Задача 36189 ...
Условие
∫L y / √(x^2 + y^2) dl, где L — дуга кардиоиды ρ = 2(1 + cos φ), 0 ≤ φ ≤ π/2.
Решение
x=ρcosφ=2*(1+ cosφ)*cosφ
y=ρsinφ=2*(1 +cosφ)*sinφ
x^2 +y^2=ρ^2=(2*(1 +cosφ))^2
sqrt(x^2+ y^2)=2*(1+ cosφ)
[b]dl= sqrt(ρ^2(φ)+ (ρ`(φ))^2)dφ [/b]
ρ`(φ)=2*(0-sin φ )
ρ^2(φ)+ (ρ`(φ))^2= (2*(1+ cos φ))^2+ (-2sin φ)^2=4+ 8cos φ + 4cos^2 φ 4sin^2 φ )=
=8 +8cos φ =8*(1+ cos φ)^2=16sin^2( φ /2)
sqrt(ρ^2(φ)+ (ρ`(φ))^2)= [b]4сos( φ /2)[/b]
∫ _(L)y/sqrt(x^2+ y^2)dl= ∫ ^(π/2)_(0) [b]([/b] 2*(1+ cosφ)*sinφ/2*(1 +cosφ) [b])[/b]* [b]4сos( φ /2)[/b]d φ =
= ∫ ^(π/2)_(0) sinφ4сos( φ /2)d φ = формула sinα * cosβ
=4 ∫ ^(π/2)_(0) ((1/2)sin(3φ/2)+ (1/2)sin(φ/2)dφ)=
=2*(2/3)*(-cos(3 φ /2))+ 2*2*(-cos( φ /2)) |^(π/2)_(0)= ...