπ/35